consider the hypothesis test h0

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4.4: Hypothesis Testing

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• Page ID 283

• David Diez, Christopher Barr, & Mine Çetinkaya-Rundel
• OpenIntro Statistics

Is the typical US runner getting faster or slower over time? We consider this question in the context of the Cherry Blossom Run, comparing runners in 2006 and 2012. Technological advances in shoes, training, and diet might suggest runners would be faster in 2012. An opposing viewpoint might say that with the average body mass index on the rise, people tend to run slower. In fact, all of these components might be influencing run time.

In addition to considering run times in this section, we consider a topic near and dear to most students: sleep. A recent study found that college students average about 7 hours of sleep per night.15 However, researchers at a rural college are interested in showing that their students sleep longer than seven hours on average. We investigate this topic in Section 4.3.4.

Hypothesis Testing Framework

The average time for all runners who finished the Cherry Blossom Run in 2006 was 93.29 minutes (93 minutes and about 17 seconds). We want to determine if the run10Samp data set provides strong evidence that the participants in 2012 were faster or slower than those runners in 2006, versus the other possibility that there has been no change. 16 We simplify these three options into two competing hypotheses :

• H 0 : The average 10 mile run time was the same for 2006 and 2012.
• H A : The average 10 mile run time for 2012 was different than that of 2006.

We call H 0 the null hypothesis and H A the alternative hypothesis.

Null and alternative hypotheses

• The null hypothesis (H 0 ) often represents either a skeptical perspective or a claim to be tested.
• The alternative hypothesis (H A ) represents an alternative claim under consideration and is often represented by a range of possible parameter values.

15 theloquitur.com/?p=1161

16 While we could answer this question by examining the entire population data (run10), we only consider the sample data (run10Samp), which is more realistic since we rarely have access to population data.

The null hypothesis often represents a skeptical position or a perspective of no difference. The alternative hypothesis often represents a new perspective, such as the possibility that there has been a change.

Hypothesis testing framework

The skeptic will not reject the null hypothesis (H 0 ), unless the evidence in favor of the alternative hypothesis (H A ) is so strong that she rejects H 0 in favor of H A .

The hypothesis testing framework is a very general tool, and we often use it without a second thought. If a person makes a somewhat unbelievable claim, we are initially skeptical. However, if there is sufficient evidence that supports the claim, we set aside our skepticism and reject the null hypothesis in favor of the alternative. The hallmarks of hypothesis testing are also found in the US court system.

Exercise \(\PageIndex{1}\)

A US court considers two possible claims about a defendant: she is either innocent or guilty. If we set these claims up in a hypothesis framework, which would be the null hypothesis and which the alternative? 17

Jurors examine the evidence to see whether it convincingly shows a defendant is guilty. Even if the jurors leave unconvinced of guilt beyond a reasonable doubt, this does not mean they believe the defendant is innocent. This is also the case with hypothesis testing: even if we fail to reject the null hypothesis, we typically do not accept the null hypothesis as true. Failing to find strong evidence for the alternative hypothesis is not equivalent to accepting the null hypothesis.

17 H 0 : The average cost is $650 per month, \(\mu\) = $650.

In the example with the Cherry Blossom Run, the null hypothesis represents no difference in the average time from 2006 to 2012. The alternative hypothesis represents something new or more interesting: there was a difference, either an increase or a decrease. These hypotheses can be described in mathematical notation using \(\mu_{12}\) as the average run time for 2012:

• H 0 : \(\mu_{12} = 93.29\)
• H A : \(\mu_{12} \ne 93.29\)

where 93.29 minutes (93 minutes and about 17 seconds) is the average 10 mile time for all runners in the 2006 Cherry Blossom Run. Using this mathematical notation, the hypotheses can now be evaluated using statistical tools. We call 93.29 the null value since it represents the value of the parameter if the null hypothesis is true. We will use the run10Samp data set to evaluate the hypothesis test.

Testing Hypotheses using Confidence Intervals

We can start the evaluation of the hypothesis setup by comparing 2006 and 2012 run times using a point estimate from the 2012 sample: \(\bar {x}_{12} = 95.61\) minutes. This estimate suggests the average time is actually longer than the 2006 time, 93.29 minutes. However, to evaluate whether this provides strong evidence that there has been a change, we must consider the uncertainty associated with \(\bar {x}_{12}\).

1 6 The jury considers whether the evidence is so convincing (strong) that there is no reasonable doubt regarding the person's guilt; in such a case, the jury rejects innocence (the null hypothesis) and concludes the defendant is guilty (alternative hypothesis).

We learned in Section 4.1 that there is fluctuation from one sample to another, and it is very unlikely that the sample mean will be exactly equal to our parameter; we should not expect \(\bar {x}_{12}\) to exactly equal \(\mu_{12}\). Given that \(\bar {x}_{12} = 95.61\), it might still be possible that the population average in 2012 has remained unchanged from 2006. The difference between \(\bar {x}_{12}\) and 93.29 could be due to sampling variation, i.e. the variability associated with the point estimate when we take a random sample.

In Section 4.2, confidence intervals were introduced as a way to find a range of plausible values for the population mean. Based on run10Samp, a 95% confidence interval for the 2012 population mean, \(\mu_{12}\), was calculated as

\[(92.45, 98.77)\]

Because the 2006 mean, 93.29, falls in the range of plausible values, we cannot say the null hypothesis is implausible. That is, we failed to reject the null hypothesis, H 0 .

Double negatives can sometimes be used in statistics

In many statistical explanations, we use double negatives. For instance, we might say that the null hypothesis is not implausible or we failed to reject the null hypothesis. Double negatives are used to communicate that while we are not rejecting a position, we are also not saying it is correct.

