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AP Statistics - PVS: P-Values from Simulations
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1
Statistical Inference > P-values from simulations
Error Rate 100%
Wrong
A marriage counselor plans a study on whether a husband or a wife is happier on a greater proportion of days when the wife has a job outside the home and the husband is a stay-at-home dad. In a random sample of married couples where the wife works and the husband is at home, the difference in proportions of happy days between husband and wife is 0.1106. The counselor, who took a required statistics class in college, is concerned that the proportions she subtracted to get 0.1106 are not from independent samples. So she runs 200 simulations of couples with the null hypothesis that there is no difference in proportions of happy days. The dotplot of resulting proportion differences is below.
Simulated Differences in Proportions
Is there sufficient evidence of a significant difference in the proportion of happy days experienced by husbands and wives in marriages where the wife works and the husband is a stay-at-home dad?
A
Yes, because 0.1106 > 0.0
B
Yes, because 0.1106 > 0.05
My Answer
C
Yes, because the distribution of simulated differences is approximately normal so the central limit theorem applies
D
No, because 0.1106 > 0.05
No, because the simulated P-value is large
Correct Answer
Explanation
The dotplot shows the differences in proportions we would expect to see by random chance if the null hypothesis of no difference was true. In \(18 + 11 + 6 + 3 + 1 + 1 = 40\) out of the 200 simulations, the difference was equal to or greater than 0.1106. The estimated P-value is \(\dfrac{40}{200} = 0.20\). With this large of a P-value, \(0.20 > 0.05\), there is not sufficient evidence to reject the null hypothesis. In other words, there is not sufficient evidence of a difference in the proportion of happy days experienced by husbands and wives in marriages where the wife works and the husband is a stay-at-home dad.
2
Statistical Inference > P-values from simulations
Error Rate 100%
Wrong
An industrial control check is as follows. Random samples are periodically gathered. If a particular statistic is significantly greater than what is expected during proper operation of the machinery, a recalibration is necessary. In a simulation of 100 such samples where the machinery is working properly, the resulting statistic is summarized in the following dotplot.
Calculations of the statistic from random samples when the machinery is operating properly
Suppose during one control check, the statistic from the random sample is 24. Is there sufficient evidence to necessitate a recalibration of the machinery?
A
No, because with the machine operating properly, the simulation gave two statistics even greater than 24
B
No, because stopping production to recalibrate is a serious, expensive decision, and given this data, the probability of a Type I error is too great
C
No, because the distribution of simulation results is roughly bell-shaped so the 68-95-99.7 rule applies
D
Yes, because the consequences of a Type II error are significant
Yes, because the estimated P-value is less than 0.05
Correct Answer
Explanation
The control check has hypotheses: \(H_0\): the machinery is operating properly, and \(H_a\): the machinery is not operating properly. In the simulation with the machine operating properly, there were 3 values out of 100 that were 24 or greater. This gives an estimated P-value of 0.03. With this small of a P-value, \(0.03 < 0.05\), there is sufficient evidence to reject \(H_0\). In other words, there is sufficient evidence to necessitate a recalibration of the machinery.
**Chi-Square Tests (pages 177–187)**