Question

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. (5 points)

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

Answer 1

Answer:

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. it is a two-tailed test. Let w be the subscript for scores of students with military scholarship and o be the subscript for scores of students without military scholarship.

Therefore, the population means would be μw and μo.

The random variable is xw - xo = difference in the sample mean scores of students with military scholarships and students without.

For students with military scholarship,

n = 8

Mean = (850 + 925 + 980 + 1080 + 1200 + 1220 + 1240 + 1300)/8

Mean = 1099.375

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

Summation(x - mean) = (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 = √(191921.875/8 = 154.89

For students without military scholarship,

n = 12

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

Mean = 1073.83

Summation(x - mean) = (820 - 1073.83)^2 + (850 - 1073.83)^2 + (980 - 1073.83)^2 + (1010 - 1073.83)^2 + (1020 - 1073.83)^2 + (1080 - 1073.83)^2 + (1100 - 1073.83)^2 + (1120 - 1073.83)^2 + (1120 - 1073.83)^2 + (1200 - 1073.83)^2 + (1220 - 1073.83)^2 + (1330 - 1073.83)^2 = 238199.4268

Standard deviation = √(238199.4268/12 = 140.89

We would set up the hypothesis.

The null hypothesis is

H0 : μw = μo H0 : μw - μo = 0

The alternative hypothesis is

Ha : μw ≠ μo Ha : μw - μo ≠ 0

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

(xw - xo)/√(sw²/nw + so²/no)

From the information given,

xw = 1099.375

xo = 1073.83

sw = 154.89

so = 140.89

nw = 8

no = 12

t = (1099.375 - 1073.83)/√(154.89²/8 + 140.89²/12)

t = 0.37

The formula for determining the degree of freedom is

df = [sw²/nw + so²/no]²/(1/nw - 1)(sw²/nw)² + (1/no - 1)(so²/no)²

df = [154.89²/8 + 140.89²/12]²/(1/8 - 1)(154.89²/8)² + (1/12 - 1)(140.89²/12)² = 21650688.37/1533492.15

df = 14

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

p value = 0.72

Since the level of significance of 0.05 < the p value of 0.72, we would not reject the null hypothesis.

Therefore, 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

z = (xw - xo) ± z ×√(sw²/nw + so²/no)

For a 95% confidence interval, the z score is 1.96

xw - xo = 1099.375 - 1073.83 = 25.55

z√(sw²/nw + so²/no) = 1.96 × √(154.89²/8 + 140.89²/12) = 1.96 × √2998.86 + 1654.17)

= 133.7

The confidence interval is

25.55 ± 133.7