estimation and hypothesis testing examples

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Statistical inference and estimation, review of introductory inference, statistical inference, model & estimation.

Recall, a statistical inference aims at learning characteristics of the population from a sample; the population characteristics are parameters and sample characteristics are statistics .

A statistical model is a representation of a complex phenomena that generated the data.

• It has mathematical formulations that describe relationships between random variables and parameters.
• It makes assumptions about the random variables, and sometimes parameters.
• A general form: data = model + residuals
• Model should explain most of the variation in the data
• Residuals are a representation of a lack-of-fit, that is of the portion of the data unexplained by the model.

Estimation represents ways or a process of learning and determining the population parameter based on the model fitted to the data.

Point estimation and interval estimation, and hypothesis testing are three main ways of learning about the population parameter from the sample statistic.

An estimator is particular example of a statistic, which becomes an estimate when the formula is replaced with actual observed sample values.

Point estimation = a single value that estimates the parameter. Point estimates are single values calculated from the sample

Confidence Intervals = gives a range of values for the parameter Interval estimates are intervals within which the parameter is expected to fall, with a certain degree of confidence.

Hypothesis tests = tests for a specific value(s) of the parameter.

In order to perform these inferential tasks, i.e., make inference about the unknown population parameter from the sample statistic, we need to know the likely values of the sample statistic. What would happen if we do sampling many times?

We need the sampling distribution of the statistic

• It depends on the model assumptions about the population distribution, and/or on the sample size.
• Standard error refers to the standard deviation of a sampling distribution.

Central Limit Theorem

Sampling distribution of the sample mean:

If numerous samples of size n are taken, the frequency curve of the sample means ( \(\bar{X}\)‘s) from those various samples is approximately bell shaped with mean μ and standard deviation, i.e. standard error \(\bar{X}/ \sim N(\mu , \sigma^2 / n)\)

• X is normally distributed
• X is NOT normal, but n is large (e.g. n >30) and μ finite.
• For continuous variables

For categorical data, the CLT holds for the sampling distribution of the sample proportion.

Proportions in Newspapers

As found in CNN in June, 2006:

The parameter of interest in the population is the proportion of U.S. adults who disapprove of how well Bush is handling Iraq, p .

The sample statistic, or point estimator is \(\hat{p}\), and an estimate, based on this sample is \(\hat{p}=0.62\).

Next question ...

If we take another poll, we are likely to get a different sample proportion, e.g. 60%, 59%,67%, etc..

So, what is the 95% confidence interval? Based on the CLT, the 95% CI is \(\hat{p}\pm 2 \ast \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\).

We often assume p = 1/2 so \(\hat{p}\pm 2 \ast \sqrt{\frac{\frac{1}{2}\ast\frac{1}{2} }{n}}=\hat{p}\pm\frac{1}{\sqrt{n}}=\hat{p}\pm\text{MOE}\).

The margin of error (MOE) is 2 × St.Dev or \(1/\sqrt{n}\).

Estimation and Hypothesis Testing

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• First Online: 14 August 2020
• Cite this living reference work entry

• Pamela A. Shaw 3 &
• Michael A. Proschan 4  

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This chapter presents basic elements of parameter estimation and hypothesis testing. The reader will learn how to form confidence intervals for the mean, and more generally, how to calculate confidence intervals for the one parameter setting and for the difference between two groups. Principles of hypothesis testing are detailed, including the choice of the null and alternative hypotheses, the significance level, and implications for choosing a one-sided versus two-sided test. The p-value is defined and a discussion of controversies that have arisen over its use are included. After reading this chapter, the reader will have a better understanding of the necessary steps to set up a hypothesis test and make valid inference about the quantity of interest. Other topics in this chapter include exact hypothesis tests, which may be preferable for small sample settings, and the choice of a parametric versus nonparametric test. The chapter also includes a brief discussion of the implications of multiple comparisons on hypothesis testing and considerations of hypothesis testing in the setting of noninferiority trials.

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National Institute of Allergy and Infectious Diseases, Bethesda, MD, USA

Michael A. Proschan

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Shaw, P.A., Proschan, M.A. (2020). Estimation and Hypothesis Testing. In: Piantadosi, S., Meinert, C.L. (eds) Principles and Practice of Clinical Trials. Springer, Cham. https://doi.org/10.1007/978-3-319-52677-5_114-1

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Hypothesis Testing

Hypothesis testing is a tool for making statistical inferences about the population data. It is an analysis tool that tests assumptions and determines how likely something is within a given standard of accuracy. Hypothesis testing provides a way to verify whether the results of an experiment are valid.

A null hypothesis and an alternative hypothesis are set up before performing the hypothesis testing. This helps to arrive at a conclusion regarding the sample obtained from the population. In this article, we will learn more about hypothesis testing, its types, steps to perform the testing, and associated examples.

What is Hypothesis Testing in Statistics?

Hypothesis testing uses sample data from the population to draw useful conclusions regarding the population probability distribution . It tests an assumption made about the data using different types of hypothesis testing methodologies. The hypothesis testing results in either rejecting or not rejecting the null hypothesis.

Hypothesis Testing Definition

Hypothesis testing can be defined as a statistical tool that is used to identify if the results of an experiment are meaningful or not. It involves setting up a null hypothesis and an alternative hypothesis. These two hypotheses will always be mutually exclusive. This means that if the null hypothesis is true then the alternative hypothesis is false and vice versa. An example of hypothesis testing is setting up a test to check if a new medicine works on a disease in a more efficient manner.

Null Hypothesis

The null hypothesis is a concise mathematical statement that is used to indicate that there is no difference between two possibilities. In other words, there is no difference between certain characteristics of data. This hypothesis assumes that the outcomes of an experiment are based on chance alone. It is denoted as \(H_{0}\). Hypothesis testing is used to conclude if the null hypothesis can be rejected or not. Suppose an experiment is conducted to check if girls are shorter than boys at the age of 5. The null hypothesis will say that they are the same height.

Alternative Hypothesis

The alternative hypothesis is an alternative to the null hypothesis. It is used to show that the observations of an experiment are due to some real effect. It indicates that there is a statistical significance between two possible outcomes and can be denoted as \(H_{1}\) or \(H_{a}\). For the above-mentioned example, the alternative hypothesis would be that girls are shorter than boys at the age of 5.

Hypothesis Testing P Value

In hypothesis testing, the p value is used to indicate whether the results obtained after conducting a test are statistically significant or not. It also indicates the probability of making an error in rejecting or not rejecting the null hypothesis.This value is always a number between 0 and 1. The p value is compared to an alpha level, \(\alpha\) or significance level. The alpha level can be defined as the acceptable risk of incorrectly rejecting the null hypothesis. The alpha level is usually chosen between 1% to 5%.

Hypothesis Testing Critical region

All sets of values that lead to rejecting the null hypothesis lie in the critical region. Furthermore, the value that separates the critical region from the non-critical region is known as the critical value.

Hypothesis Testing Formula

Depending upon the type of data available and the size, different types of hypothesis testing are used to determine whether the null hypothesis can be rejected or not. The hypothesis testing formula for some important test statistics are given below:

• z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). \(\overline{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation and n is the size of the sample.
• t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\). s is the sample standard deviation.
• \(\chi ^{2} = \sum \frac{(O_{i}-E_{i})^{2}}{E_{i}}\). \(O_{i}\) is the observed value and \(E_{i}\) is the expected value.

We will learn more about these test statistics in the upcoming section.

Types of Hypothesis Testing

Selecting the correct test for performing hypothesis testing can be confusing. These tests are used to determine a test statistic on the basis of which the null hypothesis can either be rejected or not rejected. Some of the important tests used for hypothesis testing are given below.

Hypothesis Testing Z Test

A z test is a way of hypothesis testing that is used for a large sample size (n ≥ 30). It is used to determine whether there is a difference between the population mean and the sample mean when the population standard deviation is known. It can also be used to compare the mean of two samples. It is used to compute the z test statistic. The formulas are given as follows:

• One sample: z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).
• Two samples: z = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\).

Hypothesis Testing t Test

The t test is another method of hypothesis testing that is used for a small sample size (n < 30). It is also used to compare the sample mean and population mean. However, the population standard deviation is not known. Instead, the sample standard deviation is known. The mean of two samples can also be compared using the t test.

• One sample: t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\).
• Two samples: t = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}\).

Hypothesis Testing Chi Square

The Chi square test is a hypothesis testing method that is used to check whether the variables in a population are independent or not. It is used when the test statistic is chi-squared distributed.

One Tailed Hypothesis Testing

One tailed hypothesis testing is done when the rejection region is only in one direction. It can also be known as directional hypothesis testing because the effects can be tested in one direction only. This type of testing is further classified into the right tailed test and left tailed test.

Right Tailed Hypothesis Testing

The right tail test is also known as the upper tail test. This test is used to check whether the population parameter is greater than some value. The null and alternative hypotheses for this test are given as follows:

\(H_{0}\): The population parameter is ≤ some value

\(H_{1}\): The population parameter is > some value.

If the test statistic has a greater value than the critical value then the null hypothesis is rejected

Left Tailed Hypothesis Testing

The left tail test is also known as the lower tail test. It is used to check whether the population parameter is less than some value. The hypotheses for this hypothesis testing can be written as follows:

\(H_{0}\): The population parameter is ≥ some value

\(H_{1}\): The population parameter is < some value.

The null hypothesis is rejected if the test statistic has a value lesser than the critical value.

Two Tailed Hypothesis Testing

In this hypothesis testing method, the critical region lies on both sides of the sampling distribution. It is also known as a non - directional hypothesis testing method. The two-tailed test is used when it needs to be determined if the population parameter is assumed to be different than some value. The hypotheses can be set up as follows:

\(H_{0}\): the population parameter = some value

\(H_{1}\): the population parameter ≠ some value

The null hypothesis is rejected if the test statistic has a value that is not equal to the critical value.

Hypothesis Testing Steps

Hypothesis testing can be easily performed in five simple steps. The most important step is to correctly set up the hypotheses and identify the right method for hypothesis testing. The basic steps to perform hypothesis testing are as follows:

• Step 1: Set up the null hypothesis by correctly identifying whether it is the left-tailed, right-tailed, or two-tailed hypothesis testing.
• Step 2: Set up the alternative hypothesis.
• Step 3: Choose the correct significance level, \(\alpha\), and find the critical value.
• Step 4: Calculate the correct test statistic (z, t or \(\chi\)) and p-value.
• Step 5: Compare the test statistic with the critical value or compare the p-value with \(\alpha\) to arrive at a conclusion. In other words, decide if the null hypothesis is to be rejected or not.