Example \(\PageIndex{1}\)

Next consider whether there is strong evidence that the average age of runners has changed from 2006 to 2012 in the Cherry Blossom Run. In 2006, the average age was 36.13 years, and in the 2012 run10Samp data set, the average was 35.05 years with a standard deviation of 8.97 years for 100 runners.

First, set up the hypotheses:

• H 0 : The average age of runners has not changed from 2006 to 2012, \(\mu_{age} = 36.13.\)
• H A : The average age of runners has changed from 2006 to 2012, \(\mu _{age} 6 \ne 36.13.\)

We have previously veri ed conditions for this data set. The normal model may be applied to \(\bar {y}\) and the estimate of SE should be very accurate. Using the sample mean and standard error, we can construct a 95% con dence interval for \(\mu _{age}\) to determine if there is sufficient evidence to reject H 0 :

\[\bar{y} \pm 1.96 \times \dfrac {s}{\sqrt {100}} \rightarrow 35.05 \pm 1.96 \times 0.90 \rightarrow (33.29, 36.81)\]

This confidence interval contains the null value, 36.13. Because 36.13 is not implausible, we cannot reject the null hypothesis. We have not found strong evidence that the average age is different than 36.13 years.

Exercise \(\PageIndex{2}\)

Colleges frequently provide estimates of student expenses such as housing. A consultant hired by a community college claimed that the average student housing expense was $650 per month. What are the null and alternative hypotheses to test whether this claim is accurate? 18

H A : The average cost is different than $650 per month, \(\mu \ne\) $650.

18 Applying the normal model requires that certain conditions are met. Because the data are a simple random sample and the sample (presumably) represents no more than 10% of all students at the college, the observations are independent. The sample size is also sufficiently large (n = 75) and the data exhibit only moderate skew. Thus, the normal model may be applied to the sample mean.

Exercise \(\PageIndex{3}\)

The community college decides to collect data to evaluate the $650 per month claim. They take a random sample of 75 students at their school and obtain the data represented in Figure 4.11. Can we apply the normal model to the sample mean?

If the court makes a Type 1 Error, this means the defendant is innocent (H 0 true) but wrongly convicted. A Type 2 Error means the court failed to reject H 0 (i.e. failed to convict the person) when she was in fact guilty (H A true).

Example \(\PageIndex{2}\)

The sample mean for student housing is $611.63 and the sample standard deviation is $132.85. Construct a 95% confidence interval for the population mean and evaluate the hypotheses of Exercise 4.22.

The standard error associated with the mean may be estimated using the sample standard deviation divided by the square root of the sample size. Recall that n = 75 students were sampled.

\[ SE = \dfrac {s}{\sqrt {n}} = \dfrac {132.85}{\sqrt {75}} = 15.34\]

You showed in Exercise 4.23 that the normal model may be applied to the sample mean. This ensures a 95% confidence interval may be accurately constructed:

\[\bar {x} \pm z*SE \rightarrow 611.63 \pm 1.96 \times 15.34 \times (581.56, 641.70)\]

Because the null value $650 is not in the confidence interval, a true mean of $650 is implausible and we reject the null hypothesis. The data provide statistically significant evidence that the actual average housing expense is less than $650 per month.

Decision Errors

Hypothesis tests are not flawless. Just think of the court system: innocent people are sometimes wrongly convicted and the guilty sometimes walk free. Similarly, we can make a wrong decision in statistical hypothesis tests. However, the difference is that we have the tools necessary to quantify how often we make such errors.

There are two competing hypotheses: the null and the alternative. In a hypothesis test, we make a statement about which one might be true, but we might choose incorrectly. There are four possible scenarios in a hypothesis test, which are summarized in Table 4.12.

A Type 1 Error is rejecting the null hypothesis when H0 is actually true. A Type 2 Error is failing to reject the null hypothesis when the alternative is actually true.

Exercise 4.25

In a US court, the defendant is either innocent (H 0 ) or guilty (H A ). What does a Type 1 Error represent in this context? What does a Type 2 Error represent? Table 4.12 may be useful.

To lower the Type 1 Error rate, we might raise our standard for conviction from "beyond a reasonable doubt" to "beyond a conceivable doubt" so fewer people would be wrongly convicted. However, this would also make it more difficult to convict the people who are actually guilty, so we would make more Type 2 Errors.

Exercise 4.26

How could we reduce the Type 1 Error rate in US courts? What influence would this have on the Type 2 Error rate?

To lower the Type 2 Error rate, we want to convict more guilty people. We could lower the standards for conviction from "beyond a reasonable doubt" to "beyond a little doubt". Lowering the bar for guilt will also result in more wrongful convictions, raising the Type 1 Error rate.

Exercise 4.27

How could we reduce the Type 2 Error rate in US courts? What influence would this have on the Type 1 Error rate?

A skeptic would have no reason to believe that sleep patterns at this school are different than the sleep patterns at another school.

Exercises 4.25-4.27 provide an important lesson:

If we reduce how often we make one type of error, we generally make more of the other type.

Hypothesis testing is built around rejecting or failing to reject the null hypothesis. That is, we do not reject H 0 unless we have strong evidence. But what precisely does strong evidence mean? As a general rule of thumb, for those cases where the null hypothesis is actually true, we do not want to incorrectly reject H 0 more than 5% of the time. This corresponds to a significance level of 0.05. We often write the significance level using \(\alpha\) (the Greek letter alpha): \(\alpha = 0.05.\) We discuss the appropriateness of different significance levels in Section 4.3.6.

