S.3 Hypothesis Testing
S.3 Hypothesis TestingIn reviewing hypothesis tests, we start first with the general idea. Then, we keep returning to the basic procedures of hypothesis testing, each time adding a little more detail.
The general idea of hypothesis testing involves:
 Making an initial assumption.
 Collecting evidence (data).
 Based on the available evidence (data), deciding whether to reject or not reject the initial assumption.
Every hypothesis test — regardless of the population parameter involved — requires the above three steps.
Example S.3.1
Is Normal Body Temperature Really 98.6 Degrees F?
Consider the population of many, many adults. A researcher hypothesized that the average adult body temperature is lower than the oftenadvertised 98.6 degrees F. That is, the researcher wants an answer to the question: "Is the average adult body temperature 98.6 degrees? Or is it lower?" To answer his research question, the researcher starts by assuming that the average adult body temperature was 98.6 degrees F.
Then, the researcher went out and tried to find evidence that refutes his initial assumption. In doing so, he selects a random sample of 130 adults. The average body temperature of the 130 sampled adults is 98.25 degrees.
Then, the researcher uses the data he collected to make a decision about his initial assumption. It is either likely or unlikely that the researcher would collect the evidence he did given his initial assumption that the average adult body temperature is 98.6 degrees:
 If it is likely, then the researcher does not reject his initial assumption that the average adult body temperature is 98.6 degrees. There is not enough evidence to do otherwise.
 If it is unlikely, then:
 either the researcher's initial assumption is correct and he experienced a very unusual event;
 or the researcher's initial assumption is incorrect.
In statistics, we generally don't make claims that require us to believe that a very unusual event happened. That is, in the practice of statistics, if the evidence (data) we collected is unlikely in light of the initial assumption, then we reject our initial assumption.
Example S.3.2
Criminal Trial Analogy
One place where you can consistently see the general idea of hypothesis testing in action is in criminal trials held in the United States. Our criminal justice system assumes "the defendant is innocent until proven guilty." That is, our initial assumption is that the defendant is innocent.
In the practice of statistics, we make our initial assumption when we state our two competing hypotheses  the null hypothesis (H_{0}) and the alternative hypothesis (H_{A}). Here, our hypotheses are:
 H_{0}: Defendant is not guilty (innocent)
 H_{A}: Defendant is guilty
In statistics, we always assume the null hypothesis is true. That is, the null hypothesis is always our initial assumption.
The prosecution team then collects evidence — such as finger prints, blood spots, hair samples, carpet fibers, shoe prints, ransom notes, and handwriting samples — with the hopes of finding "sufficient evidence" to make the assumption of innocence refutable.
In statistics, the data are the evidence.
The jury then makes a decision based on the available evidence:
 If the jury finds sufficient evidence — beyond a reasonable doubt — to make the assumption of innocence refutable, the jury rejects the null hypothesis and deems the defendant guilty. We behave as if the defendant is guilty.
 If there is insufficient evidence, then the jury does not reject the null hypothesis. We behave as if the defendant is innocent.
In statistics, we always make one of two decisions. We either "reject the null hypothesis" or we "fail to reject the null hypothesis."
Errors in Hypothesis Testing
Did you notice the use of the phrase "behave as if" in the previous discussion? We "behave as if" the defendant is guilty; we do not "prove" that the defendant is guilty. And, we "behave as if" the defendant is innocent; we do not "prove" that the defendant is innocent.
This is a very important distinction! We make our decision based on evidence not on 100% guaranteed proof. Again:
 If we reject the null hypothesis, we do not prove that the alternative hypothesis is true.
 If we do not reject the null hypothesis, we do not prove that the null hypothesis is true.
We merely state that there is enough evidence to behave one way or the other. This is always true in statistics! Because of this, whatever the decision, there is always a chance that we made an error.
Let's review the two types of errors that can be made in criminal trials:
Truth  
Not Guilty  Guilty  
Jury Decision  Not Guilty  OK  ERROR 
Guilty  ERROR  OK 
Table S.3.2 shows how this corresponds to the two types of errors in hypothesis testing.
Truth  
Null Hypothesis  Alternative Hypothesis  
Decision  Do not Reject Null  OK  Type II Error 
Reject Null  Type I Error  OK 
Note that, in statistics, we call the two types of errors by two different names  one is called a "Type I error," and the other is called a "Type II error." Here are the formal definitions of the two types of errors:
 Type I Error
 The null hypothesis is rejected when it is true.
 Type II Error
 The null hypothesis is not rejected when it is false.
There is always a chance of making one of these errors. But, a good scientific study will minimize the chance of doing so!
Making the Decision
Recall that it is either likely or unlikely that we would observe the evidence we did given our initial assumption. If it is likely, we do not reject the null hypothesis. If it is unlikely, then we reject the null hypothesis in favor of the alternative hypothesis. Effectively, then, making the decision reduces to determining "likely" or "unlikely."
In statistics, there are two ways to determine whether the evidence is likely or unlikely given the initial assumption:
 We could take the "critical value approach" (favored in many of the older textbooks).
 Or, we could take the "Pvalue approach" (what is used most often in research, journal articles, and statistical software).
In the next two sections, we review the procedures behind each of these two approaches. To make our review concrete, let's imagine that μ is the average grade point average of all American students who major in mathematics. We first review the critical value approach for conducting each of the following three hypothesis tests about the population mean $\mu$:
Type

