9.5: Additional Information and Full Hypothesis Test Examples
 Page ID
 16270
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left#1\right}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\) 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 thepvalue: 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?
[revealanswer q=”256590″]Click here to show solution: [/revealanswer]
[hiddenanswer a=”256590″]
 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. [/hiddenanswer]
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?
[practicearea rows=”1″][/practicearea]
[revealanswer q=”105740″]Click here to show solution:[/revealanswer]
[hiddenanswer a=”105740″]
 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. [/hiddenanswer]
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?
[revealanswer q=”590445″]Click here to show solution: [/revealanswer]
[hiddenanswer a=”590445″]
 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. [/hiddenanswer]
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?
[revealanswer q=”108750″]Show Answer[/revealanswer]
[hiddenanswer a=”108750″]
 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. [/hiddenanswer]
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?
[revealanswer q=”306871″]Show Answer[/revealanswer]
[hiddenanswer a=”306871″]
 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.
[/hiddenanswer]
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?
[revealanswer q=”293751″]Click here to show solution:[/revealanswer]
[hiddenanswer a=”293751″]
 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. [/hiddenanswer]
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.
[revealanswer q=”69231″]1. What are we testing? [/revealanswer]
[hiddenanswer a=”69231″]
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.[/hiddenanswer]
[revealanswer q=”12509″]2. What is the significance level? [/revealanswer]
[hiddenanswer a=”12509″]significance level, = 5% = 0.05[/hiddenanswer]
[revealanswer q=”540483″]3. What is the pvalue? [/revealanswer]
[hiddenanswer a=”540483″]
Graph:
= the mean time to swim the 25yard freestyle.
is normal. Population standard deviation is known: σ = 0.8
= 16,
= 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
= P
= P
= P
= 0.0187
where the sample mean in the problem is given as 16.
[/hiddenanswer]
[revealanswer q=”662892″]4. Comparison between pvalue and significance level. [/revealanswer]
[hiddenanswer a=”662892″]
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. 
[/hiddenanswer]
[revealanswer q=”715121″]5. Decision?[/revealanswer]
[hiddenanswer a=”715121″]
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}.[/hiddenanswer]
[revealanswer q=”665879″]6. Conclusion? [/revealanswer]
[hiddenanswer a=”665879″]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.[/hiddenanswer]
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, we will use TICalculator.
TiCalculator: 2nd DISTR normcdf (, 16,16.43,).
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.
[practicearea rows=”4″][/practicearea]
[revealanswer q=”205319″]StepbyStep Solution[/revealanswer]
[hiddenanswer a=”205319″]
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 = = = 11.1803
 When = 45, its corresponding zscore is 11.1803.
The area to the right when = 45
= The area to the right of zscore = 11.1803
= blue shaded area
= pvalue
=
 Because p < α, we reject the null hypothesis.
 There is sufficient evidence to suggest that the change in grip improved Marco’s throwing distance. [/hiddenanswer]
[revealanswer q=”766484″]Using TICalculator to solve: [/revealanswer]
[hiddenanswer a=”766484″]
 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. [/hiddenanswer]
Example 4
A college football coach thought that his players could bench press a mean weight of 275 pounds. It is known that the standard deviation is 55 pounds. Three of his players thought that the mean weight was more than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights were (frequencies are in parentheses) 205(3) 215(3)225(1) 241(2) 252(2) 265(2) 275(2) 313(2) 316(5) 338(2) 341(1) 345(2) 368(2) 385(1).
Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press mean is more than 275 pounds.
Solution:
[revealanswer q=”168347″]1. What are we testing?[/revealanswer]
[hiddenanswer a=”168347″]
Since the problem is about a mean weight, this is a test of a single population mean.
H_{0}: μ = 275
H_{a}: μ > 275
Significance leel, = 2.5% = 0.025
This is a righttailed test.