Hypothesis Testing Example

The best way to solve a problem on hypothesis testing is by applying the 5 steps mentioned in the previous section. Suppose a researcher claims that the mean average weight of men is greater than 100kgs with a standard deviation of 15kgs. 30 men are chosen with an average weight of 112.5 Kgs. Using hypothesis testing, check if there is enough evidence to support the researcher's claim. The confidence interval is given as 95%.

Step 1: This is an example of a right-tailed test. Set up the null hypothesis as \(H_{0}\): \(\mu\) = 100.

Step 2: The alternative hypothesis is given by \(H_{1}\): \(\mu\) > 100.

Step 3: As this is a one-tailed test, \(\alpha\) = 100% - 95% = 5%. This can be used to determine the critical value.

1 - \(\alpha\) = 1 - 0.05 = 0.95

0.95 gives the required area under the curve. Now using a normal distribution table, the area 0.95 is at z = 1.645. A similar process can be followed for a t-test. The only additional requirement is to calculate the degrees of freedom given by n - 1.

Step 4: Calculate the z test statistic. This is because the sample size is 30. Furthermore, the sample and population means are known along with the standard deviation.

z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).

\(\mu\) = 100, \(\overline{x}\) = 112.5, n = 30, \(\sigma\) = 15

z = \(\frac{112.5-100}{\frac{15}{\sqrt{30}}}\) = 4.56

Step 5: Conclusion. As 4.56 > 1.645 thus, the null hypothesis can be rejected.

Hypothesis Testing and Confidence Intervals

Confidence intervals form an important part of hypothesis testing. This is because the alpha level can be determined from a given confidence interval. Suppose a confidence interval is given as 95%. Subtract the confidence interval from 100%. This gives 100 - 95 = 5% or 0.05. This is the alpha value of a one-tailed hypothesis testing. To obtain the alpha value for a two-tailed hypothesis testing, divide this value by 2. This gives 0.05 / 2 = 0.025.

Related Articles:

• Probability and Statistics
• Data Handling

Important Notes on Hypothesis Testing

• Hypothesis testing is a technique that is used to verify whether the results of an experiment are statistically significant.
• It involves the setting up of a null hypothesis and an alternate hypothesis.
• There are three types of tests that can be conducted under hypothesis testing - z test, t test, and chi square test.
• Hypothesis testing can be classified as right tail, left tail, and two tail tests.

Examples on Hypothesis Testing

• Example 1: The average weight of a dumbbell in a gym is 90lbs. However, a physical trainer believes that the average weight might be higher. A random sample of 5 dumbbells with an average weight of 110lbs and a standard deviation of 18lbs. Using hypothesis testing check if the physical trainer's claim can be supported for a 95% confidence level. Solution: As the sample size is lesser than 30, the t-test is used. \(H_{0}\): \(\mu\) = 90, \(H_{1}\): \(\mu\) > 90 \(\overline{x}\) = 110, \(\mu\) = 90, n = 5, s = 18. \(\alpha\) = 0.05 Using the t-distribution table, the critical value is 2.132 t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\) t = 2.484 As 2.484 > 2.132, the null hypothesis is rejected. Answer: The average weight of the dumbbells may be greater than 90lbs
• Example 2: The average score on a test is 80 with a standard deviation of 10. With a new teaching curriculum introduced it is believed that this score will change. On random testing, the score of 38 students, the mean was found to be 88. With a 0.05 significance level, is there any evidence to support this claim? Solution: This is an example of two-tail hypothesis testing. The z test will be used. \(H_{0}\): \(\mu\) = 80, \(H_{1}\): \(\mu\) ≠ 80 \(\overline{x}\) = 88, \(\mu\) = 80, n = 36, \(\sigma\) = 10. \(\alpha\) = 0.05 / 2 = 0.025 The critical value using the normal distribution table is 1.96 z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\) z = \(\frac{88-80}{\frac{10}{\sqrt{36}}}\) = 4.8 As 4.8 > 1.96, the null hypothesis is rejected. Answer: There is a difference in the scores after the new curriculum was introduced.
• Example 3: The average score of a class is 90. However, a teacher believes that the average score might be lower. The scores of 6 students were randomly measured. The mean was 82 with a standard deviation of 18. With a 0.05 significance level use hypothesis testing to check if this claim is true. Solution: The t test will be used. \(H_{0}\): \(\mu\) = 90, \(H_{1}\): \(\mu\) < 90 \(\overline{x}\) = 110, \(\mu\) = 90, n = 6, s = 18 The critical value from the t table is -2.015 t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\) t = \(\frac{82-90}{\frac{18}{\sqrt{6}}}\) t = -1.088 As -1.088 > -2.015, we fail to reject the null hypothesis. Answer: There is not enough evidence to support the claim.

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FAQs on Hypothesis Testing

What is hypothesis testing.

Hypothesis testing in statistics is a tool that is used to make inferences about the population data. It is also used to check if the results of an experiment are valid.

What is the z Test in Hypothesis Testing?

The z test in hypothesis testing is used to find the z test statistic for normally distributed data . The z test is used when the standard deviation of the population is known and the sample size is greater than or equal to 30.

What is the t Test in Hypothesis Testing?

The t test in hypothesis testing is used when the data follows a student t distribution . It is used when the sample size is less than 30 and standard deviation of the population is not known.

What is the formula for z test in Hypothesis Testing?

The formula for a one sample z test in hypothesis testing is z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\) and for two samples is z = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\).

What is the p Value in Hypothesis Testing?

The p value helps to determine if the test results are statistically significant or not. In hypothesis testing, the null hypothesis can either be rejected or not rejected based on the comparison between the p value and the alpha level.

What is One Tail Hypothesis Testing?

When the rejection region is only on one side of the distribution curve then it is known as one tail hypothesis testing. The right tail test and the left tail test are two types of directional hypothesis testing.

What is the Alpha Level in Two Tail Hypothesis Testing?

To get the alpha level in a two tail hypothesis testing divide \(\alpha\) by 2. This is done as there are two rejection regions in the curve.

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• RESEARCH METHODS & STATISTICS
• 0 About these materials
• 1 Introduction & Overview
• Defining data
• Test Yourself
• Numerical summaries & display
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• Transformations: Z-Scores
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• Describing binary variables (prevalence & incidence)
• From population to sample
• An introduction to observational and experimental design
• Point estimates and population parameters
• Sampling variation and sampling distributions
• Bias and Precision
• What is a confidence interval?
• What is a hypothesis test?
• Understanding probability & the relationship with inference
• Central Limit Theorem and the Normal Distribution
• Central Limit Theorem in practice: single means and proportions
• Link between a hypothesis test & CI
• Comparing means
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• Chi squared test of independence
• Reporting results and drawing conclusions
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6 Estimation & hypothesis testing

• The Central Limit Theorem in a nutshell:  if multiple samples of the same size were drawn randomly and independently from a population, then the sampling distribution of the means of those samples would be approximately normal regardless of the shape of the underlying population distribution as long as the sample size is large enough . This result forms the basis of most statistical inference, and hence most statistical analysis you will come across. You will often hear the term normal approximation ; this refers to inference based on the central limit theorem.
• The sampling distribution is a probability distribution; it is a theoretical construct that is mathematically derived based on independent and random processes. Most statistical analyses use a sampling distribution to make inference , hence to make valid inference we also need to assume that the units of observation in our sample are independent and from a random sampling process.
• The mean and standard deviation (standard error) of sampling distributions for different point estimates are calculated differently (you will see two examples in this section- the sample mean and sample proportion), but the basic form of a confidence interval and the basic form of a test statistic for a hypothesis test are the same.
• The confidence in a confidence interval is pre-determined by us. Confidence is in the method, not in the result. Most confidence intervals take this form: estimate plus or minus a chosen number of standard errors. The chosen number is called the confidence coefficient and is selected to create the desired confidence level.
• A test statistic from a hypothesis test measures how many standard errors the observed point estimate is away from the expected null hypothesis value. If the observed value is too many standard errors from the expected value, we don't believe the null hypothesis or there is evidence against it (falsification).
• Almost all test statistics take this form:

                                        (sample statistic - null hypothesis value of parameter)/ standard error

• Confidence intervals and hypothesis tests are directly linked. Confidence intervals can be used to check the reasonableness of claims about the parameter. If someone claims the parameter is equal to 62, and 62 is not within your confidence interval, then this claim is suspect. Here's another distinction between what a p-value and CI tells us: a p-value answers the question "How surprising is my sample?" The confidence interval can answer the question "What values of the population parameter would cause me not to be surprised by the sample?"

Connections with other material

• Most of the inference for simple comparisons covered in the next theme (Simple comparisons) use the central limit theorem and so share the same principles and logic. They just have different point estimates and different standard errors.
• Inference for regression models is largely based on the normal distribution.
• In the exploratory analysis theme, we also saw how the normal probability distribution can be used to work out reference ranges for individual observations in a sample. The application of the normal distribution is the same, except here we are estimating the sampling distribution of a point estimate or sample statistic as opposed to estimating the distribution of individual values in the population for a reference range.

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In today’s data-driven world , decisions are based on data all the time. Hypothesis plays a crucial role in that process, whether it may be making business decisions, in the health sector, academia, or in quality improvement. Without hypothesis & hypothesis tests, you risk drawing the wrong conclusions and making bad decisions. In this tutorial, you will look at Hypothesis Testing in Statistics.

What Is Hypothesis Testing in Statistics?

Hypothesis Testing is a type of statistical analysis in which you put your assumptions about a population parameter to the test. It is used to estimate the relationship between 2 statistical variables.

Let's discuss few examples of statistical hypothesis from real-life - 

• A teacher assumes that 60% of his college's students come from lower-middle-class families.
• A doctor believes that 3D (Diet, Dose, and Discipline) is 90% effective for diabetic patients.

Now that you know about hypothesis testing, look at the two types of hypothesis testing in statistics.