Null

Alternative

Righttailed

H_{0} : μ = 3

H_{A} : μ > 3

Lefttailed

H_{0} : μ = 3

H_{A} : μ < 3

Twotailed

H_{0} : μ = 3

H_{A} : μ ≠ 3

In Practice

We would want to conduct the first hypothesis test if we were interested in concluding that the average grade point average of the group is more than 3.

We would want to conduct the second hypothesis test if we were interested in concluding that the average grade point average of the group is less than 3.

And, we would want to conduct the third hypothesis test if we were only interested in concluding that the average grade point average of the group differs from 3 (without caring whether it is more or less than 3).
Upon completing the review of the critical value approach, we review the Pvalue approach for conducting each of the above three hypothesis tests about the population mean \(\mu\). The procedures that we review here for both approaches easily extend to hypothesis tests about any other population parameter.
S.3.1 Hypothesis Testing (Critical Value Approach)
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 tstatistic \(t^*=\frac{\bar{x}\mu}{s/\sqrt{n}}\) which follows a tdistribution 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
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.
RightTailed
The critical value for conducting the righttailed test H_{0} : μ = 3 versus H_{A} : μ > 3 is the tvalue, 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 ttable 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.
LeftTailed
The critical value for conducting the lefttailed test H_{0} : μ = 3 versus H_{A} : μ < 3 is the tvalue, 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 ttable 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.
TwoTailed
There are two critical values for the twotailed test H_{0} : μ = 3 versus H_{A} : μ ≠ 3 — one for the lefttail denoted t_{(\(\alpha\)/2, n  1) }and one for the righttail denoted t_{(\(\alpha\)/2, n  1)}. The value t_{(\(\alpha\)/2, n  1)} is the tvalue such that the probability to the left of it is \(\alpha\)/2, and the value t_{(\(\alpha\)/2, n  1)} is the tvalue such that the probability to the right of it is \(\alpha\)/2. It can be shown using either statistical software or a ttable 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.
S.3.2 Hypothesis Testing (PValue Approach)
S.3.2 Hypothesis Testing (PValue Approach)The Pvalue approach involves determining "likely" or "unlikely" by determining the probability — assuming the null hypothesis were true — of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed. If the Pvalue is small, say less than (or equal to) \(\alpha\), then it is "unlikely." And, if the Pvalue is large, say more than \(\alpha\), then it is "likely."
If the Pvalue is less than (or equal to) \(\alpha\), then the null hypothesis is rejected in favor of the alternative hypothesis. And, if the Pvalue is greater than \(\alpha\), then the null hypothesis is not rejected.
Specifically, the four steps involved in using the Pvalue 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. Again, to conduct the hypothesis test for the population mean μ, we use the tstatistic \(t^*=\frac{\bar{x}\mu}{s/\sqrt{n}}\) which follows a tdistribution with n  1 degrees of freedom.
 Using the known distribution of the test statistic, calculate the Pvalue: "If the null hypothesis is true, what is the probability that we'd observe a more extreme test statistic in the direction of the alternative hypothesis than we did?" (Note how this question is equivalent to the question answered in criminal trials: "If the defendant is innocent, what is the chance that we'd observe such extreme criminal evidence?")
 Set the significance level, \(\alpha\), the probability of making a Type I error to be small — 0.01, 0.05, or 0.10. Compare the Pvalue to \(\alpha\). If the Pvalue is less than (or equal to) \(\alpha\), reject the null hypothesis in favor of the alternative hypothesis. If the Pvalue is greater than \(\alpha\), do not reject the null hypothesis.
Example S.3.2.1
Mean GPA
In our example concerning the mean grade point average, suppose that our random sample of n = 15 students majoring in mathematics yields a test statistic t* equaling 2.5. 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 Pvalue for conducting the righttailed test H_{0} : μ = 3 versus H_{A} : μ > 3 is the probability that we would observe a test statistic greater than t* = 2.5 if the population mean \(\mu\) really were 3. Recall that probability equals the area under the probability curve. The Pvalue is therefore the area under a t_{n  1} = t_{14} curve and to the right of the test statistic t* = 2.5. It can be shown using statistical software that the Pvalue is 0.0127. The graph depicts this visually.
