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1. The Meaning of Hypothesis Testing
Text 9.1-9.12 [9.1 – 9.12] (9.1 – 9.12), 9.16** [9.17], (9.17)
2. Steps for Testing a Hypothesis Applied to testing for a Population Mean
Text 9.20, 9.28a,c,d, 9.29 a,c,e [9.18, 9.26a,b,d,e, 9.27a,b,d,e], (9.18, 9.26a,b,d,e, 9.27a,b,d,e)
Text 9.48-9.51, 9.59 [9.46 - 9.49, 9.52] (9.44 – 9.47, 9.50)
3. The Use of p-value instead of Significance Levels.
Text 9.28b, 9.29b [9.26c, 9.27c], B7 (B6), Text 9.34-9.44 [9.32 – 9.39, 9.40*, 9.41*, 9.44] (9.26c, 9.27c, B6, 9.32 – 9.39, 9.40)
[9.40* reads ‘Suppose that in a one-tail hypothesis test where you reject only in the lower tail, you compute the value of the test statistic as 1.38, what is the p-value?’ 9.41 reads ‘ In Problem 9.40, what would be your statistical decision if you tested the null hypothesis at the 0.01 level of significance?’] B8.
Graded Assignment 2 will be posted.
4. Type One and Type Two Errors
Text 9.82-9.86 [9.86 – 9.90] (9.81 – 9.85)
5. Hypotheses about a Proportion
Text 9.66-9.69, 9.70**, 9.72**[9.62 – 9.64, 9.66*, 9.67*], (9.57 – 9.59)
6. The Sign Test
B1*, B2, B3 (B1, B2)
7. Hypothesis Test for Means - Rare Events
B4, B5 (B3, B4)
8. Hypothesis Tests for a Variance.
Text 12.45[9.80] (9.75), B6 (B5)
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Sections 4 – 6 are in this document.Note that there are some very useful sign test problem that were not assigned included in this document.
Type One and Type Two Errors
The outline says 'a Type one error is rejecting when is true (wrongly rejecting ) ; a Type two error is accepting when is true (wrongly not rejecting ).
Do not reject
/ /Reject
/ /The following problems are simply a review of the concepts covered up to now in this section. Answers to these problems come fromthe Instructor’s Solution Manual .
Exercise 9.82 [9.86 in 9th] (9.81 in 8th edition):The problem asks for difference between null and alternative hypotheses.
The null hypothesis represents the status quo or the hypothesis that is to be disproved. The null hypothesis includes an equal sign in its definition of a parameter of interest. The alternative hypothesis is the opposite of the null hypothesis, and usually represents taking an action. The alternative hypothesis includes either a less than sign, a not equal sign, or a greater than sign in its definition of a parameter of interest.
Exercise 9.83 [9.87 in 9th] (9.82 in 8th edition):The problem asks for difference between Type I and Type II errors.
A Type I error represents rejecting a true null hypothesis, while a Type II error represents not rejecting a false null hypothesis.
Exercise 9.84 [9.88 in 9th] (9.83 in 8th edition):The problem asks what the power of a test means.
The power of a test is the probability that the null hypothesis will be rejected when the null
hypothesis is false.
Exercise 9.85 [9.89 in 9th] (9.84 in 8th edition):The problem asks the difference between one-tail and two-tail tests.
In a one-tailed test for a mean or proportion, the entire rejection region is contained in one tail of the distribution. In a two-tailed test, the rejection region is split into two equal parts, one in the lower tail, and the other in the upper tail of the distribution.
Exercise 9.86 [9.90 in 9th] (9.85 in 8th edition):The problem asks for the meaning of a p-value.
The p-value (or observed significance level) is the probability of obtaining a test statistic equal to
or more extreme than the result obtained from the sample data, given that the null hypothesis is
true.
Tests for a population proportion - Remember! A proportion must be between zero and one and must also be between zero and one.
Exercise 9.66 [9.62 in 9th] (9.57 in 8th edition): Given: and find the sample proportion.
Solution:The text uses for a sample proportion. I use . Note that and .
Exercise 9.67 [9.63 in 9th] (9.58 in 8th edition):In 9.66 if the null hypothesis is , what is the value of the z test ratio?