If we use a 95% confidence interval to test a hypothesis where the null hypothesis is true, we will make an error whenever the point estimate is at least 1.96 standard errors away from the population parameter. This happens about 5% of the time (2.5% in each tail). Similarly, using a 99% con dence interval to evaluate a hypothesis is equivalent to a significance level of \(\alpha = 0.01\).

A confidence interval is, in one sense, simplistic in the world of hypothesis tests. Consider the following two scenarios:

• The null value (the parameter value under the null hypothesis) is in the 95% confidence interval but just barely, so we would not reject H 0 . However, we might like to somehow say, quantitatively, that it was a close decision.
• The null value is very far outside of the interval, so we reject H 0 . However, we want to communicate that, not only did we reject the null hypothesis, but it wasn't even close. Such a case is depicted in Figure 4.13.

In Section 4.3.4, we introduce a tool called the p-value that will be helpful in these cases. The p-value method also extends to hypothesis tests where con dence intervals cannot be easily constructed or applied.

Formal Testing using p-Values

The p-value is a way of quantifying the strength of the evidence against the null hypothesis and in favor of the alternative. Formally the p-value is a conditional probability.

definition: p-value

The p-value is the probability of observing data at least as favorable to the alternative hypothesis as our current data set, if the null hypothesis is true. We typically use a summary statistic of the data, in this chapter the sample mean, to help compute the p-value and evaluate the hypotheses.

A poll by the National Sleep Foundation found that college students average about 7 hours of sleep per night. Researchers at a rural school are interested in showing that students at their school sleep longer than seven hours on average, and they would like to demonstrate this using a sample of students. What would be an appropriate skeptical position for this research?

This is entirely based on the interests of the researchers. Had they been only interested in the opposite case - showing that their students were actually averaging fewer than seven hours of sleep but not interested in showing more than 7 hours - then our setup would have set the alternative as \(\mu < 7\).

We can set up the null hypothesis for this test as a skeptical perspective: the students at this school average 7 hours of sleep per night. The alternative hypothesis takes a new form reflecting the interests of the research: the students average more than 7 hours of sleep. We can write these hypotheses as

• H 0 : \(\mu\) = 7.
• H A : \(\mu\) > 7.

Using \(\mu\) > 7 as the alternative is an example of a one-sided hypothesis test. In this investigation, there is no apparent interest in learning whether the mean is less than 7 hours. (The standard error can be estimated from the sample standard deviation and the sample size: \(SE_{\bar {x}} = \dfrac {s_x}{\sqrt {n}} = \dfrac {1.75}{\sqrt {110}} = 0.17\)). Earlier we encountered a two-sided hypothesis where we looked for any clear difference, greater than or less than the null value.

Always use a two-sided test unless it was made clear prior to data collection that the test should be one-sided. Switching a two-sided test to a one-sided test after observing the data is dangerous because it can inflate the Type 1 Error rate.

TIP: One-sided and two-sided tests

If the researchers are only interested in showing an increase or a decrease, but not both, use a one-sided test. If the researchers would be interested in any difference from the null value - an increase or decrease - then the test should be two-sided.

TIP: Always write the null hypothesis as an equality

We will find it most useful if we always list the null hypothesis as an equality (e.g. \(\mu\) = 7) while the alternative always uses an inequality (e.g. \(\mu \ne 7, \mu > 7, or \mu < 7)\).

The researchers at the rural school conducted a simple random sample of n = 110 students on campus. They found that these students averaged 7.42 hours of sleep and the standard deviation of the amount of sleep for the students was 1.75 hours. A histogram of the sample is shown in Figure 4.14.

Before we can use a normal model for the sample mean or compute the standard error of the sample mean, we must verify conditions. (1) Because this is a simple random sample from less than 10% of the student body, the observations are independent. (2) The sample size in the sleep study is sufficiently large since it is greater than 30. (3) The data show moderate skew in Figure 4.14 and the presence of a couple of outliers. This skew and the outliers (which are not too extreme) are acceptable for a sample size of n = 110. With these conditions veri ed, the normal model can be safely applied to \(\bar {x}\) and the estimated standard error will be very accurate.

What is the standard deviation associated with \(\bar {x}\)? That is, estimate the standard error of \(\bar {x}\). 25

The hypothesis test will be evaluated using a significance level of \(\alpha = 0.05\). We want to consider the data under the scenario that the null hypothesis is true. In this case, the sample mean is from a distribution that is nearly normal and has mean 7 and standard deviation of about 0.17. Such a distribution is shown in Figure 4.15.

The shaded tail in Figure 4.15 represents the chance of observing such a large mean, conditional on the null hypothesis being true. That is, the shaded tail represents the p-value. We shade all means larger than our sample mean, \(\bar {x} = 7.42\), because they are more favorable to the alternative hypothesis than the observed mean.

We compute the p-value by finding the tail area of this normal distribution, which we learned to do in Section 3.1. First compute the Z score of the sample mean, \(\bar {x} = 7.42\):

\[Z = \dfrac {\bar {x} - \text {null value}}{SE_{\bar {x}}} = \dfrac {7.42 - 7}{0.17} = 2.47\]

Using the normal probability table, the lower unshaded area is found to be 0.993. Thus the shaded area is 1 - 0.993 = 0.007. If the null hypothesis is true, the probability of observing such a large sample mean for a sample of 110 students is only 0.007. That is, if the null hypothesis is true, we would not often see such a large mean.

We evaluate the hypotheses by comparing the p-value to the significance level. Because the p-value is less than the significance level \((p-value = 0.007 < 0.05 = \alpha)\), we reject the null hypothesis. What we observed is so unusual with respect to the null hypothesis that it casts serious doubt on H 0 and provides strong evidence favoring H A .

p-value as a tool in hypothesis testing

The p-value quantifies how strongly the data favor H A over H 0 . A small p-value (usually < 0.05) corresponds to sufficient evidence to reject H 0 in favor of H A .