The Pvalue, 0.0127, tells us it is "unlikely" that we would observe such an extreme test statistic t* in the direction of H_{A} if the null hypothesis were true. Therefore, our initial assumption that the null hypothesis is true must be incorrect. That is, since the Pvalue, 0.0127, is less than \(\alpha\) = 0.05, we reject the null hypothesis H_{0} : μ = 3 in favor of the alternative hypothesis H_{A} : μ > 3.
Note that we would not reject H_{0} : μ = 3 in favor of H_{A} : μ > 3 if we lowered our willingness to make a Type I error to \(\alpha\) = 0.01 instead, as the Pvalue, 0.0127, is then greater than \(\alpha\) = 0.01.
Left Tailed
In our example concerning the mean grade point average, suppose that our random sample of n = 15 students majoring in mathematics yields a test statistic t* instead equaling 2.5. The Pvalue for conducting the lefttailed test H_{0} : μ = 3 versus H_{A} : μ < 3 is the probability that we would observe a test statistic less than t* = 2.5 if the population mean μ really were 3. The Pvalue is therefore the area under a t_{n  1} = t_{14} curve and to the left of the test statistic t* = 2.5. It can be shown using statistical software that the Pvalue is 0.0127. The graph depicts this visually.
The Pvalue, 0.0127, tells us it is "unlikely" that we would observe such an extreme test statistic t* in the direction of H_{A }if the null hypothesis were true. Therefore, our initial assumption that the null hypothesis is true must be incorrect. That is, since the Pvalue, 0.0127, is less than α = 0.05, we reject the null hypothesis H_{0} : μ = 3 in favor of the alternative hypothesis H_{A} : μ < 3.
Note that we would not reject H_{0} : μ = 3 in favor of H_{A} : μ < 3 if we lowered our willingness to make a Type I error to α = 0.01 instead, as the Pvalue, 0.0127, is then greater than \(\alpha\) = 0.01.
Two Tailed
In our example concerning the mean grade point average, suppose again that our random sample of n = 15 students majoring in mathematics yields a test statistic t* instead equaling 2.5. The Pvalue for conducting the twotailed test H_{0} : μ = 3 versus H_{A} : μ ≠ 3 is the probability that we would observe a test statistic less than 2.5 or greater than 2.5 if the population mean μ really were 3. That is, the twotailed test requires taking into account the possibility that the test statistic could fall into either tail (and hence the name "twotailed" test). The Pvalue is therefore the area under a t_{n  1} = t_{14} curve to the left of 2.5 and to the right of the 2.5. It can be shown using statistical software that the Pvalue is 0.0127 + 0.0127, or 0.0254. The graph depicts this visually.
Note that the Pvalue for a twotailed test is always two times the Pvalue for either of the onetailed tests. The Pvalue, 0.0254, tells us it is "unlikely" that we would observe such an extreme test statistic t* in the direction of H_{A }if the null hypothesis were true. Therefore, our initial assumption that the null hypothesis is true must be incorrect. That is, since the Pvalue, 0.0254, is less than α = 0.05, we reject the null hypothesis H_{0} : μ = 3 in favor of the alternative hypothesis H_{A} : μ ≠ 3.
Note that we would not reject H_{0 }: μ = 3 in favor of H_{A} : μ ≠ 3 if we lowered our willingness to make a Type I error to α = 0.01 instead, as the Pvalue, 0.0254, is then greater than \(\alpha\) = 0.01.
Now that we have reviewed the critical value and Pvalue approach procedures for each of three possible hypotheses, let's look at three new examples — one of a righttailed test, one of a lefttailed test, and one of a twotailed test.
The good news is that, whenever possible, we will take advantage of the test statistics and Pvalues reported in statistical software, such as Minitab, to conduct our hypothesis tests in this course.
S.3.3 Hypothesis Testing Examples
S.3.3 Hypothesis Testing ExamplesAn engineer measured the Brinell hardness of 25 pieces of ductile iron that were subcritically annealed. The resulting data were:
170  167  174  179  179  187  179  183  179 
156  163  156  187  156  167  156  174  170 
183  179  174  179  170  159  187 
The engineer hypothesized that the mean Brinell hardness of all such ductile iron pieces is greater than 170. Therefore, he was interested in testing the hypotheses:
H_{0} : μ = 170
H_{A}: μ > 170
The engineer entered his data into Minitab and requested that the "onesample ttest" be conducted for the above hypotheses. He obtained the following output:
Descriptive Statistics
N  Mean  StDev  SE Mean  95% Lower Bound 

