Question

Another 503 students are selected at random from Florida. They are given a 3-hour preparation course before the test is administered. Their average score is 1019, with a standard deviation of 95. Suppose we want to test whether students in taking the 3-hour preparation course perform differently as the students who did not take the preparation course by testing the following hypothesis: H0: E(YFL,Prep) – E(YFL,NoPrep) = 0 vs H1: E(YFL,Prep) – E(YFL,NoPrep) ≠ 0Construct a 95% confidence interval for the change in average test score

associated with the prep course.

Answer 1

Answer: (1034.29, 1010.95)

Explanation:

Sample size (n) = 503

Average sample score (x) = 1019

Sample standard deviation (s) = 95

Even though our population standard deviation is unknown ( we only know of the sample standard deviation), the sample size is greater than 30, hence the critical value for the test will be that of a z test.

The 95% confidence level for population mean is given below as

u = x ± Zα/2 × s/√n

Where

Zα/2 = 1.96 = critical value for a two tailed test at a 5% level of significance ( the confidence level + level of significance = 100, hence a 95% confidence level corresponds to a 5% level of significance).

The upper limit of the interval is given as

u = x + Zα/2 × s/√n

Hence, we have that

u = 1019 + 1.96 × (95/√503)

u = 1019 + 1.96× (4.2358)

u = 1019 + 8.0481

u = 1034.29

The lower limit of the interval is given as

u = x - Zα/2 × s/√n

Hence, we have that

u = 1019 - 1.96 × (95/√503)

u = 1019 - 1.96× (4.2358)

u = 1019 - 8.0481

u = 1010.95

Hence, the 95% confidence level is (1034.29, 1010.95)