[/hiddenanswer]
[revealanswer q=”873840″]2. What is the significance level, ? [/revealanswer]
[hiddenanswer a=”873840″] = 2.5% = 0.025[/hiddenanswer]
[revealanswer q=”56192″]3. Find the pvalue.[/revealanswer]
[hiddenanswer a=”56192″]
= the mean weight (in pounds) lifted by the football players.
Distribution for the test: X is normally distributed because σ is known.
= 286.2, n = 30, σ = 55 pounds (Always use σ if you know it.)
We assume μ = 275 pounds unless our data shows us otherwise.
Calculate the pvalue using the normal distribution for a mean and using the sample mean as input.
pvalue
= P ( > 286.2 )
= P ( > )
= P (Z > )
= P (Z > )
= P (Z > 1.1153623)
= 0.1323.[/hiddenanswer]
[revealanswer q=”173168″]4. Comparison between pvalue and significance level. [/revealanswer]
[hiddenanswer a=”173168″]
Interpretation of the pvalue:
If H_{0} is true, then there is a 0.1331 probability (13.23%) that the football players can lift a mean weight of 286.2 pounds or more. Because a 13.23% chance is large enough, a mean weight lift of 286.2 pounds or more is not a rare event.
α = 0.025, pvalue = 0.1323[/hiddenanswer]
[revealanswer q=”330644″]5. Decision?[/revealanswer]
[hiddenanswer a=”330644″]Make a decision: Since α <pvalue, do not reject H_{0}.[/hiddenanswer]
[revealanswer q=”878326″]6. Conclusion?[/revealanswer]
[hiddenanswer a=”878326″]
Conclusion:
At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds.[/hiddenanswer]
The pvalue can easily be calculated.

The calculator not only calculates the pvalue (p = 0.1331, a little different from the previous calculation – in it we used the sample mean rounded to one decimal place instead of the data) but it also calculates the test statistic (zscore) for the sample mean, the sample mean, and the sample standard deviation. μ > 275 is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with z = 1.112 (test statistic) and p = 0.1331 (pvalue). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off.
Example 5
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. The data are assumed to be from a normal distribution.
He performs a hypothesis test using a 5% level of significance.
Solution:
[revealanswer q=”202884″]1. What are we testing? [/revealanswer]
[hiddenanswer a=”202884″]
Since we do not know population standard deviation, we are going to run Student’s t Test.
This is a test of a single population mean.
H_{0}: μ = 65
H_{a}: μ > 65
A 5% level of significance means that α = 0.05.
Since the instructor thinks the average score is higher, use a “>”. The “>” means the test is righttailed.[/hiddenanswer]
[revealanswer q=”510963″]2. What is the significance level, ? [/revealanswer]
[hiddenanswer a=”510963″]The significance level, = 5% = 0.05[/hiddenanswer]
[revealanswer q=”406273″]3. Find the pvalue. [/revealanswer]
[hiddenanswer a=”406273″]
Random variable: = 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 ttest.
Use t_{df}. Therefore, the distribution for the test is t_{9} where n = 10 and df = 10 – 1 = 9.
Calculate the pvalue using the Student’s tdistribution:
Given that sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data,
pvalue
= P(> 67)
= P(> )
= P( t > 1.978)
= 0.0396[/hiddenanswer]
[revealanswer q=”186204″]4. Comparison between pvalue and significance level. [/revealanswer]
[hiddenanswer a=”186204″]
Interpretation of the pvalue:
If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 65 or more.
Since α = 0.05 and pvalue = 0.0396. α > pvalue.[/hiddenanswer]
[revealanswer q=”243554″]5. Decision? [/revealanswer]
[hiddenanswer a=”243554″]Since α > pvalue, reject H_{0}.
This means you reject μ = 65. In other words, you believe the average test score is more than 65.[/hiddenanswer]
[revealanswer q=”391743″]6. Conclusion?[/revealanswer]
[hiddenanswer a=”391743″]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.[/hiddenanswer]
The pvalue can easily be calculated.

Try It
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 pvalue, state your conclusion, and identify the Type I and Type II errors.
[revealanswer q=”391557″]Show Answer[/revealanswer]
[hiddenanswer a=”391557″]
We run Student’s ttest as we do not know the population standard deviation.