Solution:From the outline, where . So and .
The Instructor’s Solution Manual also gives the following formula:
.
Exercise 9.68 [9.64in 9th] (9.59 in 8th edition):Given and ,and as computed above, what is your statistical decision?
Solution:Since this is a 2-sided test, we will reject the nullhypothesis if or
Decision rule: If < – 1.96 or > 1.96, reject H0.Make a diagram. Draw a normal cuve centered at zero and shade the areas below -1.960 and above +1.960.
Test statistic:
Decision: Sinceis between the critical bounds of , do not reject H0.
According to the outline there are three ways to do a one-sample test of a proportion if the hypotheses are and : (i) Test Ratio:, (ii) Critical Value: and (iii) Confidence Interval: where . In most problems, you should use only one method.
The following exercises appear only in one edition.
Exercise [9.66 in 9th] (Not in 8th edition):
a)If we have the following: and ,, and
, what is your statistical decision? b) Find a p-value. c) Test H0: p 0.50, H1: p > 0.50 if
now . d)Find a p-value for the result in c). e) Assume H0: p 0.55, H1: p > 0.55 and the sample
proportion is .57. So now and What is your statistical decision? f) Find a p-value in e).
Solution:a) (i) Test Ratio –This is a 2-sided test, so use and . Make a diagram with the mean at zero showing the rejection zones are below -1.960 and above 1.960.
Decision rule: If < – 1.96 or > 1.96, reject H0.
Note that .
Now compute .
If you use your diagram, you will see that this value of falls in the lower rejection zone.
Decision: Since is less than the lower critical bounds of , reject H0 and conclude that there is enough evidence to show that the percentage of all PC owners in the U.S. who rank sharing their credit card information as the number one concern in on-line shopping is not 50%.
(ii) The critical value formula is , so or .466 to .534. Make a diagram with the mean at .50 showing the rejection zones are the areas below .466 and the area above .534. Since is below .466, reject .
(iii) The confidence interval formula is where . So or .406 to .474. Make a diagram with the mean at .44 showing the confidence interval by shading the area between .406 and .474. Show the null hypothesis by marking on the same diagram. Since this point is outside the confidence interval, reject .
b)The usual way to find a p-value in a 2-sided problem is to take the probability above (or below in a left sided problem) the valueof and double it.
The probability ofobtaining a sample proportion further away from the hypothesized value of 0.50 is 0.0003 if the null hypothesis is true.
Answers in (a) and (b) differ slightly from those in the Instructor’s Solution Manual. Where did I go wrong?
c)Test H0: p 0.50, H1: p > 0.50 if now . Find a p-value.
(i) Test Ratio – This is a 1-sided upper tail test, so use . Make a diagram with the mean at zero showing the rejection zone is above 1.645.
Decision rule: If > 1.645, reject H0.
Recall that and compute .
If you use your diagram, you will see that this value of falls in the rejection zone.
Decision: Sinceis greater than the critical bound of reject H0 and conclude that there is enough evidence to show that the percentage of all PC owners in the U.S. who would pay an extra $75 for a new PC delivering a more secure on-line experience is more than 50%.
(ii) The critical value formula is , But this is a one-sided upper tail test, so use a critical value above Make a diagram with the meanat .50 showing the rejection zone as the area above .525. Since is above .525, reject .
(iii) The confidence interval formula is where . But this is a one-sided upper tail test, the alternate hypothesis is H1: p > 0.50, and we model the one-sided confidence
intervalon the alternate hypothesis, so use .
Make adiagram with the mean at .57 showing the confidence interval by shading the
area above.541. Show the null hypothesis on the same diagram by shading
an areabelow .50. These two areas will not overlap, indicating that the confidence intervalcontradicts the null hypothesis.
(d). The probability of obtaining a sample which will yield higher than 57% of all PC owners in the U.S. who would pay an extra $75 for a new PC delivering a more secure on-line experience is essentially zero if the null hypothesis is true.
Answers in (c) and (d) differ slightly from those in the Instructor’s Solution Manual. Is it really me?
9.66(e)Assume H0: p 0.55, H1: p > 0.55 and the sample proportion is .57 So now and What is your statistical decision? Find a p-value.