TIP: It is useful to First draw a picture to find the p-value

It is useful to draw a picture of the distribution of \(\bar {x}\) as though H 0 was true (i.e. \(\mu\) equals the null value), and shade the region (or regions) of sample means that are at least as favorable to the alternative hypothesis. These shaded regions represent the p-value.

The ideas below review the process of evaluating hypothesis tests with p-values:

• The null hypothesis represents a skeptic's position or a position of no difference. We reject this position only if the evidence strongly favors H A .
• A small p-value means that if the null hypothesis is true, there is a low probability of seeing a point estimate at least as extreme as the one we saw. We interpret this as strong evidence in favor of the alternative.
• We reject the null hypothesis if the p-value is smaller than the significance level, \(\alpha\), which is usually 0.05. Otherwise, we fail to reject H 0 .
• We should always state the conclusion of the hypothesis test in plain language so non-statisticians can also understand the results.

The p-value is constructed in such a way that we can directly compare it to the significance level ( \(\alpha\)) to determine whether or not to reject H 0 . This method ensures that the Type 1 Error rate does not exceed the significance level standard.

If the null hypothesis is true, how often should the p-value be less than 0.05?

About 5% of the time. If the null hypothesis is true, then the data only has a 5% chance of being in the 5% of data most favorable to H A .

Exercise 4.31

Suppose we had used a significance level of 0.01 in the sleep study. Would the evidence have been strong enough to reject the null hypothesis? (The p-value was 0.007.) What if the significance level was \(\alpha = 0.001\)? 27

27 We reject the null hypothesis whenever p-value < \(\alpha\). Thus, we would still reject the null hypothesis if \(\alpha = 0.01\) but not if the significance level had been \(\alpha = 0.001\).

Exercise 4.32

Ebay might be interested in showing that buyers on its site tend to pay less than they would for the corresponding new item on Amazon. We'll research this topic for one particular product: a video game called Mario Kart for the Nintendo Wii. During early October 2009, Amazon sold this game for $46.99. Set up an appropriate (one-sided!) hypothesis test to check the claim that Ebay buyers pay less during auctions at this same time. 28

28 The skeptic would say the average is the same on Ebay, and we are interested in showing the average price is lower.

Exercise 4.33

During early October, 2009, 52 Ebay auctions were recorded for Mario Kart.29 The total prices for the auctions are presented using a histogram in Figure 4.17, and we may like to apply the normal model to the sample mean. Check the three conditions required for applying the normal model: (1) independence, (2) at least 30 observations, and (3) the data are not strongly skewed. 30

30 (1) The independence condition is unclear. We will make the assumption that the observations are independent, which we should report with any nal results. (2) The sample size is sufficiently large: \(n = 52 \ge 30\). (3) The data distribution is not strongly skewed; it is approximately symmetric.

H 0 : The average auction price on Ebay is equal to (or more than) the price on Amazon. We write only the equality in the statistical notation: \(\mu_{ebay} = 46.99\).

H A : The average price on Ebay is less than the price on Amazon, \(\mu _{ebay} < 46.99\).

29 These data were collected by OpenIntro staff.

Example 4.34

The average sale price of the 52 Ebay auctions for Wii Mario Kart was $44.17 with a standard deviation of $4.15. Does this provide sufficient evidence to reject the null hypothesis in Exercise 4.32? Use a significance level of \(\alpha = 0.01\).

The hypotheses were set up and the conditions were checked in Exercises 4.32 and 4.33. The next step is to find the standard error of the sample mean and produce a sketch to help find the p-value.

Because the alternative hypothesis says we are looking for a smaller mean, we shade the lower tail. We find this shaded area by using the Z score and normal probability table: \(Z = \dfrac {44.17 \times 46.99}{0.5755} = -4.90\), which has area less than 0.0002. The area is so small we cannot really see it on the picture. This lower tail area corresponds to the p-value.

Because the p-value is so small - specifically, smaller than = 0.01 - this provides sufficiently strong evidence to reject the null hypothesis in favor of the alternative. The data provide statistically signi cant evidence that the average price on Ebay is lower than Amazon's asking price.

Two-sided hypothesis testing with p-values

We now consider how to compute a p-value for a two-sided test. In one-sided tests, we shade the single tail in the direction of the alternative hypothesis. For example, when the alternative had the form \(\mu\) > 7, then the p-value was represented by the upper tail (Figure 4.16). When the alternative was \(\mu\) < 46.99, the p-value was the lower tail (Exercise 4.32). In a two-sided test, we shade two tails since evidence in either direction is favorable to H A .

Exercise 4.35 Earlier we talked about a research group investigating whether the students at their school slept longer than 7 hours each night. Let's consider a second group of researchers who want to evaluate whether the students at their college differ from the norm of 7 hours. Write the null and alternative hypotheses for this investigation. 31

Example 4.36 The second college randomly samples 72 students and nds a mean of \(\bar {x} = 6.83\) hours and a standard deviation of s = 1.8 hours. Does this provide strong evidence against H 0 in Exercise 4.35? Use a significance level of \(\alpha = 0.05\).

First, we must verify assumptions. (1) A simple random sample of less than 10% of the student body means the observations are independent. (2) The sample size is 72, which is greater than 30. (3) Based on the earlier distribution and what we already know about college student sleep habits, the distribution is probably not strongly skewed.

Next we can compute the standard error \((SE_{\bar {x}} = \dfrac {s}{\sqrt {n}} = 0.21)\) of the estimate and create a picture to represent the p-value, shown in Figure 4.18. Both tails are shaded.

