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: (1034.29, 1010.95)

Step-by-step 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)