500 câu trắc nghiệm Kinh tế lượng – 11B
Use the Minitab display to test the indicated claim.
9) A manager records the production output of three employees who each work on three different machines for three different days. The sample results are given below and the Minitab results follow.
Assume that the number of items produced is not affected by an interaction between employee and machine. Using a 0.05 significance level, test the claim that the choice of employee has no effect on the number of items produced.
H0: There is no employee effect.
H1: There is an employee effect.
Test statistic: F = 0.1898. Critical value: F = 3.5546.
Fail to reject the null hypothesis. There does not appear to be an employee effect.
10) A manager records the production output of three employees who each work on three different machines for three different days. The sample results are given below and the Minitab results follow.
Using a 0.05 significance level, test the claim that the interaction between employee and machine has no effect on the number of items produced.
H0: There is no interaction effect.
H1: There is an interaction effect.
Test statistic: F = 0.7062. Critical value: F = 2.9277.
Fail to reject the null hypothesis. There does not appear to be an interaction effect.
11) A manager records the production output of three employees who each work on three different machines for three different days. The sample results are given below and the Minitab results follow.
Assume that the number of items produced is not affected by an interaction between employee and machine. Using a 0.05 significance level, test the claim that the machine has no effect on the number of items produced.
H0: There is no machine effect.
H1: There is a machine effect.
Test statistic: F = 0.0664. Critical value: F = 3.5546.
Fail to reject the null hypothesis. The type of machine does not appear to have an effect on the number of items produced
Use the data in the given table and the corresponding Minitab display to test the hypothesis.
12) The following table entries are the times in seconds for three different drivers racing on four different tracks. Assuming no effect from the interaction between driver and track, test the claim that the track has no effect on the time. Use a 0.05 significance level.
Track 1 | Track 2 | Track 3 | Track 4 | |
Driver 1 | 72 | 70 | 68 | 71 |
Driver 2 | 74 | 71 | 66 | 72 |
Driver 3 | 76 | 69 | 64 | 70 |
Source | DF | SS | MS | F | p |
Driver | 2 | 2 | 1 | 0.33 | 0.729 |
Track | 3 | 98.25 | 32.75 | 10.92 | 0.00763 |
Error | 6 | 18 | 3 | ||
Total | 11 | 118.25 |
H0: There is no track effect. H1: There is a track effect. The P-value is 0.00763, which is less than 0.05.
We reject the null hypothesis; it appears that the track does effect the racing times.
13) The following table entries are test scores for males and females at different times of day. Assuming no effect from the interaction between gender and test time, test the claim that time of day does not affect test scores. Use a 0.05 significance level.
6 a.m. – 9 a.m. | 9 a.m. – 12 p.m. | 12 p.m. – 3 p.m. | 3 p.m. – 6 p.m. | |
Male | 87 | 89 | 92 | 85 |
Female | 72 | 84 | 94 | 89 |
Source | DF | SS | MS | F | p |
Gender | 1 | 24.5 | 24.5 | 0.6652 | 0.4745 |
Time | 3 | 183 | 61 | 1.6561 | 0.3444 |
Error | 3 | 110.5 | 36.83 | ||
Total | 7 | 318 | |||
H0: There is no effect due to the time of day. H1: There is an effect due to the time of day. The
P-value is 0.3444, which is greater than 0.05. We fail to reject the null hypothesis; it appears that the scores are not affected by time of day.
Provide an appropriate response.
14) The following results are from a statistics software package in which all of the F values and P-values are given. Is there a significant effect from the interaction? Should you test to see if there is a significant effect due to either A or B? If the answer is yes, is there a significant effect due to either A or B?
ANOVA Table
Source | DF | Sum squares | Mean square | F test | P-value |
A | 2 | 164.020 | 82.010 | 25.010 | <.0001 |
B | 4 | 230.786 | 57.697 | 18.002 | <.0001 |
Interaction | 8 | 80.879 | 10.110 | 3.154 | .0031 |
Error | 101 | 323.708 | 3.205 | ||
Total | 115 | 799.393 |
Since P = 0.0031 for the interaction, you reject the null hypothesis that there is no effect due to the interaction. No, it is not appropriate to see if there is a significant effect due to either A or B. Do not consider the effects of either factor without considering the effects of the other.
15) The following data show annual income, in thousands of dollars, categorized according to the two factors of gender and level of education. Test the null hypothesis of no interaction between gender and level of education at a significance level of 0.05.
Female | Male | |
High school | 23, 27, 24, 26 | 25, 26, 22, 24 |
College | 28, 36, 31, 33 | 35, 32, 39, 28 |
Advanced degree | 41, 38, 43, 49 | 35, 50, 47, 44 |
H0: There is no interaction between gender and level of education. H1: There is an interaction between gender and level of education. The test statistic is F = 0.177472, and the corresponding P-value is 0.838832. Because the P-value is greater than 0.05, we fail to reject the null hypothesis of no interaction between gender and level of education
16) The following data contains task completion times, in minutes, categorized according to the gender of the machine operator and the machine used.
Male | Female | |
Machine 1 | 15, 17 | 16, 17 |
Machine 2 | 14, 13 | 15, 13 |
Machine 3 | 16, 18 | 17, 19 |
Assume that two-way ANOVA is used to analyze the data. How are the ANOVA results affected if the first sample value in the first cell is changed to 30 minutes?
If the first sample value is changed to 30 minutes, the ANOVA results are changed. The null hypothesis of no interaction between machine and gender is still not rejected. The null hypothesis of no effect from gender is still not rejected. However, the null hypothesis of no effect from machine is now accepted instead of rejected.
Before the change, the F test statistic = 9.7222. After the change, the F statistic = 2.5956. The F critical
value at (2,6), \(\alpha \) = 0.05 is 5.1433.