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
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