One-way ANOVA
F=2.2239835933 p=0.1246311696 Factor df=2 SS=317.802408 Down MS=158.901204
One-way ANOVA
UP MS=158.901204 Error df=32 SS=2286.36512 MS=71.4489101 Sxp=8.45274571
What is the conclusion for this hypothesis test?
a Reject H0. There is insufficient evidence to warrant the rejection of the claim that the three books have the same mean Flesch Reading Ease score.
b Fail to reject H0. There is sufficient evidence to warrant the rejection of the claim that three books have the same mean Flesch Reading Ease score.
c Fail to reject H0. There is insufficient evidence to warrant the rejection of the claim that three books have the same mean Flesch Reading Ease score.
d Reject H0. There is sufficient evidence to warrant the rejection of the claim that the three books have the same mean Flesch Reading Ease score.
2. Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean.
Identify the value of the test statistic.
Source DF SS MS F p Factor 3 30 10.00 1.6 0.264 Error 8 50 6.25 Total 11 80
a 10.00
b 1.6
c 0.264
d 30
3. Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean.
Identify the p-value
Source DF SS MS F p Factor 3 30 10.00 1.6 0.264 Error 8 50 6.25 Total 11 80
a 1.6
b 10.00
c 6.25
d 0.264
4. Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean.
Find the critical value.
Source DF SS MS F p
Factor 3 30 10.00 1.6 0.264
Error 8 50 6.25
Total 11 80
5. Given below are the anlysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean.
What can you conclude about the equality of the population means?
Source DF SS MS F p Factor 3 30 10.00 1.6 0.264 Error 8 50 6.25 Total 11 80
a Reject the null hypothesis since the p-value is greater than the significance level.
b Accept the null hypothesis since the p-value is greater than the significance level.
c Reject the null hypothesis since the p-value is less than the significance level.
d Accept the null hypothesis since the p-value is less than the significance level.
6. If we use the amounts (in millions of dollars) grossed by movies in categories with PG, PG-13, and R ratings, we obtain the SPSS analysis of variance results shown below. Use a 0.05 significance level to test the claim that PG movies, PG-13 movies, and R movies have the same mean gross amount.
Gross Sum of Squares df Mean Square F Sig Between Groups 59110.874 2 29555.437 3.705 .035
Within Groups 271224.0 34 7977.176 Total 330334.9 36
What is the conclusion for this hypothesis test?
a Fail to reject H0. There is insufficient evidence to warrant the rejection of the claim that PG movies, PG-13 movies, and R movies have the same mean gross amount.
b Fail to reject H0. There is sufficient evidence to warrant the rejection of the claim that PG movies, PG-13 movies, and R movies have the same mean gross amount.
c Reject H0. There is sufficient evidence to warrant the rejection of the claim that PG movies, PG-13 movies, and R movies have the same mean gross amount.
d Reject H0. There is insufficient evidence to warrant the rejection of the claim that PG movies, PG-13 movies, and R movies have the same mean gross amount.
7. The data was obtained from car crash experiments. The table values are the chest deceleration data (g) for the dummy in the driver's seat. Use a 0.05 significance level to test the null hypothesis that the different weight categories have the same mean. Do the data suggest that larger cars are safer?
Subcompact Compact Midsize Full-size
59 53 55 44
59 55 47 45
58 46 50 42
55 46 54 42
59 51 53 53
a The results of an ANOVA test are insufficient to make such a claim.
b No, because the null hypothesis is not rejected.
c Yes, because the null hypothesis is rejected.
d Yes, because the null hypothesis is not rejected.
8. Refer to data table below, which shows the amounts of nicotine (mg per cigarette) in king size, 100mm menthol, and 100mm nonmenthol cigarettes. The king size cigarettes are nonfiltered, while the 100mm menthol and the 100mm nonmenthol cigarettes are filtered. Use a 0.05 significance level ot test the claim that three categories of cigarettes yield the same mean amount of nicotine. Given that only the king size cigarettes are not filtered, do the filters appear to make a difference?