(i) Test Ratio – This is a 1-sided upper tail test, so use . Make a diagram with the mean at zero showing the rejection zone is above 1.645.
Decision rule: If > 1.645, reject H0.
Compute and compute .
If you use your diagram, you will see that this value of does not fall in the rejection zone.
Decision: Sincefalls below the critical bound of 1.645, do not reject H0 and conclude that there is not enough evidence to show that the percentage of all PC owners in the U.S. who would pay an extra $75 for a new PC delivering a more secure on-line experience is more than 55%.
(ii) The critical value formula is , But this is a one-sided upper tail test, so use a critical value above Make a diagram with the mean at .55 showing the rejection zone as the area above .579. Since is not above .525,do not reject .
(iii) The confidence interval formula is where . But this is a one-sided upper tail test, the alternate hypothesis is H1: p > 0.50, and we model the one-sided confidence
intervalon the alternate hypothesis, so use .Make adiagram with the mean at .57 showing the confidence interval by shading the
area above.541. Show the null hypothesis on the same diagram by shading
an areabelow .55. These two areas will overlap, indicating that the confidence interval
does notcontradict the null hypothesis.
(f). The probability of obtaining a sample which will yield a higher percentage than 57% of all PC owners in the U.S. who would pay an extra $75 for a new PC delivering a more secure on-line experience is 0.1261 if the null hypothesis is true.
Answers in (e) and (f) differ slightly from those in the Instructor’s Solution Manual and I’m getting annoyed.
For (e) the Instructor’s Solution Manual has “Test statistic: =1.1258.” I don’t like to work this hard,
Butlet’s try
Exercise[9.67 in 9th] (Not in 8thor 10thedition):The statement to be evaluated is that less than half of all
companies are delaying data-storage expenditures because of an economic slowdown. In a) for example, the
resulting hypotheses are tested after a survey of 50 firms shows that 19 have delayed. In c) the survey is
repeated with 100 firms and similar results.
a) Given the following, what is your statistical decision?H0: p 0.5, H1: p < 0.5,,
and b) Find a p-value for a). c) Given thefollowing, what is your statistical decision? H0: p 0.5,
H1: p < 0.5,, and d)Find a p-value for c). e) What is the effect of the
larger sample size? f) What are theconsequences of failing to state a sample size when results are stated?
9.67(a)Given the following, what is your statistical decision? H0: p 0.5, H1: p < 0.5,, and Find a p-value.
(i) Test Ratio – This is a 1-sided lower tail test, so use Make a diagram with the mean at zero showing the rejection zone is below -2.327.
Decision rule: If < -2.327, reject H0.
Compute and compute .
If you use your diagram, you will see that this value of does not fall in the rejection zone.
Decision: Since is not below the critical bound of do not reject H0. There is not enough evidence to show less than half of all companies are delaying data-storage expenditures.
(ii) The critical value formula is , But this is a one-sided lower tail test, so use a critical value below Make a diagram with the mean at .50 showing the rejection zone as the area below .335. Since is above .335, do not reject .
(iii) The confidence interval formula is where . But this is a one-sided lower tail test, the alternate hypothesis is H1: p < 0.5, and we model the one-sided confidence
intervalon the alternate hypothesis, so use .Make adiagram with the mean at .38 showing the confidence interval by shading the
area below.539. Show the null hypothesis on the same diagram by shading
an areaabove .50. These two areas will overlap, indicating that the confidence interval does notcontradict the null hypothesis.
(b). The probability of observing 38% or less of the companies surveyed having delayed expenditures on storage deployments because of economic slowdown is 0.0446 if the null hypothesis is true.
(c)Given the following, what is your statistical decision? H0: p 0.5,H1: p < 0.5,, and Find a p-value.
i) Test Ratio – This is a 1-sided lower tail test, so use Make a diagram with the mean at zero showing the rejection zone is below -2.327.
Decision rule: If < -2.327, reject H0.
Compute and compute . If you use your diagram, you will see that this value of falls in the rejection zone.
Decision: Since is below the critical bound of reject H0. There is enough evidence to show less than half of all companies are delaying data-storage expenditures.
(ii) The critical value formula is Make a diagram with the mean at .50 showing the rejection zone as the area below .384. Since is below .384, reject .