H_{0}: μ = 5
H_{a}: μ < 5
p = 0.0082
Because p < α, 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).
[/hiddenanswer]
Example 6
Solution:
[revealanswer q=”877865″]1. What are we testing? [/revealanswer]
[hiddenanswer a=”877865″]
This is a Normal test of a single population proportion.
H_{0}: p = 0.50
H_{a}: p ≠ 0.50
The 1% level of significance means that α = 0.01.
The words “is different from” tell you this is a twotailed test. [/hiddenanswer]
[revealanswer q=”544872″]2. What is the significance level, ?[/revealanswer]
[hiddenanswer a=”544872″]the significance level, = 1% = 0.01[/hiddenanswer]
[revealanswer q=”774681″]3. Find the pvalue. [/revealanswer]
[hiddenanswer a=”774681″]This is a twotailed test. We will include both left tail and right tail in the hypothesis test.
P′ = the percent of of firsttime brides who are younger than their grooms.
The proportion of firsttime brides who reply that they are younger than their grooms in sample of 100 brides = = 0.53.
0.53 is the right tail of this test as 0.53 is larger than 0.5 (the population mean).
How about the left tail?
Since 0.53 is on the right side of 0.50, the left tail will fall on the left side of 0.50.
Moreover, 0.03 is the difference between 0.53 and 0.50. Hence the distance between the left tail and 0.50 is equal to 0.03 as well.
0.50 – 0.03 = 0.47.
The left tail of this test is 0.47.
Given that p = 0.50, q = 1 − p = 0.50, and n = 100,
pvalue
= area to the right of right tail + area to the left of left tail
= area to the right of 0.53 + area to the left of 0.47
= P (p’ > 0.53) + P (p’ < 0.47)
= P ( > ) + P ( > )
= P ( Z > 0.6 ) + P(Z < 0.6 )
= 0.27425 + 0.27425
= 0.5485
[/hiddenanswer]
[revealanswer q=”970068″]4. Comparison between pvalue and significance level. [/revealanswer]
[hiddenanswer a=”970068″]
pvalue = 0.5485, significance level = 0.01 [/hiddenanswer]
[revealanswer q=”505115″]5. Decision?[/revealanswer]
[hiddenanswer a=”505115″]Since pvalue > significance level, we do not reject H_{0}.[/hiddenanswer]
[revealanswer q=”666927″]6. Conclusion? [/revealanswer]
[hiddenanswer a=”666927″]There is no sufficient evidence to suggest that the percentage is different than 50%. [/hiddenanswer]
[hiddenanswer a=”71421″]
 Press
STAT
and arrow over toTESTS
.  Press
5:1PropZTest
. Enter .5 for p_{0}, 53 for x and 100 for n.  Arrow down to
Prop
and arrow tonot equals
p_{0}. PressENTER
.  Arrow down to
Calculate
and pressENTER
.  The calculator calculates the pvalue (p = 0.5485) and the test statistic (zscore).
Prop not equals
.5 is the alternate hypothesis.
Do this set of instructions again except arrow to Draw
(instead of Calculate
). Press ENTER
. A shaded graph appears with z = 0.6 (test statistic) and p = 0.5485 (pvalue). Make sure when you use Draw
that no other equations are highlighted in Y = and the plots are turned off.[/hiddenanswer]
Try It
A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.
First, determine what type of test this is, set up the hypothesis test, find the pvalue, sketch the graph, and state your conclusion.
[revealanswer q=”401155″]Show Answer[/revealanswer]
[hiddenanswer a=”401155″]
Since the problem is about percentages, this is a test of single population proportions.
H_{0} : p = 0.85
H_{a}: p ≠ 0.85
p = 0.7554
Because p > α, we fail to reject the null hypothesis.
There is not sufficient evidence to suggest that the proportion of students that want to go to the zoo is not 85%.[/hiddenanswer]
Example 7
Suppose a consumer group suspects that the proportion of households that have three cell phones is 30%. A cell phone company survey 150 households with the result that 43 of the households have three cell phones. They believe that the proportion is less than 30%. Conduct a hypothesis test to check their claim at 5% significance level.