King Size Menthol Filtered 100mm Nonmenthold
Brand Nicotine Brand Nicotine Brand Nicotine
1 1.6 1 1.1 1 0.7
2 1.0 2 0.9 2 1.1
3 1.0 3 1.1 3 0.5
4 1.1 4 0.8 4 1.1
5 1.5 5 1.2 5 1.1
6 1.3 6 1.4 6 0.8
7 1.2 7 1.0 7 1.1
8 1.0 8 1.2 8 1.2
9 1.2 9 1.4 9 0.8
10 1.1 10 0.9 10 1.0
Do the filters appear to make a difference?
a The results are inconclusive because the king size cigarettes are a different size than the filtered cigarettes.
b No, the filters do not appear to make a difference because there is insufficient evidence to warrant the rejection of the claim.
c No, the filters do not appear to make a difference because there is sufficient evidence to warrant the rejection of the claim.
d Given that the king size cigarettes have the largest mean, it appears that the filters do make a difference (although this conclusion is not justified by the results from analysis of variance).
9. The following table shows the two-way ANOVA output for the weights of poplar trees. Test the hypothesis that the weights of poplar trees are not affected by an interaction between site and treatment.
Source DF SS MF F P
Site 3 2.4776 0.825876 3.11 0.037
Treatment 1 0.3718 0.371777 1.40 0.243
Interaction 3 2.6768 0.892265 3.36 0.028
Error 40 10.6222 0.265555
Total 47 16.1484
Is there evidence to support the claim of interaction? (Assume a 0.05 significance level.)
a Since thee P-value for interaction is large, there is evidence of interaction.
b Since the P-value for interaction is small, there is no evidence of interaction.
c Since the P-value for interaction is large, there is no evidence of interaction.
d Since the P-value for interaction is small, there is evidence of interaction.
10. The following table shows the two-way ANOVA output for SAT scores. Assuming no interaction between gender and type of test (verbal/match), is there sufficient evidence to support the claim that gender has an effect on SAT scores?
Source DF SS MS F P
Gender 1 64094 64093.6 6.39 0.016
Verb/match 1 5818 5817.6 0.58 0.452
Interaction 1 31295 31294.5 3.12 0.086
Error 36 361091 10030.3
Total 39 462298
Is there any evidence that gender has an effect? (Assume a 0.05 significance level.)
a Since the P-value for gender is small, there is no evidence to support the claim.
b Since the P-value for gender is large, there is no evidence to support the claim.
c Since the P-value for gender is small, there is evidence to support the claim.
d Since the P-value for gender is large, there is evidence to support the claim.
11. Fill in the missing entries in the following partially completed one-way ANOVA table.
a Source df SS MS=SS/df F-statistic
Treatment 3 11.16
Error 13.72 0.686
Total
b Source df SS MS=SS/df F-statistic
Treatment 3 22.97 7.66 11.16
Error 20 13.72 0.686
Total 23 36.69
c Source df SS MS=SS/df F-statistic
Treatment 3 2.55 7.66 11.16
Error 20 13.72 0.686
Total 23 16.27
d Source df SS MS=SS/df F-statistic
Treatment 3 0.184 0.061 11.16
Error 20 13.72 0.686
Total 23 13.90
e Source df SS MS=SS/df F-statistic
d Treatment 3 48.80 16.27 11.16
Error 20 13.72 0.686
Total 23 62.52
12. Assume that matched pairs of data result in the given the number of signs when the second variable is subtracted from the corresponding value of the first variable. Use the sign test with a 0.05 significance to test the null hypothesis of no difference.
Positive signs: 15; negative signs: 7; ties: 1
a Yes, because the test statistic is greater than the critical value.
b No, because the test statistic is greater than the critical value.
c Yes, because the test statistic is less than or equal to the critical value.
d No, because the test statistic is less than or equal to the critical value.
13. Use a 0.05 significance level to test the claim that when the 13th of the month falls on a Friday, the number of storms is not affected.
A random Friday Friday the 13th
12 9
15 21
12 13
8 7
2 3
6 8
Should we reject the null hypothesis that when the 13th of the month is on a Friday, the number of storms is not affected at a significance level of a=0.05?
a No, because the test statistic is less than or equal to the critical value.
b No, because the test statistic is greater than the critical value.
c Yes, because the test statistic is greater than the critical value.
d Yes, because the test statistic is less than or equal to the critical value.