25  172.52  10.31  2.06  168.99 
$\mu$: mean of Brinelli
Test
Null hypothesis H₀: $\mu$ = 170
Alternative hypothesis H₁: $\mu$ > 170
TValue  PValue 

1.22  0.117 
The output tells us that the average Brinell hardness of the n = 25 pieces of ductile iron was 172.52 with a standard deviation of 10.31. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 10.31 by the square root of n = 25, is 2.06). The test statistic t* is 1.22, and the Pvalue is 0.117.
If the engineer set his significance level α at 0.05 and used the critical value approach to conduct his hypothesis test, he would reject the null hypothesis if his test statistic t* were greater than 1.7109 (determined using statistical software or a ttable):
Since the engineer's test statistic, t* = 1.22, is not greater than 1.7109, the engineer fails to reject the null hypothesis. That is, the test statistic does not fall in the "critical region." There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean Brinell hardness of all such ductile iron pieces is greater than 170.
If the engineer used the Pvalue approach to conduct his hypothesis test, he would determine the area under a t_{n  1} = t_{24} curve and to the right of the test statistic t* = 1.22:
In the output above, Minitab reports that the Pvalue is 0.117. Since the Pvalue, 0.117, is greater than \(\alpha\) = 0.05, the engineer fails to reject the null hypothesis. There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean Brinell hardness of all such ductile iron pieces is greater than 170.
Note that the engineer obtains the same scientific conclusion regardless of the approach used. This will always be the case.
A biologist was interested in determining whether sunflower seedlings treated with an extract from Vinca minor roots resulted in a lower average height of sunflower seedlings than the standard height of 15.7 cm. The biologist treated a random sample of n = 33 seedlings with the extract and subsequently obtained the following heights:
11.5  11.8  15.7  16.1  14.1  10.5  9.3  15.0  11.1 
15.2  19.0  12.8  12.4  19.2  13.5  12.2  13.3  
16.5  13.5  14.4  16.7  10.9  13.0  10.3  15.8  
15.1  17.1  13.3  12.4  8.5  14.3  12.9  13.5 
The biologist's hypotheses are:
H_{0} : μ = 15.7
H_{A}: μ < 15.7
The biologist entered her data into Minitab and requested that the "onesample ttest" be conducted for the above hypotheses. She obtained the following output:
Descriptive Statistics
N  Mean  StDev  SE Mean  95% Upper Bound 

33  13.664  2.544  0.443  14.414 
$\mu$: mean of Height
Test
Null hypothesis H₀: $\mu$ = 15.7
Alternative hypothesis H₁: $\mu$ < 15.7
TValue  PValue 

