45 9.5 Additional Information and Full Hypothesis Test Examples
 In a hypothesis test problem, you may see words such as “the level of significance is 1%.” The “1%” is the preconceived or preset α.
 The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data.
 If no level of significance is given, a common standard to use is α = 0.05.
 When you calculate the pvalue and draw the picture, the pvalue is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.
 The alternative hypothesis, Ha, tells you if the test is left, right, or twotailed. It is the key to conducting the appropriate test.
 H_{a} never has a symbol that contains an equal sign.
 Thinking about the meaning of the pvalue: A data analyst (and anyone else) should have more confidence that he made the correct decision to reject the null hypothesis with a smaller pvalue (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large pvalue such as 0.4, as opposed to a pvalue of 0.056 (alpha = 0.05 is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules.
The following examples illustrate a left, right, and twotailed test.
Example 1
H_{o}: μ = 5
H_{a}: μ < 5
Significance level = 5%
Assume the pvalue is 0.0243.
 What type of test is this?
 Determine if the test is left, right, or twotailed.
 Draw the picture of the pvalue.
 Do we reject null hypothesis, H_{0}: μ = 5?
 Do we have enough evidence to conclude that μ < 5?
Click here to show solution:
 Test of a single population mean.
 H_{a} tells you the test is lefttailed.
 The picture of the pvalue is as follows:
 Since pvalue < significance level, we reject null hypothesis, H_{o}: μ = 5.
 We have enough evidence to conclude that H_{a}: μ < 5.
Try It
H_{0}: μ = 10
H_{a}: μ < 10
Significance level = 5% = 0.05
Assume the pvalue is 0.0435.
 What type of test is this?
 Determine if the test is left, right, or twotailed.
 Draw the picture of the pvalue.
 Do we reject null hypothesis, H_{0}: μ = 10?
 Do we have enough evidence to conclude that μ < 10?
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Click here to show solution:
 Test of a single population mean.
 lefttailed test
 The picture of the pvalue is as follows:
 Since pvalue < significance level, we reject null hypothesis, H_{0}: μ = 10.
 We have enough evidence to conclude H_{a}: μ < 10.
Example 2
H_{a}: p > 0.2
Significance level = 0.05
Assume the pvalue is 0.0719.
 What type of test is this?
 Determine if the test is left, right, or twotailed.
 Draw the picture of the pvalue.
 Do we reject null hypothesis, H_{0}: p ≤ 0.2?
 Do we have enough evidence to conclude that p > 0.2?
Click here to show solution:
 This is a test of a single population proportion.
 H_{a} tells you the test is righttailed.
 The picture of the pvalue is as follows:
 Since pvalue > significance level, we do not reject null hypothesis (H_{0}: p ≤ 0.2).
 We do not have enough evidence to conclude H_{a}: p > 0.2.
Try It
H_{0}: μ ≤ 1
H_{a}: μ > 1
Significance level = 1%
Assume the pvalue is 0.1243.
 What type of test is this?
 Determine if the test is left, right, or twotailed.
 Draw the picture of the pvalue.
 Do we reject null hypothesis, H_{0}: μ ≤ 1?
 Do we have enough evidence to conclude that μ > 1?
Show Answer
 Test of a population mean.
 righttailed test
 The picture of the pvalue is as follows:
 Since pvalue > significance level, we do not reject null hypothesis, H_{0}: μ ≤ 1.
 We do not have enough evidence to conclude that μ > 1.
Example 3
H_{a}: p ≠ 50
Significance level = 1%
Assume the pvalue is 0.0005
 What type of test is this?
 Determine if the test is left, right, or twotailed.
 Draw the picture of the pvalue.
 Do we reject null hypothesis, H_{0}: p = 50?
 Do we have enough evidence to conclude that p ≠ 50?
Show Answer
 This is a test of a single population mean.
 Ha tells you the test is twotailed.
 The picture of the pvalue is as follows:
 Since pvalue < significance level, we reject null hypothesis, H_{0}: p = 50.
 We have enough evidence to conclude H_{a}: p ≠ 50.
Try It
H_{0}: p = 0.5
H_{a}: p ≠ 0.5
Significance level = 0.05
Assume the pvalue is 0.2564.
 What type of test is this?
 Determine if the test is left, right, or twotailed.
 Draw the picture of the pvalue.
 Do we reject null hypothesis, H_{0}: p = 0.5?
 Do we have enough evidence to conclude that p ≠ 0.5?
Click here to show solution:
 Hypothesis test of a single population proportion.
 twotailed test
 The picture of the pvalue is as follows:
 Since pvalue > significance level, we do not reject null hypothesis ( H_{0}: p = 0.5).
 We do not have enough evidence to conclude H_{a}: p ≠ 0.5.
Steps to set up a hypothesis test:

Full Hypothesis Test Examples
Jeffrey, as an eightyear old, established a mean time of 16.43 seconds for swimming the 25yard freestyle, with a standard deviation of 0.8 seconds. His dad, Frank, thought that Jeffrey could swim the 25yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25yard freestyle swims.
For the 15 swims, Jeffrey’s mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds.