[revealanswer q=”828285″]Show Answer[/revealanswer]
[hiddenanswer a=”828285″]
Ha : p < 0.3
5% significance level means = 0.05.
= P ( < 0.30)
= P ( < )
= P ( Z < – 0.3563)
[/hiddenanswer]
Try It
Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the pvalue, state your conclusion, and identify the Type I and Type II errors.
[revealanswer q=”33832″]Click here to show solution: [/revealanswer]
[hiddenanswer a=”33832″]
H_{0}: p = 0.92
H_{a}: p < 0.92
pvalue = 0.0046
Because pvalue < 0.05, we reject the null hypothesis.
There is sufficient evidence to conclude that fewer than 92% of American adults own cell phones.
Type I Error: To conclude that fewer than 92% of American adults own cell phones when, in fact, 92% of American adults do own cell phones (reject the null hypothesis when the null hypothesis is true).
Type II Error: To conclude that 92% of American adults own cell phones when, in fact, fewer than 92% of American adults own cell phones (do not reject the null hypothesis when the null hypothesis is false).[/hiddenanswer]
Example 8
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. (Assume the population is normal. )
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 1?
Use a significance level of 0.05.
[revealanswer q=”104041″]Stepbystep Solution:[/revealanswer]
[hiddenanswer a=”104041″]
Significance level is 5% ( = 0.05)
= 0.0359
Since pvalue < significance level, we reject null hypothesis.
[revealanswer q=”236792″]Using TICalculator to solve:[/revealanswer]
[hiddenanswer a=”236792″]
Significance level is 5% ( = 0.05)
We will input the sample data into the TI83 as follows.
pvalue = 0.03586
Since pvalue < significance level, we reject null hypothesis.
[/hiddenanswer]
Try It
In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for noncell phone users (the rate of brain cancer for noncell phone users is 0.0340%). (Since this is a critical issue, use a 0.5% significance level to run the hypothesis test.
[revealanswer q=”919196″]Stepbystep solution:[/revealanswer]
[hiddenanswer a=”919196″]
Significance level is 0.5% ( = 0.005)
= 0.00728
Since pvalue < significance level, we reject null hypothesis.
[revealanswer q=”115189″]Using TICalculator to solve:[/revealanswer]
[hiddenanswer a=”115189″]
Significance level is 0.5% ( = 0.005)
pvalue = 0.00728
Since pvalue < significance level, we reject null hypothesis.
We have sufficient evidence to conclude that cell phone users developed brain cancer at a greater rate.
[/hiddenanswer]
Example 9
According to the US Census there are approximately 268,608,618 residents aged 12 and older. Statistics from the Rape, Abuse, and Incest National Network indicate that, on average, 207,754 rapes occur each year (male and female) for persons aged 12 and older. This translates into a percentage of sexual assaults of 0.07734%. In Daviess County, KY, there were reported 11 rapes for a population of 37,937. Conduct an appropriate hypothesis test to determine if there is a statistically significant difference between the local sexual assault percentage and the national sexual assault percentage. Use 1% significance level.
[revealanswer q=”907316″]Show Answer[/revealanswer]
[hiddenanswer a=”907316″]
Significance level is 1% ( = 0.001)
pvalue = 0.0006288
Since pvalue < significance level, we reject null hypothesis.
Concept Review
The hypothesis test itself has an established process. This can be summarized as follows:
Notice that in performing the hypothesis test, you use α and not β. β is needed to help determine the sample size of the data that is used in calculating the pvalue. Remember that the quantity 1 – β is called thePower 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.
 Introductory Statistics . Authored by: Barbara Illowski, Susan Dean. Provided by: Open Stax. Located at: http://cnx.org/contents/3018944269984686ac05ed152b91b9de@18.54. License: CC BY: Attribution. License Terms: Download for free at http://cnx.org/contents/30189442699...2b91b9de@18.54.