31 Because the researchers are interested in any difference, they should use a two-sided setup: H 0 : \(\mu\) = 7, H A : \(\mu \ne 7.\)

An estimate of 7.17 or more provides at least as strong of evidence against the null hypothesis and in favor of the alternative as the observed estimate, \(\bar {x} = 6.83\).

We can calculate the tail areas by rst nding the lower tail corresponding to \(\bar {x}\):

\[Z = \dfrac {6.83 - 7.00}{0.21} = -0.81 \xrightarrow {table} \text {left tail} = 0.2090\]

Because the normal model is symmetric, the right tail will have the same area as the left tail. The p-value is found as the sum of the two shaded tails:

\[ \text {p-value} = \text {left tail} + \text {right tail} = 2 \times \text {(left tail)} = 0.4180\]

This p-value is relatively large (larger than \(\mu\)= 0.05), so we should not reject H 0 . That is, if H 0 is true, it would not be very unusual to see a sample mean this far from 7 hours simply due to sampling variation. Thus, we do not have sufficient evidence to conclude that the mean is different than 7 hours.

Example 4.37 It is never okay to change two-sided tests to one-sided tests after observing the data. In this example we explore the consequences of ignoring this advice. Using \(\alpha = 0.05\), we show that freely switching from two-sided tests to onesided tests will cause us to make twice as many Type 1 Errors as intended.

Suppose the sample mean was larger than the null value, \(\mu_0\) (e.g. \(\mu_0\) would represent 7 if H 0 : \(\mu\) = 7). Then if we can ip to a one-sided test, we would use H A : \(\mu > \mu_0\). Now if we obtain any observation with a Z score greater than 1.65, we would reject H 0 . If the null hypothesis is true, we incorrectly reject the null hypothesis about 5% of the time when the sample mean is above the null value, as shown in Figure 4.19.

Suppose the sample mean was smaller than the null value. Then if we change to a one-sided test, we would use H A : \(\mu < \mu_0\). If \(\bar {x}\) had a Z score smaller than -1.65, we would reject H 0 . If the null hypothesis is true, then we would observe such a case about 5% of the time.

By examining these two scenarios, we can determine that we will make a Type 1 Error 5% + 5% = 10% of the time if we are allowed to swap to the "best" one-sided test for the data. This is twice the error rate we prescribed with our significance level: \(\alpha = 0.05\) (!).

Caution: One-sided hypotheses are allowed only before seeing data

After observing data, it is tempting to turn a two-sided test into a one-sided test. Avoid this temptation. Hypotheses must be set up before observing the data. If they are not, the test must be two-sided.

Choosing a Significance Level

Choosing a significance level for a test is important in many contexts, and the traditional level is 0.05. However, it is often helpful to adjust the significance level based on the application. We may select a level that is smaller or larger than 0.05 depending on the consequences of any conclusions reached from the test.

• If making a Type 1 Error is dangerous or especially costly, we should choose a small significance level (e.g. 0.01). Under this scenario we want to be very cautious about rejecting the null hypothesis, so we demand very strong evidence favoring H A before we would reject H 0 .
• If a Type 2 Error is relatively more dangerous or much more costly than a Type 1 Error, then we should choose a higher significance level (e.g. 0.10). Here we want to be cautious about failing to reject H 0 when the null is actually false. We will discuss this particular case in greater detail in Section 4.6.

Significance levels should reflect consequences of errors

The significance level selected for a test should reflect the consequences associated with Type 1 and Type 2 Errors.

Example 4.38

A car manufacturer is considering a higher quality but more expensive supplier for window parts in its vehicles. They sample a number of parts from their current supplier and also parts from the new supplier. They decide that if the high quality parts will last more than 12% longer, it makes nancial sense to switch to this more expensive supplier. Is there good reason to modify the significance level in such a hypothesis test?

The null hypothesis is that the more expensive parts last no more than 12% longer while the alternative is that they do last more than 12% longer. This decision is just one of the many regular factors that have a marginal impact on the car and company. A significancelevel of 0.05 seems reasonable since neither a Type 1 or Type 2 error should be dangerous or (relatively) much more expensive.

Example 4.39

The same car manufacturer is considering a slightly more expensive supplier for parts related to safety, not windows. If the durability of these safety components is shown to be better than the current supplier, they will switch manufacturers. Is there good reason to modify the significance level in such an evaluation?

The null hypothesis would be that the suppliers' parts are equally reliable. Because safety is involved, the car company should be eager to switch to the slightly more expensive manufacturer (reject H 0 ) even if the evidence of increased safety is only moderately strong. A slightly larger significance level, such as \(\mu = 0.10\), might be appropriate.

Exercise 4.40

A part inside of a machine is very expensive to replace. However, the machine usually functions properly even if this part is broken, so the part is replaced only if we are extremely certain it is broken based on a series of measurements. Identify appropriate hypotheses for this test (in plain language) and suggest an appropriate significance level. 32

Module 9: Hypothesis Testing With One Sample

Null and alternative hypotheses, learning outcomes.

• Describe hypothesis testing in general and in practice

The actual test begins by considering two  hypotheses . They are called the null hypothesis and the alternative hypothesis . These hypotheses contain opposing viewpoints.

H 0 : The null hypothesis: It is a statement about the population that either is believed to be true or is used to put forth an argument unless it can be shown to be incorrect beyond a reasonable doubt.

H a : The alternative hypothesis : It is a claim about the population that is contradictory to H 0 and what we conclude when we reject H 0 .

Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data.