Hypothesis Testing Formula

Z = ( x̅ – μ0 ) / (σ /√n)

• Here, x̅ is the sample mean,
• μ0 is the population mean,
• σ is the standard deviation,
• n is the sample size.

How Hypothesis Testing Works?

An analyst performs hypothesis testing on a statistical sample to present evidence of the plausibility of the null hypothesis. Measurements and analyses are conducted on a random sample of the population to test a theory. Analysts use a random population sample to test two hypotheses: the null and alternative hypotheses.

The null hypothesis is typically an equality hypothesis between population parameters; for example, a null hypothesis may claim that the population means return equals zero. The alternate hypothesis is essentially the inverse of the null hypothesis (e.g., the population means the return is not equal to zero). As a result, they are mutually exclusive, and only one can be correct. One of the two possibilities, however, will always be correct.

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Null Hypothesis and Alternate Hypothesis

The Null Hypothesis is the assumption that the event will not occur. A null hypothesis has no bearing on the study's outcome unless it is rejected.

H0 is the symbol for it, and it is pronounced H-naught.

The Alternate Hypothesis is the logical opposite of the null hypothesis. The acceptance of the alternative hypothesis follows the rejection of the null hypothesis. H1 is the symbol for it.

Let's understand this with an example.

A sanitizer manufacturer claims that its product kills 95 percent of germs on average. 

To put this company's claim to the test, create a null and alternate hypothesis.

H0 (Null Hypothesis): Average = 95%.

Alternative Hypothesis (H1): The average is less than 95%.

Another straightforward example to understand this concept is determining whether or not a coin is fair and balanced. The null hypothesis states that the probability of a show of heads is equal to the likelihood of a show of tails. In contrast, the alternate theory states that the probability of a show of heads and tails would be very different.

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Hypothesis Testing Calculation With Examples

Let's consider a hypothesis test for the average height of women in the United States. Suppose our null hypothesis is that the average height is 5'4". We gather a sample of 100 women and determine that their average height is 5'5". The standard deviation of population is 2.

To calculate the z-score, we would use the following formula:

z = ( x̅ – μ0 ) / (σ /√n)

z = (5'5" - 5'4") / (2" / √100)

z = 0.5 / (0.045)

 We will reject the null hypothesis as the z-score of 11.11 is very large and conclude that there is evidence to suggest that the average height of women in the US is greater than 5'4".

Steps of Hypothesis Testing

Step 1: specify your null and alternate hypotheses.

It is critical to rephrase your original research hypothesis (the prediction that you wish to study) as a null (Ho) and alternative (Ha) hypothesis so that you can test it quantitatively. Your first hypothesis, which predicts a link between variables, is generally your alternate hypothesis. The null hypothesis predicts no link between the variables of interest.

Step 2: Gather Data

For a statistical test to be legitimate, sampling and data collection must be done in a way that is meant to test your hypothesis. You cannot draw statistical conclusions about the population you are interested in if your data is not representative.

Step 3: Conduct a Statistical Test

Other statistical tests are available, but they all compare within-group variance (how to spread out the data inside a category) against between-group variance (how different the categories are from one another). If the between-group variation is big enough that there is little or no overlap between groups, your statistical test will display a low p-value to represent this. This suggests that the disparities between these groups are unlikely to have occurred by accident. Alternatively, if there is a large within-group variance and a low between-group variance, your statistical test will show a high p-value. Any difference you find across groups is most likely attributable to chance. The variety of variables and the level of measurement of your obtained data will influence your statistical test selection.

Step 4: Determine Rejection Of Your Null Hypothesis

Your statistical test results must determine whether your null hypothesis should be rejected or not. In most circumstances, you will base your judgment on the p-value provided by the statistical test. In most circumstances, your preset level of significance for rejecting the null hypothesis will be 0.05 - that is, when there is less than a 5% likelihood that these data would be seen if the null hypothesis were true. In other circumstances, researchers use a lower level of significance, such as 0.01 (1%). This reduces the possibility of wrongly rejecting the null hypothesis.

Step 5: Present Your Results 

The findings of hypothesis testing will be discussed in the results and discussion portions of your research paper, dissertation, or thesis. You should include a concise overview of the data and a summary of the findings of your statistical test in the results section. You can talk about whether your results confirmed your initial hypothesis or not in the conversation. Rejecting or failing to reject the null hypothesis is a formal term used in hypothesis testing. This is likely a must for your statistics assignments.

Types of Hypothesis Testing

To determine whether a discovery or relationship is statistically significant, hypothesis testing uses a z-test. It usually checks to see if two means are the same (the null hypothesis). Only when the population standard deviation is known and the sample size is 30 data points or more, can a z-test be applied.

A statistical test called a t-test is employed to compare the means of two groups. To determine whether two groups differ or if a procedure or treatment affects the population of interest, it is frequently used in hypothesis testing.

Chi-Square 

You utilize a Chi-square test for hypothesis testing concerning whether your data is as predicted. To determine if the expected and observed results are well-fitted, the Chi-square test analyzes the differences between categorical variables from a random sample. The test's fundamental premise is that the observed values in your data should be compared to the predicted values that would be present if the null hypothesis were true.

Hypothesis Testing and Confidence Intervals

Both confidence intervals and hypothesis tests are inferential techniques that depend on approximating the sample distribution. Data from a sample is used to estimate a population parameter using confidence intervals. Data from a sample is used in hypothesis testing to examine a given hypothesis. We must have a postulated parameter to conduct hypothesis testing.

Bootstrap distributions and randomization distributions are created using comparable simulation techniques. The observed sample statistic is the focal point of a bootstrap distribution, whereas the null hypothesis value is the focal point of a randomization distribution.

A variety of feasible population parameter estimates are included in confidence ranges. In this lesson, we created just two-tailed confidence intervals. There is a direct connection between these two-tail confidence intervals and these two-tail hypothesis tests. The results of a two-tailed hypothesis test and two-tailed confidence intervals typically provide the same results. In other words, a hypothesis test at the 0.05 level will virtually always fail to reject the null hypothesis if the 95% confidence interval contains the predicted value. A hypothesis test at the 0.05 level will nearly certainly reject the null hypothesis if the 95% confidence interval does not include the hypothesized parameter.

Simple and Composite Hypothesis Testing

Depending on the population distribution, you can classify the statistical hypothesis into two types.

Simple Hypothesis: A simple hypothesis specifies an exact value for the parameter.

Composite Hypothesis: A composite hypothesis specifies a range of values.

A company is claiming that their average sales for this quarter are 1000 units. This is an example of a simple hypothesis.

Suppose the company claims that the sales are in the range of 900 to 1000 units. Then this is a case of a composite hypothesis.

One-Tailed and Two-Tailed Hypothesis Testing

The One-Tailed test, also called a directional test, considers a critical region of data that would result in the null hypothesis being rejected if the test sample falls into it, inevitably meaning the acceptance of the alternate hypothesis.

In a one-tailed test, the critical distribution area is one-sided, meaning the test sample is either greater or lesser than a specific value.

In two tails, the test sample is checked to be greater or less than a range of values in a Two-Tailed test, implying that the critical distribution area is two-sided.

If the sample falls within this range, the alternate hypothesis will be accepted, and the null hypothesis will be rejected.

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Right Tailed Hypothesis Testing

If the larger than (>) sign appears in your hypothesis statement, you are using a right-tailed test, also known as an upper test. Or, to put it another way, the disparity is to the right. For instance, you can contrast the battery life before and after a change in production. Your hypothesis statements can be the following if you want to know if the battery life is longer than the original (let's say 90 hours):

• The null hypothesis is (H0 <= 90) or less change.
• A possibility is that battery life has risen (H1) > 90.

The crucial point in this situation is that the alternate hypothesis (H1), not the null hypothesis, decides whether you get a right-tailed test.

Left Tailed Hypothesis Testing

Alternative hypotheses that assert the true value of a parameter is lower than the null hypothesis are tested with a left-tailed test; they are indicated by the asterisk "<".

Suppose H0: mean = 50 and H1: mean not equal to 50

According to the H1, the mean can be greater than or less than 50. This is an example of a Two-tailed test.

In a similar manner, if H0: mean >=50, then H1: mean <50

Here the mean is less than 50. It is called a One-tailed test.

Type 1 and Type 2 Error

A hypothesis test can result in two types of errors.

Type 1 Error: A Type-I error occurs when sample results reject the null hypothesis despite being true.

Type 2 Error: A Type-II error occurs when the null hypothesis is not rejected when it is false, unlike a Type-I error.

Suppose a teacher evaluates the examination paper to decide whether a student passes or fails.

H0: Student has passed

H1: Student has failed

Type I error will be the teacher failing the student [rejects H0] although the student scored the passing marks [H0 was true]. 

Type II error will be the case where the teacher passes the student [do not reject H0] although the student did not score the passing marks [H1 is true].

Level of Significance

The alpha value is a criterion for determining whether a test statistic is statistically significant. In a statistical test, Alpha represents an acceptable probability of a Type I error. Because alpha is a probability, it can be anywhere between 0 and 1. In practice, the most commonly used alpha values are 0.01, 0.05, and 0.1, which represent a 1%, 5%, and 10% chance of a Type I error, respectively (i.e. rejecting the null hypothesis when it is in fact correct).

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A p-value is a metric that expresses the likelihood that an observed difference could have occurred by chance. As the p-value decreases the statistical significance of the observed difference increases. If the p-value is too low, you reject the null hypothesis.

Here you have taken an example in which you are trying to test whether the new advertising campaign has increased the product's sales. The p-value is the likelihood that the null hypothesis, which states that there is no change in the sales due to the new advertising campaign, is true. If the p-value is .30, then there is a 30% chance that there is no increase or decrease in the product's sales.  If the p-value is 0.03, then there is a 3% probability that there is no increase or decrease in the sales value due to the new advertising campaign. As you can see, the lower the p-value, the chances of the alternate hypothesis being true increases, which means that the new advertising campaign causes an increase or decrease in sales.

Why is Hypothesis Testing Important in Research Methodology?

