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 a = 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. (5 points) (10
points)

Answer 1

Data provide convincing evidence of a difference in SAT scores between students with and without a military scholarship is explained below in details.

Explanation:

This is a quiz of 2 autonomous groups. The population model differences are not understood. it is a two-tailed examination. Let w be the index for scores of students with army research and o be the index for scores of students without army research.

Therefore, the population means would be μw and μo.

The irregular variable is x w - xo = variation in the sample mean records of students with military accomplishments and students without.

For students with military accomplishments,

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 recognized, we would analyis the examination statistic by using the t examination. 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 get the probability count from the t test calculator. It becomes

p value = 0.72

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

Therefore, these data do not present an acceptable indication of a difference in SAT scores between students with and without a military scholarship.

Part B

The formula for getting 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