Two AP Stats Questions

Question

QUESTION 1: Twenty students from Sherman High School were accepted at Wallaby University. Of those students, eight were offered military scholarships and 12 were not. The newly accepted student SAT scores are shown here.

Military scholarship: 850, 925, 980, 1080, 1200, 1220, 1240, 1300

No military scholarship: 820, 850, 980, 1010, 1020, 1080, 1100, 1120, 1120, 1200, 1220, 1330

Part A: Do these data provide convincing evidence of a difference in SAT scores between students with and without a military scholarship? Carry out an appropriate test at the α = 0.05 significance level.

Part B: Create and interpret a 95% confidence interval for the difference in SAT scores between students with and without a military scholarship.

QUESTION 2: Researchers fed mice a specific amount of Aldrin, a poisonous pesticide, and studied their nervous systems to find out why Aldrin causes convulsions. The absolute refractory period, time required for nerves to recover after a stimulus, was measured and varies Normally. The measurements, in milliseconds, for five mice were 2.1, 2.3, 2.3, 2.4, and 2.5.

Part A: Find the mean refractory period and the standard error of the mean.

Part B: Calculate a 95% confidence interval for the mean absolute refractory period for all mice when subjected to the same treatment.

Part C: Suppose the mean absolute refractory period for unpoisoned mice is known to be 2.1 milliseconds. Aldrin poisoning should slow nerve recovery and therefore increase this period. Do the data give good evidence to support this theory? What can you conclude from a hypothesis test? Justify your response with statistical reasoning.

Answer 1

Question 1:

Compute the mean (MEAN1, MEAN2) and standard deviation (SD1, SD2) for both samples of size (N1, N2).

Ho: means are equal

Ha: means are not equal

this is to be evaluated by a two sided t-test.

calculated t statistic (tcalc) = (mean1 - mean2)/sqrt(SD1^2/N1 + SD^2/N2)

the degrees of freedom is min(N1-1,N2-1) = 7

critical value (tcrit) for test is t score for 0.025 tail probability and 7 DOF = 2.365

if |tcalc| > tcrit, then reject Ho.

95% CI is (M1 - M2) +/- tcrit * sqrt(SD1^2/N1 + SD^2/N2)

Question 2:

Find the mean (MEAN) and standard deviation (SD) of the data.

Standard error (SE) is the standard deviation/sqrt(sample size)

the degrees of freedom for the single same = 5 - 1 = 4

critical value (tcrit) for test is t score for 0.025 tail probability and 4 DOF = 2.776

95% CI is MEAN +/- tcrit * SE

Since the minimum period is 2.1 the data would appear to support that the period is increased.

Ho: mean = 2.1

Ha: mean > 2.1

this is a one-sided (right tail) test

t-calc = (MEAN - 2.1)/SE

critical value (tcrit) for test is t score for 0.05 tail probability and 4 DOF = 2.132

if |tcalc| > tcrit, then reject Ho