Hypothesis testing is crucial in research methodology for several reasons:

• Provides evidence-based conclusions: It allows researchers to make objective conclusions based on empirical data, providing evidence to support or refute their research hypotheses.
• Supports decision-making: It helps make informed decisions, such as accepting or rejecting a new treatment, implementing policy changes, or adopting new practices.
• Adds rigor and validity: It adds scientific rigor to research using statistical methods to analyze data, ensuring that conclusions are based on sound statistical evidence.
• Contributes to the advancement of knowledge: By testing hypotheses, researchers contribute to the growth of knowledge in their respective fields by confirming existing theories or discovering new patterns and relationships.

Limitations of Hypothesis Testing

Hypothesis testing has some limitations that researchers should be aware of:

• It cannot prove or establish the truth: Hypothesis testing provides evidence to support or reject a hypothesis, but it cannot confirm the absolute truth of the research question.
• Results are sample-specific: Hypothesis testing is based on analyzing a sample from a population, and the conclusions drawn are specific to that particular sample.
• Possible errors: During hypothesis testing, there is a chance of committing type I error (rejecting a true null hypothesis) or type II error (failing to reject a false null hypothesis).
• Assumptions and requirements: Different tests have specific assumptions and requirements that must be met to accurately interpret results.

After reading this tutorial, you would have a much better understanding of hypothesis testing, one of the most important concepts in the field of Data Science . The majority of hypotheses are based on speculation about observed behavior, natural phenomena, or established theories.

If you are interested in statistics of data science and skills needed for such a career, you ought to explore Simplilearn’s Post Graduate Program in Data Science.

If you have any questions regarding this ‘Hypothesis Testing In Statistics’ tutorial, do share them in the comment section. Our subject matter expert will respond to your queries. Happy learning!

1. What is hypothesis testing in statistics with example?

Hypothesis testing is a statistical method used to determine if there is enough evidence in a sample data to draw conclusions about a population. It involves formulating two competing hypotheses, the null hypothesis (H0) and the alternative hypothesis (Ha), and then collecting data to assess the evidence. An example: testing if a new drug improves patient recovery (Ha) compared to the standard treatment (H0) based on collected patient data.

2. What is hypothesis testing and its types?

Hypothesis testing is a statistical method used to make inferences about a population based on sample data. It involves formulating two hypotheses: the null hypothesis (H0), which represents the default assumption, and the alternative hypothesis (Ha), which contradicts H0. The goal is to assess the evidence and determine whether there is enough statistical significance to reject the null hypothesis in favor of the alternative hypothesis.

Types of hypothesis testing:

• One-sample test: Used to compare a sample to a known value or a hypothesized value.
• Two-sample test: Compares two independent samples to assess if there is a significant difference between their means or distributions.
• Paired-sample test: Compares two related samples, such as pre-test and post-test data, to evaluate changes within the same subjects over time or under different conditions.
• Chi-square test: Used to analyze categorical data and determine if there is a significant association between variables.
• ANOVA (Analysis of Variance): Compares means across multiple groups to check if there is a significant difference between them.

3. What are the steps of hypothesis testing?

The steps of hypothesis testing are as follows:

• Formulate the hypotheses: State the null hypothesis (H0) and the alternative hypothesis (Ha) based on the research question.
• Set the significance level: Determine the acceptable level of error (alpha) for making a decision.
• Collect and analyze data: Gather and process the sample data.
• Compute test statistic: Calculate the appropriate statistical test to assess the evidence.
• Make a decision: Compare the test statistic with critical values or p-values and determine whether to reject H0 in favor of Ha or not.
• Draw conclusions: Interpret the results and communicate the findings in the context of the research question.

4. What are the 2 types of hypothesis testing?

• One-tailed (or one-sided) test: Tests for the significance of an effect in only one direction, either positive or negative.
• Two-tailed (or two-sided) test: Tests for the significance of an effect in both directions, allowing for the possibility of a positive or negative effect.

The choice between one-tailed and two-tailed tests depends on the specific research question and the directionality of the expected effect.

5. What are the 3 major types of hypothesis?

The three major types of hypotheses are:

• Null Hypothesis (H0): Represents the default assumption, stating that there is no significant effect or relationship in the data.
• Alternative Hypothesis (Ha): Contradicts the null hypothesis and proposes a specific effect or relationship that researchers want to investigate.
• Nondirectional Hypothesis: An alternative hypothesis that doesn't specify the direction of the effect, leaving it open for both positive and negative possibilities.

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9.E: Hypothesis Testing with One Sample (Exercises)

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These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.

9.1: Introduction

9.2: null and alternative hypotheses.

Some of the following statements refer to the null hypothesis, some to the alternate hypothesis.

State the null hypothesis, \(H_{0}\), and the alternative hypothesis. \(H_{a}\), in terms of the appropriate parameter \((\mu \text{or} p)\).

• The mean number of years Americans work before retiring is 34.
• At most 60% of Americans vote in presidential elections.
• The mean starting salary for San Jose State University graduates is at least $100,000 per year.
• Twenty-nine percent of high school seniors get drunk each month.
• Fewer than 5% of adults ride the bus to work in Los Angeles.
• The mean number of cars a person owns in her lifetime is not more than ten.
• About half of Americans prefer to live away from cities, given the choice.
• Europeans have a mean paid vacation each year of six weeks.
• The chance of developing breast cancer is under 11% for women.
• Private universities' mean tuition cost is more than $20,000 per year.
• \(H_{0}: \mu = 34; H_{a}: \mu \neq 34\)
• \(H_{0}: p \leq 0.60; H_{a}: p > 0.60\)
• \(H_{0}: \mu \geq 100,000; H_{a}: \mu < 100,000\)
• \(H_{0}: p = 0.29; H_{a}: p \neq 0.29\)
• \(H_{0}: p = 0.05; H_{a}: p < 0.05\)
• \(H_{0}: \mu \leq 10; H_{a}: \mu > 10\)
• \(H_{0}: p = 0.50; H_{a}: p \neq 0.50\)
• \(H_{0}: \mu = 6; H_{a}: \mu \neq 6\)
• \(H_{0}: p ≥ 0.11; H_{a}: p < 0.11\)
• \(H_{0}: \mu \leq 20,000; H_{a}: \mu > 20,000\)

Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin? The alternative hypothesis is:

• \(p < 0.30\)
• \(p \leq 0.30\)
• \(p \geq 0.30\)
• \(p > 0.30\)

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is:

• \(p = 0.20\)
• \(p > 0.20\)
• \(p < 0.20\)
• \(p \leq 0.20\)

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The null and alternative hypotheses are:

• \(H_{0}: \bar{x} = 4.5, H_{a}: \bar{x} > 4.5\)
• \(H_{0}: \mu \geq 4.5, H_{a}: \mu < 4.5\)
• \(H_{0}: \mu = 4.75, H_{a}: \mu > 4.75\)
• \(H_{0}: \mu = 4.5, H_{a}: \mu > 4.5\)

9.3: Outcomes and the Type I and Type II Errors

State the Type I and Type II errors in complete sentences given the following statements.

• The mean number of cars a person owns in his or her lifetime is not more than ten.
• Private universities mean tuition cost is more than $20,000 per year.
• Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years.
• Type I error: We conclude that more than 60% of Americans vote in presidential elections, when the actual percentage is at most 60%.Type II error: We conclude that at most 60% of Americans vote in presidential elections when, in fact, more than 60% do.
• Type I error: We conclude that the mean starting salary is less than $100,000, when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000.
• Type I error: We conclude that the proportion of high school seniors who get drunk each month is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who get drunk each month is 29% when, in fact, it is not 29%.
• Type I error: We conclude that fewer than 5% of adults ride the bus to work in Los Angeles, when the percentage that do is really 5% or more. Type II error: We conclude that 5% or more adults ride the bus to work in Los Angeles when, in fact, fewer that 5% do.
• Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in his or her lifetime is not more than 10 when, in fact, it is more than 10.
• Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half.
• Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not.
• Type I error: We conclude that the proportion is less than 11%, when it is really at least 11%. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11%, when in fact it is less than 11%.
• Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than $20,000.

For statements a-j in Exercise 9.109 , answer the following in complete sentences.

• State a consequence of committing a Type I error.
• State a consequence of committing a Type II error.

When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is “the drug is unsafe.” What is the Type II Error?

• To conclude the drug is safe when in, fact, it is unsafe.
• Not to conclude the drug is safe when, in fact, it is safe.
• To conclude the drug is safe when, in fact, it is safe.
• Not to conclude the drug is unsafe when, in fact, it is unsafe.

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. The Type I error is to conclude that the percent of EVC students who attended is ________.

• at least 20%, when in fact, it is less than 20%.
• 20%, when in fact, it is 20%.
• less than 20%, when in fact, it is at least 20%.
• less than 20%, when in fact, it is less than 20%.

It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average?

The Type II error is not to reject that the mean number of hours of sleep LTCC students get per night is at least seven when, in fact, the mean number of hours

• is more than seven hours.
• is at most seven hours.
• is at least seven hours.
• is less than seven hours.

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test, the Type I error is:

• to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher
• to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same
• to conclude that the mean hours per week currently is 4.5, when in fact, it is higher
• to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher

9.4: Distribution Needed for Hypothesis Testing

It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average? The distribution to be used for this test is \(\bar{X} \sim\) ________________

• \(N\left(7.24, \frac{1.93}{\sqrt{22}}\right)\)
• \(N\left(7.24, 1.93\right)\)

9.5: Rare Events, the Sample, Decision and Conclusion

The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population.

• Is this a test of one mean or proportion?
• State the null and alternative hypotheses. \(H_{0}\) : ____________________ \(H_{a}\) : ____________________
• Is this a right-tailed, left-tailed, or two-tailed test?
• What symbol represents the random variable for this test?
• In words, define the random variable for this test.
• \(x =\) ________________
• \(n =\) ________________
• \(p′ =\) _____________
• Calculate \(\sigma_{x} =\) __________. Show the formula set-up.
• State the distribution to use for the hypothesis test.
• Find the \(p\text{-value}\).
• Reason for the decision:
• Conclusion (write out in a complete sentence):

9.6: Additional Information and Full Hypothesis Test Examples

For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in [link] . Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

If you are using a Student's \(t\) - distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.)

A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using \(\alpha = 0.05\), is the data highly inconsistent with the claim?

