Fifteen students from Poppy High School were accepted at Branch University. Of those students, six were offered academic scholarships and nine were not. Mrs. Bergen believes Branch University may be accepting students with lower ACT scores if they have an academic scholarship. The newly accepted student ACT scores are shown here.

Academic scholarship: 25, 24, 23, 21, 22, 20
No academic scholarship: 23, 25, 30, 32, 29, 26, 27, 29, 27

Part A: Do these data provide convincing evidence of a difference in ACT scores between students with and without an academic scholarship? Carry out an appropriate test at the α = 0.02 significance level. (5 points)

Part B: Create and interpret a 98% confidence interval for the difference in the ACT scores between students with and without an academic scholarship. (5 points)

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.