Conduct a hypothesis test using 5% significance level. Assume that the swim times for the 25yard freestyle are normal.
Solution:
mean = 16.43 seconds, standard deviation = 0.8 seconds.
Since the problem is about a mean, this is a test of a single population mean.
1. What are we testing?
H_{0}: μ = 16.43
H_{a}: μ < 16.43
For Jeffrey to swim faster, his time will be less than 16.43 seconds. The “<” tells you this is lefttailed.
2. What is the significance level?
significance level, [latex]\alpha[/latex] = 5% = 0.05
3. What is the pvalue?
Graph:
[latex]\overline{X}[/latex] = the mean time to swim the 25yard freestyle.
[latex]\overline{X}[/latex] is normal. Population standard deviation is known: σ = 0.8
[latex]\overline{X}[/latex] = 16,
[latex]\mu[/latex] = 16.43 (comes from H_{0} and not the data.)
σ = 0.8, and
n = 15.
Calculate the pvalue using the normal distribution for a mean:
pvalue
= P[latex]\left(\overline{X}<{16}\right)[/latex]
= P[latex]\left(\frac{\overline{X}  {\mu}}{\frac{\sigma}{\sqrt{n}}}<\frac{16  {\mu}}{\frac{\sigma}{\sqrt{n}}}\right)[/latex]
= P[latex]\left( Z <\frac{16  {\mu}}{\frac{\sigma}{\sqrt{n}}}\right)[/latex]
= P[latex]\left( Z <\frac{16 16.43}{\frac{0.8}{\sqrt{15}}}\right)[/latex]
= 0.0187
where the sample mean in the problem is given as 16.
4. Comparison between pvalue and significance level.
pvalue = 0.0187 (This is called the actual level of significance.)
The pvalue is the area to the left of the sample mean is given as 16.
pvalue = 0.0187, α = 0.05
Therefore, α > pvalue.
Interpretation of the pvalue: If H_{0} is true, there is a 0.0187 probability (1.87%)that Jeffrey’s mean time to swim the 25yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event. 
5. Decision?
This means that you reject μ = 16.43. In other words, you do not think Jeffrey swims the 25yard freestyle in 16.43 seconds but faster with the new goggles.
Make a decision: Since α > pvalue, reject H_{0}.
6. Conclusion?
At the 5% significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey’s mean time to swim the 25yard freestyle is less than 16.43 seconds.
Using TICalculator to find pvalue:
The calculator not only calculates the pvalue (p = 0.0187), but it also calculates the test statistic (zscore) for the sample mean. Do this set of instructions again except arrow to Draw (instead of Calculate ) and press ENTER . A shaded graph appears with z = 2.08 (test statistic) and p = 0.0187 (pvalue). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. 
To find P[latex]\left(\overline{x}<{16}\right)[/latex], we will use TICalculator. TiCalculator: 2nd DISTR normcdf ([latex]{10}^{99}[/latex], 16,16.43,[latex]\frac{{0.8}}{{\sqrt{15}}}[/latex]).
The Type I and Type II errors for this problem are as follows:
The Type I error is to conclude that Jeffrey swims the 25yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25yard freestyle, on average, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)
The Type II error is that there is not evidence to conclude that Jeffrey swims the 25yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25yard freestyle, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)
Try It
The mean throwing distance of a football for a Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of 2 yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset α = 0.05. Assume the throw distances for footballs are normal.
First, determine what type of test this is, set up the hypothesis test, find the pvalue, sketch the graph, and state your conclusion.
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StepbyStep Solution
Since the problem is about a mean, this is a test of a single population mean.
 H_{0} : μ = 40
 H_{a} : μ > 40
 α = 0.05
 Z = [latex]\frac{\overline{X}\mu}{\frac{\sigma}{\sqrt{n}}}[/latex] = [latex]\frac{45  40}{\frac{2}{\sqrt{20}}}[/latex] = 11.1803
 When [latex]\overline{X}[/latex] = 45, its corresponding zscore is 11.1803.
The area to the right when [latex]\overline{X}[/latex] = 45
= The area to the right of zscore = 11.1803
= blue shaded area
= pvalue
= [latex]2.6115*10^{29}[/latex]
 Because p < α, we reject the null hypothesis.
 There is sufficient evidence to suggest that the change in grip improved Marco’s throwing distance.
Using TICalculator to solve:
 Press STAT and arrow over to TESTS.
 Press 1:ZTest.
 Arrow over to Stats and press ENTER.
 Arrow down and enter 40 for μ0 (null hypothesis), 2 for σ, 45 for the sample mean, and 20 for n.
 Arrow down to μ: (alternative hypothesis) and set it either as <, ≠, or >.
 Press ENTER.
 Arrow down to Calculate and press ENTER.
The pvalue = 0.0062. α = 0.05.
Because p < α, we reject the null hypothesis.
There is sufficient evidence to suggest that the change in grip improved Marco’s throwing distance.
The calculator not only calculates the pvalue but it also calculates the test statistic (zscore) for the sample mean. Select <, ≠, or >; for the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with test statistic and pvalue. Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off.