• \(H_{0}: \mu \geq 50,000\)
• \(H_{a}: \mu < 50,000\)
• Let \(\bar{X} =\) the average lifespan of a brand of tires.
• normal distribution
• \(z = -2.315\)
• \(p\text{-value} = 0.0103\)
• Check student’s solution.
• alpha: 0.05
• Decision: Reject the null hypothesis.
• Reason for decision: The \(p\text{-value}\) is less than 0.05.
• Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.
• \((43,537, 49,463)\)

From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?

The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1% level?

• \(H_{0}: \mu = $1.00\)
• \(H_{a}: \mu \neq $1.00\)
• Let \(\bar{X} =\) the average cost of a daily newspaper.
• \(z = –0.866\)
• \(p\text{-value} = 0.3865\)
• \(\alpha: 0.01\)
• Decision: Do not reject the null hypothesis.
• Reason for decision: The \(p\text{-value}\) is greater than 0.01.
• Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1.
• \(($0.84, $1.06)\)

An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?

The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let \(x =\) the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten?

• \(H_{0}: \mu = 10\)
• \(H_{a}: \mu \neq 10\)
• Let \(\bar{X}\) the mean number of sick days an employee takes per year.
• Student’s t -distribution
• \(t = –1.12\)
• \(p\text{-value} = 0.300\)
• \(\alpha: 0.05\)
• Reason for decision: The \(p\text{-value}\) is greater than 0.05.
• Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten.
• \((4.9443, 11.806)\)

In 1955, Life Magazine reported that the 25 year-old mother of three worked, on average, an 80 hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level?

Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think?

• \(H_{0}: p \geq 0.6\)
• \(H_{a}: p < 0.6\)
• Let \(P′ =\) the proportion of students who feel more enriched as a result of taking Elementary Statistics.
• normal for a single proportion
• \(p\text{-value} = 0.1308\)
• Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched.

The “plus-4s” confidence interval is \((0.411, 0.648)\)

A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief.

Refer to Exercise 9.119 . Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout’s mean I.Q. is not four.

• \(H_{0}: \mu = 4\)
• \(H_{a}: \mu \neq 4\)
• Let \(\bar{X}\) the average I.Q. of a set of brown trout.
• two-tailed Student's t-test
• \(t = 1.95\)
• \(p\text{-value} = 0.076\)
• Reason for decision: The \(p\text{-value}\) is greater than 0.05
• Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four.
• \((3.8865,5.9468)\)

According to an article in Newsweek , the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7% girls). Suppose you don’t believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percent of girls born in China is 46.7?

A poll done for Newsweek found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent’s survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results.

• \(H_{a}: p < 0.13\)
• Let \(P′ =\) the proportion of Americans who have seen or sensed angels
• –2.688
• \(p\text{-value} = 0.0036\)
• Reason for decision: The \(p\text{-value}\)e is less than 0.05.
• Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%.

The“plus-4s” confidence interval is (0.0022, 0.0978)

The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it’s shorter. She asks ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours?

Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55.

Use the “Lap time” data for Lap 4 (see [link] ) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all twenty races given.

• \(H_{0}: \mu \geq 129\)
• \(H_{a}: \mu < 129\)
• Let \(\bar{X} =\) the average time in seconds that Terri finishes Lap 4.
• Student's t -distribution
• \(t = 1.209\)
• Conclusion: There is insufficient evidence to conclude that Terri’s mean lap time is less than 129 seconds.
• \((128.63, 130.37)\)

Use the “Initial Public Offering” data (see [link] ) to test the claim that the mean offer price was $18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices.

The following questions were written by past students. They are excellent problems!

"Asian Family Reunion," by Chau Nguyen

Every two years it comes around.

We all get together from different towns.

In my honest opinion,

It's not a typical family reunion.

Not forty, or fifty, or sixty,

But how about seventy companions!

The kids would play, scream, and shout

One minute they're happy, another they'll pout.

The teenagers would look, stare, and compare

From how they look to what they wear.

The men would chat about their business

That they make more, but never less.

Money is always their subject

And there's always talk of more new projects.

The women get tired from all of the chats

They head to the kitchen to set out the mats.

Some would sit and some would stand

Eating and talking with plates in their hands.

Then come the games and the songs

And suddenly, everyone gets along!

With all that laughter, it's sad to say

That it always ends in the same old way.

They hug and kiss and say "good-bye"

And then they all begin to cry!

I say that 60 percent shed their tears

But my mom counted 35 people this year.

She said that boys and men will always have their pride,

So we won't ever see them cry.

I myself don't think she's correct,

So could you please try this problem to see if you object?

• \(H_{0}: p = 0.60\)
• \(H_{a}: p < 0.60\)
• Let \(P′ =\) the proportion of family members who shed tears at a reunion.
• –1.71
• Reason for decision: \(p\text{-value} < \alpha\)
• Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the \(p\text{-value}\) and alpha are quite close, so other tests should be done.
• We are 95% confident that between 38.29% and 61.71% of family members will shed tears at a family reunion. \((0.3829, 0.6171)\). The“plus-4s” confidence interval (see chapter 8) is \((0.3861, 0.6139)\)

Note that here the “large-sample” \(1 - \text{PropZTest}\) provides the approximate \(p\text{-value}\) of 0.0438. Whenever a \(p\text{-value}\) based on a normal approximation is close to the level of significance, the exact \(p\text{-value}\) based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course.

"The Problem with Angels," by Cyndy Dowling

Although this problem is wholly mine,

The catalyst came from the magazine, Time.

On the magazine cover I did find

The realm of angels tickling my mind.

Inside, 69% I found to be

In angels, Americans do believe.

Then, it was time to rise to the task,

Ninety-five high school and college students I did ask.

Viewing all as one group,

Random sampling to get the scoop.

So, I asked each to be true,

"Do you believe in angels?" Tell me, do!

Hypothesizing at the start,

Totally believing in my heart

That the proportion who said yes

Would be equal on this test.

Lo and behold, seventy-three did arrive,

Out of the sample of ninety-five.

Now your job has just begun,

Solve this problem and have some fun.

"Blowing Bubbles," by Sondra Prull

Studying stats just made me tense,

I had to find some sane defense.

Some light and lifting simple play

To float my math anxiety away.

Blowing bubbles lifts me high

Takes my troubles to the sky.

POIK! They're gone, with all my stress

Bubble therapy is the best.

The label said each time I blew

The average number of bubbles would be at least 22.

I blew and blew and this I found

From 64 blows, they all are round!

But the number of bubbles in 64 blows

Varied widely, this I know.

20 per blow became the mean

They deviated by 6, and not 16.

From counting bubbles, I sure did relax

But now I give to you your task.

Was 22 a reasonable guess?

Find the answer and pass this test!

• \(H_{0}: \mu \geq 22\)
• \(H_{a}: \mu < 22\)
• Let \(\bar{X} =\) the mean number of bubbles per blow.
• –2.667
• \(p\text{-value} = 0.00486\)
• Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22.
• \((18.501, 21.499)\)

"Dalmatian Darnation," by Kathy Sparling

A greedy dog breeder named Spreckles

Bred puppies with numerous freckles

The Dalmatians he sought

Possessed spot upon spot

The more spots, he thought, the more shekels.

His competitors did not agree

That freckles would increase the fee.

They said, “Spots are quite nice

But they don't affect price;

One should breed for improved pedigree.”

The breeders decided to prove

This strategy was a wrong move.

Breeding only for spots

Would wreak havoc, they thought.

His theory they want to disprove.

They proposed a contest to Spreckles

Comparing dog prices to freckles.

In records they looked up

One hundred one pups:

Dalmatians that fetched the most shekels.

They asked Mr. Spreckles to name

An average spot count he'd claim

To bring in big bucks.

Said Spreckles, “Well, shucks,

It's for one hundred one that I aim.”

Said an amateur statistician

Who wanted to help with this mission.

“Twenty-one for the sample

Standard deviation's ample:

They examined one hundred and one

Dalmatians that fetched a good sum.

They counted each spot,

Mark, freckle and dot

And tallied up every one.

Instead of one hundred one spots

They averaged ninety six dots

Can they muzzle Spreckles’

Obsession with freckles

Based on all the dog data they've got?

"Macaroni and Cheese, please!!" by Nedda Misherghi and Rachelle Hall

As a poor starving student I don't have much money to spend for even the bare necessities. So my favorite and main staple food is macaroni and cheese. It's high in taste and low in cost and nutritional value.

One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! $2.02 !!! Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most $1, but now I wasn't so sure. However, I was determined to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook:

Price per box of Mac and Cheese:

• 5 stores @ $2.02
• 15 stores @ $0.25
• 3 stores @ $1.29
• 6 stores @ $0.35
• 4 stores @ $2.27
• 7 stores @ $1.50
• 5 stores @ $1.89
• 8 stores @ 0.75.

I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most $1, then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. (After all, as a poor starving student I can't be expected to feed our class of animals!)

• \(H_{0}: \mu \leq 1\)
• \(H_{a}: \mu > 1\)
• Let \(\bar{X} =\) the mean cost in dollars of macaroni and cheese in a certain town.
• Student's \(t\)-distribution
• \(t = 0.340\)
• \(p\text{-value} = 0.36756\)
• Conclusion: The mean cost could be $1, or less. At the 5% significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than $1.
• \((0.8291, 1.241)\)

"William Shakespeare: The Tragedy of Hamlet, Prince of Denmark," by Jacqueline Ghodsi

THE CHARACTERS (in order of appearance):

• HAMLET, Prince of Denmark and student of Statistics
• POLONIUS, Hamlet’s tutor
• HOROTIO, friend to Hamlet and fellow student

Scene: The great library of the castle, in which Hamlet does his lessons

(The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is reprimanding Hamlet regarding the latter’s recent experience. Horatio is seated at the large table at right stage.)

POLONIUS: My Lord, how cans’t thou admit that thou hast seen a ghost! It is but a figment of your imagination!

HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin damn’d, be their intents wicked or charitable.

POLONIUS If thou doest insist upon thy wretched vision then let me invest your time; be true to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns to Horatio.) Did not Hamlet himself say, “What piece of work is man, how noble in reason, how infinite in faculties? Then let not this foolishness persist. Go, Horatio, make a survey of three-and-sixty and discover what the true proportion be. For my part, I will never succumb to this fantasy, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one hundred hold true.

HORATIO (to Hamlet): What should we do, my Lord?

