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alexira [117]
3 years ago
9

Twenty students from Sherman High School were accepted at Wallaby University. Of

Mathematics
1 answer:
nydimaria [60]3 years ago
6 0

Answer:

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

Step-by-step 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

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