4.60  0.000 
The output tells us that the average height of the n = 33 sunflower seedlings was 13.664 with a standard deviation of 2.544. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 13.664 by the square root of n = 33, is 0.443). The test statistic t* is 4.60, and the Pvalue, 0.000, is to three decimal places.
Minitab Note. Minitab will always report Pvalues to only 3 decimal places. If Minitab reports the Pvalue as 0.000, it really means that the Pvalue is 0.000....something. Throughout this course (and your future research!), when you see that Minitab reports the Pvalue as 0.000, you should report the Pvalue as being "< 0.001."
If the biologist set her significance level \(\alpha\) at 0.05 and used the critical value approach to conduct her hypothesis test, she would reject the null hypothesis if her test statistic t* were less than 1.6939 (determined using statistical software or a ttable):s33
Since the biologist's test statistic, t* = 4.60, is less than 1.6939, the biologist rejects the null hypothesis. That is, the test statistic falls in the "critical region." There is sufficient evidence, at the α = 0.05 level, to conclude that the mean height of all such sunflower seedlings is less than 15.7 cm.
If the biologist used the Pvalue approach to conduct her hypothesis test, she would determine the area under a t_{n  1} = t_{32} curve and to the left of the test statistic t* = 4.60:
In the output above, Minitab reports that the Pvalue is 0.000, which we take to mean < 0.001. Since the Pvalue is less than 0.001, it is clearly less than \(\alpha\) = 0.05, and the biologist rejects the null hypothesis. There is sufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean height of all such sunflower seedlings is less than 15.7 cm.
Note again that the biologist obtains the same scientific conclusion regardless of the approach used. This will always be the case.
A manufacturer claims that the thickness of the spearmint gum it produces is 7.5 onehundredths of an inch. A quality control specialist regularly checks this claim. On one production run, he took a random sample of n = 10 pieces of gum and measured their thickness. He obtained:
7.65  7.60  7.65  7.70  7.55 
7.55  7.40  7.40  7.50  7.50 
The quality control specialist's hypotheses are:
H_{0} : μ = 7.5
H_{A}: μ ≠ 7.5
The quality control specialist entered his data into Minitab and requested that the "onesample ttest" be conducted for the above hypotheses. He obtained the following output:
Descriptive Statistics
N  Mean  StDev  SE Mean  95% CI for $\mu$ 

10  7.550  0.1027  0.0325  (7.4765, 7.6235) 
$\mu$: mean of Thickness
Test
Null hypothesis H₀: $\mu$ = 7.5
Alternative hypothesis H₁: $\mu \ne$ 7.5
TValue  PValue 

1.54  0.158 
The output tells us that the average thickness of the n = 10 pieces of gums was 7.55 onehundredths of an inch with a standard deviation of 0.1027. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 0.1027 by the square root of n = 10, is 0.0325). The test statistic t* is 1.54, and the Pvalue is 0.158.
If the quality control specialist sets his significance level \(\alpha\) at 0.05 and used the critical value approach to conduct his hypothesis test, he would reject the null hypothesis if his test statistic t* were less than 2.2616 or greater than 2.2616 (determined using statistical software or a ttable):
Since the quality control specialist's test statistic, t* = 1.54, is not less than 2.2616 nor greater than 2.2616, the quality control specialist fails to reject the null hypothesis. That is, the test statistic does not fall in the "critical region." There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean thickness of all of the manufacturer's spearmint gum differs from 7.5 onehundredths of an inch.
If the quality control specialist used the Pvalue approach to conduct his hypothesis test, he would determine the area under a t_{n  1} = t_{9} curve, to the right of 1.54 and to the left of 1.54:
In the output above, Minitab reports that the Pvalue is 0.158. Since the Pvalue, 0.158, is greater than \(\alpha\) = 0.05, the quality control specialist fails to reject the null hypothesis. There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean thickness of all pieces of spearmint gum differs from 7.5 onehundredths of an inch.
Note that the quality control specialist obtains the same scientific conclusion regardless of the approach used. This will always be the case.
In closing
In our review of hypothesis tests, we have focused on just one particular hypothesis test, namely that concerning the population mean \(\mu\). The important thing to recognize is that the topics discussed here — the general idea of hypothesis tests, errors in hypothesis testing, the critical value approach, and the Pvalue approach — generally extend to all of the hypothesis tests you will encounter.