HAMLET: Go to thy purpose, Horatio.

HORATIO: To what end, my Lord?

HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no.

(Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.)

(The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and places some books upon the table just a moment before Horatio enters.)

POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations?

HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might not this believe, without the sensible and true avouch of mine own eyes.

POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look to’t I charge you, my Lord. Come Horatio, let us go together, for this is not our test. (Horatio and Polonius leave together.)

HAMLET: To reject, or not reject, that is the question: whether ‘tis nobler in the mind to suffer the slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing, end them. (Hamlet resignedly attends to his task.)

(Curtain falls)

"Untitled," by Stephen Chen

I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1%.

So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I'm right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right?

• \(H_{0}: p = 0.01\)
• \(H_{a}: p > 0.01\)
• Let \(P′ =\) the proportion of errors generated
• Normal for a single proportion
• Decision: Reject the null hypothesis
• Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01.

The“plus-4s” confidence interval is \((0.004, 0.144)\).

"Japanese Girls’ Names"

by Kumi Furuichi

It used to be very typical for Japanese girls’ names to end with “ko.” (The trend might have started around my grandmothers’ generation and its peak might have been around my mother’s generation.) “Ko” means “child” in Chinese characters. Parents would name their daughters with “ko” attaching to other Chinese characters which have meanings that they want their daughters to become, such as Sachiko—happy child, Yoshiko—a good child, Yasuko—a healthy child, and so on.

However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names which end with “ko.” More and more, parents seem to have become creative, modernized, and, sometimes, westernized in naming their children.

I have a feeling that, while 70 percent or more of my mother’s generation would have names with “ko” at the end, the proportion has dropped among my peers. I wrote down all my Japanese friends’, ex-classmates’, co-workers, and acquaintances’ names that I could remember. Following are the names. (Some are repeats.) Test to see if the proportion has dropped for this generation.

Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko.

"Phillip’s Wish," by Suzanne Osorio

My nephew likes to play

Chasing the girls makes his day.

He asked his mother

If it is okay

To get his ear pierced.

She said, “No way!”

To poke a hole through your ear,

Is not what I want for you, dear.

He argued his point quite well,

Says even my macho pal, Mel,

Has gotten this done.

It’s all just for fun.

C’mon please, mom, please, what the hell.

Again Phillip complained to his mother,

Saying half his friends (including their brothers)

Are piercing their ears

And they have no fears

He wants to be like the others.

She said, “I think it’s much less.

We must do a hypothesis test.

And if you are right,

I won’t put up a fight.

But, if not, then my case will rest.”

We proceeded to call fifty guys

To see whose prediction would fly.

Nineteen of the fifty

Said piercing was nifty

And earrings they’d occasionally buy.

Then there’s the other thirty-one,

Who said they’d never have this done.

So now this poem’s finished.

Will his hopes be diminished,

Or will my nephew have his fun?

• \(H_{0}: p = 0.50\)
• \(H_{a}: p < 0.50\)
• Let \(P′ =\) the proportion of friends that has a pierced ear.
• –1.70
• \(p\text{-value} = 0.0448\)
• Reason for decision: The \(p\text{-value}\) is less than 0.05. (However, they are very close.)
• Conclusion: There is sufficient evidence to support the claim that less than 50% of his friends have pierced ears.
• Confidence Interval: \((0.245, 0.515)\): The “plus-4s” confidence interval is \((0.259, 0.519)\).

"The Craven," by Mark Salangsang

Once upon a morning dreary

In stats class I was weak and weary.

Pondering over last night’s homework

Whose answers were now on the board

This I did and nothing more.

While I nodded nearly napping

Suddenly, there came a tapping.

As someone gently rapping,

Rapping my head as I snore.

Quoth the teacher, “Sleep no more.”

“In every class you fall asleep,”

The teacher said, his voice was deep.

“So a tally I’ve begun to keep

Of every class you nap and snore.

The percentage being forty-four.”

“My dear teacher I must confess,

While sleeping is what I do best.

The percentage, I think, must be less,

A percentage less than forty-four.”

This I said and nothing more.

“We’ll see,” he said and walked away,

And fifty classes from that day

He counted till the month of May

The classes in which I napped and snored.

The number he found was twenty-four.

At a significance level of 0.05,

Please tell me am I still alive?

Or did my grade just take a dive

Plunging down beneath the floor?

Upon thee I hereby implore.

Toastmasters International cites a report by Gallop Poll that 40% of Americans fear public speaking. A student believes that less than 40% of students at her school fear public speaking. She randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test to determine if the percent at her school is less than 40%.

• \(H_{0}: p = 0.40\)
• \(H_{a}: p < 0.40\)
• Let \(P′ =\) the proportion of schoolmates who fear public speaking.
• –1.01
• \(p\text{-value} = 0.1563\)
• Conclusion: There is insufficient evidence to support the claim that less than 40% of students at the school fear public speaking.
• Confidence Interval: \((0.3241, 0.4240)\): The “plus-4s” confidence interval is \((0.3257, 0.4250)\).

Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California, was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68% represents California. NOTE: For more accurate results, use more California community colleges and this past year's data.

According to an article in Bloomberg Businessweek , New York City's most recent adult smoking rate is 14%. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if the rate is still 14% or if it has decreased.

• \(H_{0}: p = 0.14\)
• \(H_{a}: p < 0.14\)
• Let \(P′ =\) the proportion of NYC residents that smoke.
• –0.2756
• \(p\text{-value} = 0.3914\)
• At the 5% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14.
• Confidence Interval: \((0.0502, 0.2070)\): The “plus-4s” confidence interval (see chapter 8) is \((0.0676, 0.2297)\).

The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. She randomly surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test.

Registered nurses earned an average annual salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110 for California nurses. The sample average was $71,121 with a sample standard deviation of $7,489. Conduct a hypothesis test.

• \(H_{0}: \mu = 69,110\)
• \(H_{0}: \mu > 69,110\)
• Let \(\bar{X} =\) the mean salary in dollars for California registered nurses.
• \(t = 1.719\)
• \(p\text{-value}: 0.0466\)
• Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110.
• \(($68,757, $73,485)\)

La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old.

Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin?

After conducting the test, your decision and conclusion are

• Reject \(H_{0}\): There is sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
• Do not reject \(H_{0}\): There is not sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.
• Do not reject \(H_{0}\): There is not sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
• Reject \(H_{0}\): There is sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing.

At a 1% level of significance, an appropriate conclusion is:

• There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
• There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is more than 20%.
• There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
• There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is at least 20%.

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test.

At a significance level of \(a = 0.05\), what is the correct conclusion?

• There is enough evidence to conclude that the mean number of hours is more than 4.75
• There is enough evidence to conclude that the mean number of hours is more than 4.5
• There is not enough evidence to conclude that the mean number of hours is more than 4.5
• There is not enough evidence to conclude that the mean number of hours is more than 4.75

Instructions: For the following ten exercises,

Hypothesis testing: For the following ten exercises, answer each question.

State the null and alternate hypothesis.

State the \(p\text{-value}\).

State \(\alpha\).

What is your decision?

Write a conclusion.

Answer any other questions asked in the problem.

According to the Center for Disease Control website, in 2011 at least 18% of high school students have smoked a cigarette. An Introduction to Statistics class in Davies County, KY conducted a hypothesis test at the local high school (a medium sized–approximately 1,200 students–small city demographic) to determine if the local high school’s percentage was lower. One hundred fifty students were chosen at random and surveyed. Of the 150 students surveyed, 82 have smoked. Use a significance level of 0.05 and using appropriate statistical evidence, conduct a hypothesis test and state the conclusions.

A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate?

• \(H_{0}: p = 0.488\) \(H_{a}: p \neq 0.488\)
• \(p\text{-value} = 0.0114\)
• \(\alpha = 0.05\)
• Reject the null hypothesis.
• At the 5% level of significance, there is enough evidence to conclude that 48.8% of families own stocks.
• The survey does not appear to be accurate.

Driver error can be listed as the cause of approximately 54% of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using \(\alpha = 0.05\), is the AAA proportion accurate?

The US Department of Energy reported that 51.7% of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the \(\alpha = 0.05\) level in Kentucky? Are the results applicable across the country? Why?

• \(H_{0}: p = 0.517\) \(H_{0}: p \neq 0.517\)
• \(p\text{-value} = 0.9203\).
• \(\alpha = 0.05\).
• Do not reject the null hypothesis.
• At the 5% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517.
• However, we cannot generalize this result to the entire nation. First, the sample’s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation.

For Americans using library services, the American Library Association claims that at most 67% of patrons borrow books. The library director in Owensboro, Kentucky feels this is not true, so she asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY? Use \(\alpha = 0.01\) level of significance. What is the possible proportion of patrons that do borrow books from the Owensboro Library?

The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the \(\alpha = 0.05 level\), can it be concluded that the mean rainfall was below the reported average? What if \(\alpha = 0.01\)? Assume the amount of summer rainfall follows a normal distribution.

• \(H_{0}: \mu \geq 11.52\) \(H_{a}: \mu < 11.52\)
• \(p\text{-value} = 0.000002\) which is almost 0.
• At the 5% significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average.
• We would make the same conclusion if alpha was 1% because the \(p\text{-value}\) is almost 0.

A survey in the N.Y. Times Almanac finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX chamber of commerce feels that Austin’s commute time is less and wants to publicize this fact. The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the \(\alpha = 0.10\) level, is the Austin, TX commute significantly less than the mean commute time for the 15 largest US cities?

A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals

3; 2; 1; 3; 7; 2; 9; 4; 6; 6; 8; 0; 5; 6; 4; 2; 1; 3; 4; 1

At the \(\alpha = 0.05\) level can it be concluded that the sample mean is higher than 5.8 visits per year?

• \(H_{0}: \mu \leq 5.8\) \(H_{a}: \mu > 5.8\)
• \(p\text{-value} = 0.9987\)
• At the 5% level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year.

According to the N.Y. Times Almanac the mean family size in the U.S. is 3.18. A sample of a college math class resulted in the following family sizes:

5; 4; 5; 4; 4; 3; 6; 4; 3; 3; 5; 5; 6; 3; 3; 2; 7; 4; 5; 2; 2; 2; 3; 2

At \(\alpha = 0.05\) level, is the class’ mean family size greater than the national average? Does the Almanac result remain valid? Why?