After you have determined which hypothesis the sample supports, you make adecision. There are two options for a  decision . They are “reject H 0 ” if the sample information favors the alternative hypothesis or “do not reject H 0 ” or “decline to reject H 0 ” if the sample information is insufficient to reject the null hypothesis.

Mathematical Symbols Used in  H 0 and H a :

H 0 always has a symbol with an equal in it. H a never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. However, be aware that many researchers (including one of the co-authors in research work) use = in the null hypothesis, even with > or < as the symbol in the alternative hypothesis. This practice is acceptable because we only make the decision to reject or not reject the null hypothesis.

H 0 : No more than 30% of the registered voters in Santa Clara County voted in the primary election. p ≤ 30

H a : More than 30% of the registered voters in Santa Clara County voted in the primary election. p > 30

A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25%. State the null and alternative hypotheses.

H 0 : The drug reduces cholesterol by 25%. p = 0.25

H a : The drug does not reduce cholesterol by 25%. p ≠ 0.25

We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are:

H 0 : μ = 2.0

H a : μ ≠ 2.0

We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. H 0 : μ __ 66 H a : μ __ 66

• H 0 : μ = 66
• H a : μ ≠ 66

We want to test if college students take less than five years to graduate from college, on the average. The null and alternative hypotheses are:

H 0 : μ ≥ 5

H a : μ < 5

We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. H 0 : μ __ 45 H a : μ __ 45

• H 0 : μ ≥ 45
• H a : μ < 45

In an issue of U.S. News and World Report , an article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third pass. The same article stated that 6.6% of U.S. students take advanced placement exams and 4.4% pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6%. State the null and alternative hypotheses.

H 0 : p ≤ 0.066

H a : p > 0.066

On a state driver’s test, about 40% pass the test on the first try. We want to test if more than 40% pass on the first try. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses. H 0 : p __ 0.40 H a : p __ 0.40

• H 0 : p = 0.40
• H a : p > 0.40

Concept Review

In a  hypothesis test , sample data is evaluated in order to arrive at a decision about some type of claim. If certain conditions about the sample are satisfied, then the claim can be evaluated for a population. In a hypothesis test, we: Evaluate the null hypothesis , typically denoted with H 0 . The null is not rejected unless the hypothesis test shows otherwise. The null statement must always contain some form of equality (=, ≤ or ≥) Always write the alternative hypothesis , typically denoted with H a or H 1 , using less than, greater than, or not equals symbols, i.e., (≠, >, or <). If we reject the null hypothesis, then we can assume there is enough evidence to support the alternative hypothesis. Never state that a claim is proven true or false. Keep in mind the underlying fact that hypothesis testing is based on probability laws; therefore, we can talk only in terms of non-absolute certainties.

Formula Review

H 0 and H a are contradictory.

• OpenStax, Statistics, Null and Alternative Hypotheses. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:58/Introductory_Statistics . License : CC BY: Attribution
• Introductory Statistics . Authored by : Barbara Illowski, Susan Dean. Provided by : Open Stax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]
• Simple hypothesis testing | Probability and Statistics | Khan Academy. Authored by : Khan Academy. Located at : https://youtu.be/5D1gV37bKXY . License : All Rights Reserved . License Terms : Standard YouTube License

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S.3.1 hypothesis testing (critical value approach).

The critical value approach involves determining "likely" or "unlikely" by determining whether or not the observed test statistic is more extreme than would be expected if the null hypothesis were true. That is, it entails comparing the observed test statistic to some cutoff value, called the " critical value ." If the test statistic is more extreme than the critical value, then the null hypothesis is rejected in favor of the alternative hypothesis. If the test statistic is not as extreme as the critical value, then the null hypothesis is not rejected.

Specifically, the four steps involved in using the critical value approach to conducting any hypothesis test are:

• Specify the null and alternative hypotheses.
• Using the sample data and assuming the null hypothesis is true, calculate the value of the test statistic. To conduct the hypothesis test for the population mean μ , we use the t -statistic \(t^*=\frac{\bar{x}-\mu}{s/\sqrt{n}}\) which follows a t -distribution with n - 1 degrees of freedom.
• Determine the critical value by finding the value of the known distribution of the test statistic such that the probability of making a Type I error — which is denoted \(\alpha\) (greek letter "alpha") and is called the " significance level of the test " — is small (typically 0.01, 0.05, or 0.10).
• Compare the test statistic to the critical value. If the test statistic is more extreme in the direction of the alternative than the critical value, reject the null hypothesis in favor of the alternative hypothesis. If the test statistic is less extreme than the critical value, do not reject the null hypothesis.

Example S.3.1.1

Mean gpa section  .

In our example concerning the mean grade point average, suppose we take a random sample of n = 15 students majoring in mathematics. Since n = 15, our test statistic t * has n - 1 = 14 degrees of freedom. Also, suppose we set our significance level α at 0.05 so that we have only a 5% chance of making a Type I error.

Right-Tailed

The critical value for conducting the right-tailed test H 0 : μ = 3 versus H A : μ > 3 is the t -value, denoted t \(\alpha\) , n - 1 , such that the probability to the right of it is \(\alpha\). It can be shown using either statistical software or a t -table that the critical value t 0.05,14 is 1.7613. That is, we would reject the null hypothesis H 0 : μ = 3 in favor of the alternative hypothesis H A : μ > 3 if the test statistic t * is greater than 1.7613. Visually, the rejection region is shaded red in the graph.

Left-Tailed

The critical value for conducting the left-tailed test H 0 : μ = 3 versus H A : μ < 3 is the t -value, denoted -t ( \(\alpha\) , n - 1) , such that the probability to the left of it is \(\alpha\). It can be shown using either statistical software or a t -table that the critical value -t 0.05,14 is -1.7613. That is, we would reject the null hypothesis H 0 : μ = 3 in favor of the alternative hypothesis H A : μ < 3 if the test statistic t * is less than -1.7613. Visually, the rejection region is shaded red in the graph.

