Twenty students from Sherman High School were accepted at Wallaby University. Of those students, eight were offered military scholarships and 12 were not. Mr. Dory believes Wallaby University may be accepting students with lower SAT scores if they have a military scholarship. 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.

Summation(x - mean)² = (850 - 1099.375)^2 + (925 - 1099.375)^2 + (980 - 1099.375)^2 + (1080 - 1099.375)^2 + (1200 - 1099.375)^2 + (1220 - 1099.375)^2 + (1240 - 1099.375)^2 + (1300 - 1099.375)^2 = 191921.875

Standard deviation, s1 = √(191921.875/8

s1 = 154.89

For no military scholarship,

Mean, x2 = (820 + 850 + 980 + 1010 + 1020 + 1080 + 1100 + 1120 + 1120 + 1200 + 1220 + 1330)/12

x2 = 1070.83

Standard deviation = √(summation(x - mean)²/n

n2 = 12

Summation(x - mean)² = (820 - 1070.83)^2 + (850 - 1070.83)^2 + (980 - 1070.83)^2 + (1010 - 1070.83)^2 + (1020 - 1070.83)^2 + (1080 - 1070.83)^2 + (1100 - 1070.83)^2 + (1120 - 1070.83)^2 + (1120 - 1070.83)^2 + (1200 - 1070.83)^2 + (1220 - 1070.83)^2 + (1330 - 1070.83)^2 = 238091.6668

Standard deviation, s2 = =√(238091.6668/12

s2 = 140.86

Part A)

This is a test of 2 independent groups. The population standard deviations are not known. Let μ1 be the mean score of students with military scholarship and μ2 be the mean score of students without military scholarship.

The random variable is μ1 - μ2 = difference in the mean score between students with military scholarship and without military scholarship

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0

This is a two tailed test

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

t = (1099.375 - 1070.83)/√(154.89²/8 + 140.86²/12)

t = 0.42

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [154.89²/8 + 140.86²/12]²/[(1/8 - 1)(154.89²/8)² + (1/12 - 1)(140.86²/12)²] = 21644133.914878543/1533280.3458504018

df = 14

We would determine the probability value from the t test calculator. It becomes

p value = 0.68

Since alpha, 0.05 < than the p value, 0.68, then we would fail to reject the null hypothesis. Therefore, at a significance level of 5%, these data do not provide convincing evidence of a difference in SAT scores between students with and without a military scholarship

Part B)

The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

For a 95% confidence interval, we would determine the z score from the t distribution table because the number of samples are small

Degree of freedom =

(n1 - 1) + (n2 - 1) = (8 - 1) + (12 - 1) = 18

z = 2.101

x1 - x2 = 1099.375 - 1070.83 = 28.545

Margin of error = 2.101√(154.89²/8 + 140.86²/12) = 143.3

The 95% confidence interval is 28.545 ± 143.3