The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At α = 0.01 level, is the student academic group’s claim correct?

• \(H_{0}: \mu \geq 150\) \(H_{0}: \mu < 150\)
• \(p\text{-value} = 0.0622\)
• \(\alpha = 0.01\)
• At the 1% significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average.
• The student academic group’s claim appears to be correct.

9.7: Hypothesis Testing of a Single Mean and Single Proportion

PoD in Public Health: Estimation & Hypothesis Testing

Maria kamenetsky, ms, summer 2021, central limit theorem, estimation of the variance of a distribution, one-tailed tests, two-tailed tests, equal variance, unequal variance, equal vs. unequal variance test, additional resources.

Statistical inference can be divided indo two parts: 1) estimation and 2) hypothesis testing . Estimation estimates values of specific population parameters; hypothesis testing tests whether the value of a population parameter is equal to a specified value. In order to make statistical inference about population parameters, samples must be taken. A random sample is a selection of some members of a population of interest such that each member is independently chosen and has a known non-zero probability of being selected. A simple random sample is a random sample in which each group member has the same probability of being selected. The study population is the group of interest to be studied, from which the random sample is taken.

Estimation of the Mean of a Distribution

Suppose we have a random sample \(x_1, x_2, \dots, x_n\) which we will use to estimate the population parameters \(\mu\) (mean) and \(\sigma^2\) (variance) of the underlying distribution. An estimator of a parameter, for example \(\theta\) is denoted as \(\hat{\theta}\) ; the estimator is unbiased if the expectation of \(\hat{\theta}\) is equal to \(\theta\) ( \(E(\hat{\theta}) = \theta\) ). This means that the average value of \(\hat{\theta}\) over a large number of \(n\) repeated samples is \(\theta\) , the true population parameter. We have already been working with an unbiased estimator, \(\bar{X}\) , where

\[\bar{X} = \sum_{i=1}^n X_i/n\]

When the underlying distribution of the population is normal, it can be shown that the unbiased estimator of \(\mu\) with the smallest variance is \(\bar{X}\) . The sampling distribution of \(\bar{X}\) is the distribution of values of \(\bar{x}\) over all possible samples of size \(n\) that could have been drawn from the study population.

The standard error of the mean (SEM) or standard error is the estimated standard deviation from a set of sample means from repeated samples from the study population of size \(n\) with a population with underlying variance \(\sigma^2\) . Formally, let \(X_1, X_2, \dots, X_n\) be a random sample from a population with underlying mean \(\mu\) and variance \(\sigma^2\) , Then the set of sample means from repeated random samples of size \(n\) from this population will have variance \(\sigma^2/n\) , where the standard deviation of this set of sample means is:

\[SE(\bar{X})=\sigma/\sqrt{n}\]

Note that the standard error is not the standard deviation of an individual observation, but instead the standard deviation of the sample mean \(\bar{X}\) .

A sample proportion is a mean of independent Bernoulli (0 or 1) trials. If the population proportion, \(\pi\) , is known, then the standard error of the proportion is:

\[SE(\pi) = \sqrt{\frac{\pi(1-\pi)}{n}}\]

We can estimate the population proportion ( \(\pi\) ) by the sample proportion \(p\) , giving:

\[SE(p) = \sqrt{\frac{p(1-p)}{n}}\]

If an underlying distribution is normal, then the sample mean is normally distributed with mean \(\mu\) and variance \(\sigma^2/n\) ( \(\bar{X} \sim N(\mu, \sigma^2/n)\) ). If an underlying distribution is not normally distributed, the Central Limit Theorem (CLT) may be applied in certain situations to enable use to perform statistical inference based on the approximate normality of the sample mean. Let \(X_1, \dots, X_n\) be a random sample from a study population with mean \(\mu\) and variance \(\sigma^2\) . Then as \(n \rightarrow \infty\) , \(\bar{X} \sim N(\mu, \sigma^2/n)\) even if the underlying distribution of individual observations in the population is not normally distributed.

Suppose you have a series of basal body temperature measurements from a participant: 97.2,97.8, 98.1, 96.9, 97.4, 98.3, 97.7, 97.1, 96.8, 97.3. What is the best estimate of this individual’s body temperature ( \(\mu\) ) and how precise is this estimate?

We start by creating a vector called temps in R by combining the 10 measurements using the concatenate ( c() ) function:

The best estimate of this individual’s body temperature will be the sample mean, \(\bar{x}\) which we calculate using the mean() function:

The mean temperature is 97.46 degrees Fahrenheit. The precision of this estimate is given to us by its standard error, which is \(s/\sqrt{10}\) , where \(s\) is the standard deviation. We first calculate the standard deviation using the sd() function and divide it by 10, which can get programatically with length(temps) :

If we wanted to, we could now calculate a 95% confidence interval:

The 95% confidence interval for the mean is (97.3616955839129, 97.5583044160871).

Suppose you have a study with 22000 participants - 3200 of these participants self-report as ever-smokers. Calculate the sample proportion and standard error of the proportion.

First, we calculate the sample proportion of ever-smokers and store the result into the object prop_smokers :

Sixteen percent of respondents report being ever-smokers. Next, we calculate the standard error of the proportion:

The standard error is 0. If we assume a normal sampling distribution due to the large sample size of the study, we can calculate a 95% confidence interval:

The 95% confidence interval by normal approximation for the proportion is (0.140795725103603, 0.150113365805488).

The sample variance is defined as:

\[s^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i -\bar{x})^2\]

The sample variance, \(S^2\) is an unbiased estimator of \(\sigma^2\) over all possible random samples of size \(n\) possibly drawn from a study population. In other words, \(E(S^2) = \sigma^2\) .

Hypothesis Testing

The null hypothesis ( \(H_0\) ) is the hypothesis to be tested. The alternative hypothesis ( \(H_A\) ) is the hypothesis that contradicts the null. With hypothesis testing, you can reject or fail to reject \(H_0\) . If we were to know ground truth, there are four possible outcomes that can occur:

• Fail to reject \(H_0\) when \(H_0\) is really true.
• Fail to reject \(H_0\) when \(H_A\) is really true.
• Reject \(H_0\) when \(H_0\) is really true.
• Reject \(H_0\) when \(H_A\) is really true.

It is important to note that the null hypothesis cannot be accepted - we only have evidence to reject or fail to reject the null hypothesis.

Note the discordant pairs in the table above: \(H_0\) is true and \(H_0\) is rejected and \(H_A\) is true and \(H_0\) is not rejected . In these cases, an error has been made. There are two types of errors that can be made:

• Type I Error : The probability of rejecting the null hypothesis ( \(H_0\) ) when \(H_0\) is really true.
• Type II Error : The probability of failing to reject the null hypothesis when \(H_A\) is true.

The probability of a Type I error is denoted by \(\alpha\) and is commonly referred to as the significance level of a test.

The probability of a Type II error is denoted as \(\beta\) . The power of a test is \(1-\beta=1 - \text{probability of a type II error} = Pr(\text{rejecting } H_0 | H_A \text{ is true})\) .

The acceptance region or non-rejection region is the range of values of \(\bar{x}\) for which \(H_0\) is not rejected. Similarly, the rejection region is the range of \(\bar{x}\) values for which \(H_0\) is rejected.

A one-tailed test is a test in which the values of the parameter being studies (for example, \(\mu\) ) under the alternative hypothesis are allowed to either be greater than or less than the values of the parameter under the null hypothesis (for example, \(\mu_0\) ), but not both (this would be a two-tailed test).

We can test the following hypothesis:

\[H_0: \mu = \mu_0\] \[H_A: \mu < \mu_0\]

with \(\sigma\) unknown and significance level \(\alpha\) . We first calculate a test statistic:

\[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\]

• if the test statistic \(t\) (above) is less than the critical value \(t_{n-1,\alpha}\) , then we reject \(H_0\) .
• if the test statistic \(t\) is \(\geq t_{n-1,\alpha}\) , then we fail to reject \(H_0\) . Figure 1 visualizes the critical value approach to hypothesis testing:

Visualization of hypothesis testing approach for one-tailed test (lower tail).

Consider the following example. From a single hospital, you have a random sample of 200 consecutive births, with the mean birthweight ( \(\bar{x}\) ) is 112 oz and the sample standard deviation ( \(s\) ) is 20 oz. From a nationwide study of millions of births, it is known that the mean birthweight in the united states is 120 oz.

Using a one-sample \(t\) -test, test the hypothesis that \(H_0: \mu = 112\) vs.  \(H_A: \mu < 112\)

First, we calculate the test statistics :

To find the critical value , the degrees of freedom for the t-distribution will be \(n-1\) or \(200-1=199\) . We use the function qt() or quantile function with arguments p=0.05 (the significance level) and df=199 (the degrees of freedom):

The test statistic (-5.6568542) is much smaller than the critical value (-1.6525467). Thus, we can reject \(H_0\) in favor of \(H_A\) at the 0.05 significance level. The birth weights at this particular hospital are lower than the mean population birth weight. Figure 2 visualizes the components to the critical value testing approach to hypothesis testing:

Visualization of hypothesis testing approach to one-tailed test for the birth weight example.

If instead of having just the summary statistics of the mean and standard deviation we had the actual data (as a vector from a data frame), we could use the t.test() function. To demonstrate the function, I will first generate 200 samples from a normal distribution with mean 112 and standard deviation 20:

The result here differs slightly from our calculations (t-statistic -6.1263 vs -5.6568542). However this is due to the variation of the random samples drawn from the distribution with mean 112 and sd 20. From above, we can see that though the observed sample mean of the samples is near 120 (111.8285911) and the standard deviation of the samples is near 20 (18.8631976), there is some slight variation. The final conclusion of rejecting \(H_0\) , however, is consistent.

From the test above, we also have a reported p-value. The p-value is the probability of obtaining a test statistic as extreme or more extreme than the observed test statistic. In the above results, the p-value is $2.365 ^{-9} $, very very small! There is very strong evidence against the null hypothesis.

Guidance on interpreting p-values and the strength of evidence against the null hypothesis can be summarized in the table below:

With a two-tailed test, the values of the parameter being studied (for example, \(\mu\) ) under the alternative hypothesis are allowed to be either greater than or less than the values of the parameter under the null hypothesis (for example, \(\mu_0\) ). In other words, the alternative is that the values of the parameter are not equal to that of the values under the null.