14. An institute conducted a clinical trial of its methods for gender selection. The results showed that 275 of 330 babies born to parents using a specific gender-selection method were boys. Use the sign test and a 0.05 significance level to test the claim that the method used is effective in increasing the likelihood of a boy.
a Since the test statistic falls within the critical region, reject the null hypothesis. There is sufficient evidence that the method used is effective in increasing the likelihood of a boy.
b Since the test statistic does not fall within the critical region, fail to reject the null hypothesis. There is insufficient evidence that the method used is effective in increasing the likelihood of a boy.
c Since the test statistic falls within the critical region, fail to reject the null hypothesis. There is sufficient evidence that the method used is effective in increasing the likelihood of a boy.
d Since the test statistic does not fall within the critical region, reject the null hypothesis. There is insufficient evidence that the method used is effective in increasing the likelihood of a boy.
15. The data is the accompanying table give the weights (in g) of randomly selected quarters that were minted after 1964. The quarters are supposed to have a median weight of 5.670g. Use the sign test and a 0.01 significance level to test the claim that the median is equal to 5.670g.
Post-1964 Quarters
5.7321 5.6652 5.5635 5.6746
5.5834 5.6968 5.5509 5.5772
5.5976 5.7233 5.6592 5.7455
5.7176 5.6642 5.6774 5.5575
5.6885 5.6823 5.7330 5.7228
5.7099 5.6669 5.6385 5.6141
5.6293 5.7447 5.5336 5.6408
5.5995 5.6436 5.6374 5.5646
5.6701 5.5803 5.6596 5.7023
5.7149 5.5809 5.5749 5.6636
a Since the test statistic does fall within the critical region, fail to reject H0.
b Since the test statistic does fall within the critical region, reject H0.
c Since the test statistic does not fall within the critical region, fail to reject H0.
d Since the test statistic does not fall within the critical region, reject H0.
16. The same sample data are used for the sign test and for the Wilcoxon signed-ranks test, but different conclusions result. Which conclusion is likely to be better?
a The sign test
b The Wilcoxon signed-ranks test
c Both tests
d Neither test
17. Researchers collected data on the number of hospital admissions resulting from motor vehicle crashes, and results are given for Friday the 6th and Friday the 13th in the same month. Use the Wilcoxon signed-ranks test to test the claim that the matched pairs have differences that come from a population with median equal to zero at a significance level of a=0.1.
Friday 6th Friday 13th
9 10
8 9
7 12
13 6
11 6
6 12
17. Can the null hypothesis be rejected? Why?
a Yes, because the test statistic is greater than the critical value.
b Yes, because the test statistic is less than or equal to the critical value.
c No, because the test statistic is less than or equal to the critical value.
d No, because the test statistic is greater than the critical value.
18. The weights of randomly selected quarters that were minted after 1964 are given below. The quarters are supposed to have a median weight of 5.680g. Use the Wilcoxon signed-ranks test and a 0.01 significance level to test the claim that the median is equal to 5.680g. Do quarters appear to be minted according to specifications?
Post-1964 Quarters
5.7027 5.6234 5.5591 5.5636
5.7495 5.5928 5.5864 5.6485
5.7050 5.6486 5.6872 5.6703
5.5941 5.6661 5.6274 5.6848
5.7247 5.5361 5.6157 5.5609
5.6114 5.5491 5.6668 5.7344
5.6160 5.7239 5.7198 5.6449
5.5999 5.6555 5.6694 5.5804
5.7790 5.6063 5.5454 5.6010
5.6841 5.5709 5.6646 5.6022
a Fail to reject H0. There is sufficient evidence to warrant rejection of the claim that the memdian is 5.680. The quarters do not appear to be minted according to specifications.
b Reject H0. There is sufficient evidence to warrant rejection of the claim that the memdian is 5.680. The quarters do not appear to be minted according to specifications.
c Reject H0. There is insufficient evidence to warrant rejection of the claim that the memdian is 5.680. The quarters do not appear to be minted according to specifications.
d Fail to reject H0. There is insufficient evidence to warrant rejection of the claim that the memdian is 5.680. The quarters do not appear to be minted according to specifications.
19. Which of the following distribution-free tests has no parametric counterpart?
a rank correlation test
b sign test
c runs test
d Kruskal-Wallis test
20. Which of the following tests could detect some nonlinear relationships between two variables?
a rank correlation test
b Wilcoxon signed-ranks test
c sign test
d Wilcoxon rank-sum test