There are two critical values for the two-tailed test H 0 : μ = 3 versus H A : μ ≠ 3 — one for the left-tail denoted -t ( \(\alpha\) / 2, n - 1) and one for the right-tail denoted t ( \(\alpha\) / 2, n - 1) . The value - t ( \(\alpha\) /2, n - 1) is the t -value such that the probability to the left of it is \(\alpha\)/2, and the value t ( \(\alpha\) /2, n - 1) is the t -value such that the probability to the right of it is \(\alpha\)/2. It can be shown using either statistical software or a t -table that the critical value -t 0.025,14 is -2.1448 and the critical value t 0.025,14 is 2.1448. That is, we would reject the null hypothesis H 0 : μ = 3 in favor of the alternative hypothesis H A : μ ≠ 3 if the test statistic t * is less than -2.1448 or greater than 2.1448. Visually, the rejection region is shaded red in the graph.

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AP®︎/College Statistics

Course: ap®︎/college statistics   >   unit 10.

• Idea behind hypothesis testing
• Examples of null and alternative hypotheses
• Writing null and alternative hypotheses
• P-values and significance tests
• Comparing P-values to different significance levels
• Estimating a P-value from a simulation
• Estimating P-values from simulations

Using P-values to make conclusions

• (Choice A)   Fail to reject H 0 ‍   A Fail to reject H 0 ‍  
• (Choice B)   Reject H 0 ‍   and accept H a ‍   B Reject H 0 ‍   and accept H a ‍  
• (Choice C)   Accept H 0 ‍   C Accept H 0 ‍  
• (Choice A)   The evidence suggests that these subjects can do better than guessing when identifying the bottled water. A The evidence suggests that these subjects can do better than guessing when identifying the bottled water.
• (Choice B)   We don't have enough evidence to say that these subjects can do better than guessing when identifying the bottled water. B We don't have enough evidence to say that these subjects can do better than guessing when identifying the bottled water.
• (Choice C)   The evidence suggests that these subjects were simply guessing when identifying the bottled water. C The evidence suggests that these subjects were simply guessing when identifying the bottled water.
• (Choice A)   She would have rejected H a ‍   . A She would have rejected H a ‍   .
• (Choice B)   She would have accepted H 0 ‍   . B She would have accepted H 0 ‍   .
• (Choice C)   She would have rejected H 0 ‍   and accepted H a ‍   . C She would have rejected H 0 ‍   and accepted H a ‍   .
• (Choice D)   She would have reached the same conclusion using either α = 0.05 ‍   or α = 0.10 ‍   . D She would have reached the same conclusion using either α = 0.05 ‍   or α = 0.10 ‍   .
• (Choice A)   The evidence suggests that these bags are being filled with a mean amount that is different than 7.4  kg ‍   . A The evidence suggests that these bags are being filled with a mean amount that is different than 7.4  kg ‍   .
• (Choice B)   We don't have enough evidence to say that these bags are being filled with a mean amount that is different than 7.4  kg ‍   . B We don't have enough evidence to say that these bags are being filled with a mean amount that is different than 7.4  kg ‍   .
• (Choice C)   The evidence suggests that these bags are being filled with a mean amount of 7.4  kg ‍   . C The evidence suggests that these bags are being filled with a mean amount of 7.4  kg ‍   .
• (Choice A)   They would have rejected H a ‍   . A They would have rejected H a ‍   .
• (Choice B)   They would have accepted H 0 ‍   . B They would have accepted H 0 ‍   .
• (Choice C)   They would have failed to reject H 0 ‍   . C They would have failed to reject H 0 ‍   .
• (Choice D)   They would have reached the same conclusion using either α = 0.05 ‍   or α = 0.01 ‍   . D They would have reached the same conclusion using either α = 0.05 ‍   or α = 0.01 ‍   .

Ethics and the significance level α ‍  

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8.6: Hypothesis Test of a Single Population Mean with Examples

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Steps for performing Hypothesis Test of a Single Population Mean

Step 1: State your hypotheses about the population mean. Step 2: Summarize the data. State a significance level. State and check conditions required for the procedure

• Find or identify the sample size, n, the sample mean, \(\bar{x}\) and the sample standard deviation, s .

The sampling distribution for the one-mean test statistic is, approximately, T- distribution if the following conditions are met

• Sample is random with independent observations .
• Sample is large. The population must be Normal or the sample size must be at least 30.

Step 3: Perform the procedure based on the assumption that \(H_{0}\) is true

• Find the Estimated Standard Error: \(SE=\frac{s}{\sqrt{n}}\).
• Compute the observed value of the test statistic: \(T_{obs}=\frac{\bar{x}-\mu_{0}}{SE}\).
• Check the type of the test (right-, left-, or two-tailed)
• Find the p-value in order to measure your level of surprise.

Step 4: Make a decision about \(H_{0}\) and \(H_{a}\)

• Do you reject or not reject your null hypothesis?

Step 5: Make a conclusion

• What does this mean in the context of the data?

The following examples illustrate a left-, right-, and two-tailed test.

Example \(\pageindex{1}\).