We can test the following two-tailed hypothesis:

\[H_0: \mu = \mu_0\] \[H_A: \mu \neq \mu_0\]

with significance level \(\alpha\) . We calculate the test statistic:

• if the test statistic \(|t|> t_{n-1, 1-\alpha/2}\) , then \(H_0\) is rejected.
• if \(|t| \leq t_{n-1, 1-\alpha/2}\) , then we fail to reject \(H_0\) .

Visualization of hypothesis testing approach for two-tailed test.

Consider the following example: We would like to test the hypothesis that the mean cholesterol level of female immigrants from Country A are different from the mean cholesterol of the U.S. population. Suppose that you are given that the mean cholesterol level for women in the United States is 190 mg/dL. You conduct a study where from a random sample of women from Country A, you take blood samples and find the mean level in this group to be 180.23 mg/DL with a standard deviation of 40 mg/DL.

For a two-sided one-sample test , we first calculate the test statistic:

\[t = \frac{180.23 - 190}{40/\sqrt{100}} = -2.4425\] We can do this in R :

To find the critical value , the degrees of freedom for the t-distribution will be \(n-1\) or \(100-1=99\) . We use the function qt() or quantile function with arguments p=0.05 (the significance level) and df=99 (the degrees of freedom). The two-sided critical values are \(c_1 = t_{99, 0.025}\) and \(c_2 = t_{99, 0.975}\)

The test statistic (-2.4425) is much smaller than the critical value (-1.984217). Thus, we can reject \(H_0\) in favor of \(H_A\) at the 0.05 significance level. The mean cholesterol of women from Country A is lower than the mean cholesterol for U.S. women. Figure 3 visualizes the components to the critical value testing approach to hypothesis testing:

Visualization of hypothesis testing approach for two-tailed test for mean cholesterol in the U.S. example.

If instead of having just the summary statistics of the mean and standard deviation we had the actual data (as a vector from a data frame), we could use the t.test() function. To demonstrate the function, I will first generate 100 samples from a normal distribution with mean 180.23 and standard deviation 40:

Two-Sample Tests

When you have underlying parameters of two different populations (neither of which values are known) being compared, you have a two-sample hypothesis-testing problem .

We assume there are two normally-distributed samples where the first sample is a random sample of size \(n_1\) from a \(N(\mu_1, \sigma^2_1)\) distribution and the second sample is a random sample of size \(n_2\) from a \(N(\mu_2, \sigma^2_2)\) distribution, and \(\sigma^2_1 = \sigma^2_2 = \sigma^2_2\) .

Suppose we conduct a study where you are interested in comparing systolic blood pressure (SBP) in pre-menopausal oral contraceptive (OC) users and non-users. We can formulate the hypothesis as:

\[H_0: \mu_1 = \mu_2\] \[H_A: \mu_1 \neq \mu_2\]

If we assume that the variance of the users and non-users groups are the same ( \(\sigma^2_1 = \sigma^2_2 = \sigma^2\) ), then we have a two sample test for independent samples with equal variances . The test statistic is given by:

\[t = \frac{\bar{x}_1 - \bar{x}_2}{s \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}},\]

where \[s = \sqrt{\frac{(n_1 -1)s_1^2 + (n_2 -1)s_2^2}{n_1 +n_2 -2}}\]

The degrees of freedom will be \(n_1 + n_2 -2\) . \(H_0\) will be rejected if \(t > t_{n_1 + n_2 -2, 1-\alpha/2}\) or when \(t < t_{n_1 +n_2 -2, \alpha/2}\) .

Let’s continue with the OC users vs. non-users example. Let the mean SBP of a sample of 10 OC users be 132.03 mm Hg (standard deviation of 13.42) and the mean SBP of a sample of 19 non-OC users be 127.93 mm Hg (standard deviation of 15.32). Are these two groups different?

The test statistic 0.8946672 is greater than \(t_{n_1 +n_2 -2, \alpha/2}\) (-2.0518305) and less than \(t_{n_1 + n_2 -2, 1-\alpha/2}\) (2.0518305). We fail to reject \(H_0\) at \(\alpha=0.05\) .

We calculate the p-value by 2 times taking the distribution at the test statistic value ( \(2 \times P(t_{n_1+n_2-2} > t) = 2 \times 1- P(t_{n_1+n_2-2} \leq t)\) ):

There is no evidence against \(H_0\) .

Let us go back to the Melanoma data set from the R package MASS , introduced in the Descriptive Statistics notes.

Recall, the Melanoma data set contains data collected on 205 patients in Denmark on malignant melanoma. The variables are defined as below:

• time : survival time in days (possibly censored)
• status : 1=died from melanoma; 2=alive; 3=dead from other causes
• sex : 1=male; 0=female
• age : age in years
• year : year of operation
• thickness : tumor thickness in \(mm\)
• ulcer : 1=presence; 0=absence

Suppose we want to use a two-sided two-sample t-test whether thickness is the same across males and females ( sex ). We use the same function t.test() and add the argument var.equal=TRUE . We take the response thickness and use the ~ operator to specify that the two samples are sex .

From the results above, we see that the there is moderate evidence to reject \(H_0\) at the \(\alpha=0.05\) level. There is a difference in thickness between males and females.

If you have two samples where the two variances are not equal , then you can use a two-sample test for unequal variances . This is also known as Satterthwaite’s Method . We assume there are two normally-distributed samples where the first sample is a random sample of size \(n_1\) from a \(N(\mu_1, \sigma^2_1)\) distribution and the second sample is a random sample of size \(n_2\) from a \(N(\mu_2, \sigma^2_2)\) distribution, and \(\sigma^2_1 \neq \sigma^2_2\) .

The hypothesis is still:

The test statistic is given by:

\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]

The approximate degrees of freedom ( \(d'\) ) are given by: \[d' = \frac{(s_1^2/n_1 + s_2^2/n_2)^2}{(s_1^2/n_1)^2/(n_1 -1) + (s_2^2/n_2)^2/(n_2-1)}\]

The same rejection rules apply as before.

Consider a study of cholesterol. Suppose that you have conducted a study of cholesterol levels from 100 children (aged 2-14 years old) whose biological fathers had died of heart disease (group 1). You also collect a second sample of cholesterol levels from 90 children whose biological fathers had no history or diagnosis of heart disease and were still living as controls (group 2). In the first group, the mean cholesterol level is 210.3 mg/dL (standard deviation 32.9 mg/DL). In the second group, the mean cholesterol level is 190.1 mg/DL (standard deviation 22.3 mg/DL). Perform a two-sample \(t\) test for unequal variances.

The test statistic 4.9957252 is greater than (-1.9736036). We reject \(H_0\) at \(\alpha=0.05\) .

The p-value is 1.4108816^{-6}. There is very strong evidence against \(H_0\) .

Suppose we want to use a two-sided two-sample t-test whether thickness is the same across males and females ( sex ). However this time, we want to use a two-sided \(t\) test for unequal variance. We use the same function t.test() , but add in the additional argument var.equal=FALSE (the default).

For the two-sided \(t\) test with equal variances, we assume a common variance \(\sigma^2\) and take a weighted average of the individual sample variances. To determine if this is a reasonable approach and to determine whether or not to use the equal vs. unequal variance \(t\) test, we can develop a hypothesis. Namely, that:

\[H_0: \sigma^2_1 = \sigma^2_2\]

\[H_A: \sigma^2_1 \neq \sigma^2_2\]

where the two samples are assumed to be independent random samples from \(N(\mu_1, \sigma^2_1)\) and from \(N(\mu_2, \sigma^2_2)\) . We use the sample variances \(s_1^2\) and \(s^2_2\) , which are the estimators of \(\sigma^2_1\) and \(\sigma^2_2\) . We want to use the relative magnitudes of the sample variances or their ratio, \(s^2_1/s_2^2\) , where \(H_0\) would be rejected if the ratio is too large or too small. A rule of thumb , is that if the ratio \(s^2_1/s_2^2\) > 2 or \(s^2_1/s_2^2\) < 1/2, then the variances are not equal and you would reject \(H_0\) . Formally, we can use the F test for the equality of two variances . The test statistic is:

\[F = s^2_1/s_2^2\]

If \(F< F_{n_1-1,n_2-1, \alpha/2}\) or \(F > F_{n_1-1,n_2-1, 1-\alpha/2}\) , then \(H_0\) is rejected. If \(F_{n_1-1,n_2-1, \alpha/2} \leq F \leq F_{n_1-1,n_2-1, 1-\alpha/2}\) , then we fail to reject \(H_0\) . \(F\) here is the F-distribution .

In the above examples, we used both the equal and unequal variances \(t\) test to test if there was a differences in thickness between females and males. We will use the F-test for the equality of two variances to identify which test we should use.

The F-statistic 0.7619462 is greater than critical value 1 (0.6764302) and less than critical value 2 (1.506064). We therefore fail to reject the \(H_0\) . There is no evidence that the two variances are not equal to each other and we can proceed with the test for equal variances.

We can also use the function var.test() to more easily perform the same F-test for equal variances:

Notice that we obtain the same F-statistic and reach the same conclusions.

• Palta, Mari. PHS 551: Introduction to Biostatistics Lecture Notes, Fall 2016.
• Rosner, Bernard, Fundamentals of Biostatistics, Seventh Edition . Chapter 6 . Epidemiology , Fifth Edition. Chapter3, Chapter 9-Chapter 14. --> Epidemiology: Beyond the Basics . *Chapter 1: Basic Study Designs in Analytical Epidemiology*. -->
• Examples adapted from (2), (3), (5), (6).
• Data Carpentry, Introduction to R
• Data Carpentry, Data Visualization with ggplot2

Copyright (c) 2021 Maria Kamenetsky

Confidence Interval/Hypothesis Testing for the Difference of Means Cal

How does the confidence interval/hypothesis testing for the difference of means calculator work.

Free Confidence Interval/Hypothesis Testing for the Difference of Means Calculator - Given two large or two small distriutions, this will determine a (90-99)% estimation of confidence interval for the difference of means for small or large sample populations. Also performs hypothesis testing including standard error calculation. This calculator has 7 inputs.

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