(iii) The confidence interval formula uses,
so .Make a diagram with the mean at .38
showing the confidence interval by shading the
area below .493. Show the null hypothesis on the same diagram by shading
an area above .50. These two areas will not overlap, indicating that the confidence
intervalcontradicts the null hypothesis.
(d). (The remainder of this problem is quoted fromthe Instructor’s Solution Manual.) The probability of observing 38% or less of the companies surveyed having delayed expenditures on storage deployments because of economic slowdown is 0.0082 if the null hypothesis is true.
(e)What is the effect of the larger sample size? The larger sample size of 100 responses reduces the standard error (variation) of the sample proportion and reduces the probability of observing 38% or less of the companies surveyed having delayed expenditures on storage deployments because of economic slowdown if the null hypothesis is true.
(f)What are the consequences of failing to state a sample size when results are stated? When one fails to report the sample size used in a survey, the readers cannot accurately evaluate the sampling error and, hence, there is a potential that the study can promote a viewpoint that might otherwise be truly insignificant.
The following exercises appear only in the 10th edition.
Exercise 9.69: In a survey, 593 out of 1040 respondents indicated that they would rather have $100 than a day off. a) Test the statement that more than half of workers would prefer the money to a day off. Use a 95% confidence level. b) Find a p-value.
Solution:(a) and ,, and .
(i) Test Ratio – This is a 1-sided test, so use .Make a diagram with the mean at zero showing the rejection zone above 1.645.
Decision rule: If > 1.645, reject H0.
Note that .
Now compute .
If you use your diagram, you will see that this value of falls in the rejectionzone.
Decision: Since is more than the critical bound of ,reject H0 and conclude that there is enough evidence to show that more than half of workers would prefer $100 to a day off.
(ii) The critical value formula is , so for a right sided test we use . Make a diagram with the mean at .50 showing the rejection zone is are the area above .5255. Since is above .5255, reject .
(iii) The confidence interval formula is where .But, since the alternate hypothesis is we must use the one-sided interval with the same direction as the alternate hypothesis, which is . This becomes . Make a diagram with the middle at .5702 showing theconfidence interval by shading the area above .5449. Show the null hypothesisby marking or shading the area below .50(differently) on the same diagram. Since .50 (and the area below it) is outside the confidence interval, reject .
(b) This is a right sided test, since we will reject the null hypothesis if the sample proportion is significantly larger than .50. No matter what the observed proportion is, the p-value will be the probability that the observed proportion is as large or larger than the actually observed proportion. Since is essentially Normally distributed, we can say
Exercise 9.70:We are testing to see if the proportion of employers who plan to recruit new employees is above last year’s level of 43%. Out of a sample of 362 potential employers, 181 plan to hire. . a) State and test the null and alternative hypotheses. b) find a p-value.
(a) and ,, and . Useonly one of the following methods in a test situation.
(i) Test Ratio – This is a 1-sided test, so use . Make a diagram with the mean at zero showing the rejection zone above 1.645.
Decision rule: If > 1.645, reject H0.
Note that .
Now compute .
If you use your diagram, you will see that this value of falls in the rejection zone.
Decision: Since is more than the critical bound of ,reject H0 and conclude that there is enough evidence to show that more than 43% plan to hire.
If we want the p-value instead, the p-value will be . Since this is below 5%, we reject the null hypothesis.
(ii) The critical value formula is , so for a right sided test we use . Make a diagram with the mean at .50 showing the rejection zone is are the area above .4728. Since is above .4728, reject .
(iii) The confidence interval formula is where .But, since the alternate hypothesis is we must use the one-sided interval with the same direction as the alternate hypothesis, which is . This becomes . Make a diagram with the mean at .5000 showing the confidence interval by shading the area above .4568. Show the null hypothesis by marking or shading the area below .43 (differently) on the same diagram. Since .43 (and the area below it) is outside the confidence interval, reject .
(b) This is a right sided test, since we will reject the null hypothesis if the sample proportion is significantly larger than .50. No matter what the observed proportion is, the p-value will be the probability that the observed proportion is as large as or larger than the actually observed proportion. Since is essentially Normally distributed, we can say