\(H_{0}: \mu = 5, H_{a}: \mu < 5\)

Test of a single population mean. \(H_{a}\) tells you the test is left-tailed. The picture of the \(p\)-value is as follows:

Exercise \(\PageIndex{1}\)

\(H_{0}: \mu = 10, H_{a}: \mu < 10\)

Assume the \(p\)-value is 0.0935. What type of test is this? Draw the picture of the \(p\)-value.

left-tailed test

Example \(\PageIndex{2}\)

\(H_{0}: \mu \leq 0.2, H_{a}: \mu > 0.2\)

This is a test of a single population proportion. \(H_{a}\) tells you the test is right-tailed . The picture of the p -value is as follows:

Exercise \(\PageIndex{2}\)

\(H_{0}: \mu \leq 1, H_{a}: \mu > 1\)

Assume the \(p\)-value is 0.1243. What type of test is this? Draw the picture of the \(p\)-value.

right-tailed test

Example \(\PageIndex{3}\)

\(H_{0}: \mu = 50, H_{a}: \mu \neq 50\)

This is a test of a single population mean. \(H_{a}\) tells you the test is two-tailed . The picture of the \(p\)-value is as follows.

Exercise \(\PageIndex{3}\)

\(H_{0}: \mu = 0.5, H_{a}: \mu \neq 0.5\)

Assume the p -value is 0.2564. What type of test is this? Draw the picture of the \(p\)-value.

two-tailed test

Full Hypothesis Test Examples

Example \(\pageindex{4}\).

Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores 65 65 70 67 66 63 63 68 72 71. He performs a hypothesis test using a 5% level of significance. The data are assumed to be from a normal distribution.

Set up the hypothesis test:

A 5% level of significance means that \(\alpha = 0.05\). This is a test of a single population mean .

\(H_{0}: \mu = 65  H_{a}: \mu > 65\)

Since the instructor thinks the average score is higher, use a "\(>\)". The "\(>\)" means the test is right-tailed.

Determine the distribution needed:

Random variable: \(\bar{X} =\) average score on the first statistics test.

Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given . You are only given \(n = 10\) sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a student's \(t\).

Use \(t_{df}\). Therefore, the distribution for the test is \(t_{9}\) where \(n = 10\) and \(df = 10 - 1 = 9\).

The sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data.

Calculate the \(p\)-value using the Student's \(t\)-distribution:

\[t_{obs} = \dfrac{\bar{x}-\mu_{\bar{x}}}{\left(\dfrac{s}{\sqrt{n}}\right)}=\dfrac{67-65}{\left(\dfrac{3.1972}{\sqrt{10}}\right)}\]

Use the T-table or Excel's t_dist() function to find p-value:

\(p\text{-value} = P(\bar{x} > 67) =P(T >1.9782 )= 1-0.9604=0.0396\)

Interpretation of the p -value: If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 65 or more.

Compare \(\alpha\) and the \(p-\text{value}\):

Since \(α = 0.05\) and \(p\text{-value} = 0.0396\). \(\alpha > p\text{-value}\).

Make a decision: Since \(\alpha > p\text{-value}\), reject \(H_{0}\).

This means you reject \(\mu = 65\). In other words, you believe the average test score is more than 65.

Conclusion: At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.

The \(p\text{-value}\) can easily be calculated.

Put the data into a list. Press STAT and arrow over to TESTS . Press 2:T-Test . Arrow over to Data and press ENTER . Arrow down and enter 65 for \(\mu_{0}\), the name of the list where you put the data, and 1 for Freq: . Arrow down to \(\mu\): and arrow over to \(> \mu_{0}\). Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the \(p\text{-value}\) (p = 0.0396) but it also calculates the test statistic ( t -score) for the sample mean, the sample mean, and the sample standard deviation. \(\mu > 65\) is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with \(t = 1.9781\) (test statistic) and \(p = 0.0396\) (\(p\text{-value}\)). Make sure when you use Draw that no other equations are highlighted in \(Y =\) and the plots are turned off.

Exercise \(\PageIndex{4}\)

It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, find the p -value, state your conclusion, and identify the Type I and Type II errors.

• \(H_{0}: \mu = 5\)
• \(H_{a}: \mu < 5\)
• \(p = 0.0082\)

Because \(p < \alpha\), we reject the null hypothesis. There is sufficient evidence to suggest that the stock price of the company grows at a rate less than $5 a week.

• Type I Error: To conclude that the stock price is growing slower than $5 a week when, in fact, the stock price is growing at $5 a week (reject the null hypothesis when the null hypothesis is true).
• Type II Error: To conclude that the stock price is growing at a rate of $5 a week when, in fact, the stock price is growing slower than $5 a week (do not reject the null hypothesis when the null hypothesis is false).

Example \(\PageIndex{5}\)

The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.

1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98 1.02; .95; .95

Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05. Assume the population is normal.

Let’s follow a four-step process to answer this statistical question.

• \(H_{0}: \mu \leq 1\)
• \(H_{a}: \mu > 1\)
• Plan : We are testing a sample mean without a known population standard deviation. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal.
• Do the calculations : \(p\text{-value} ( = 0.036)\)

4. State the Conclusions : Since the \(p\text{-value} (= 0.036)\) is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.

The hypothesis test itself has an established process. This can be summarized as follows:

• Determine \(H_{0}\) and \(H_{a}\). Remember, they are contradictory.
• Determine the random variable.
• Determine the distribution for the test.
• Draw a graph, calculate the test statistic, and use the test statistic to calculate the \(p\text{-value}\). (A t -score is an example of test statistics.)
• Compare the preconceived α with the p -value, make a decision (reject or do not reject H 0 ), and write a clear conclusion using English sentences.

Notice that in performing the hypothesis test, you use \(\alpha\) and not \(\beta\). \(\beta\) is needed to help determine the sample size of the data that is used in calculating the \(p\text{-value}\). Remember that the quantity \(1 – \beta\) is called the Power of the Test . A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same.If the power is low, the null hypothesis might not be rejected when it should be.

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• Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed June 27, 2013).
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