Question

The admissions officer at a small college compares the scores on the Scholastic Aptitude Test (SAT) for the school's male and female applicants. A random sample of 12 male applicants results in a SAT scoring mean of 1053 with a standard deviation of 30. A random sample of 18 female applicants results in a SAT scoring mean of 1155 with a standard deviation of 42. Using this data, find the 90% confidence interval for the true mean difference between the scoring mean for male applicants and female applicants. Assume that the population variances are not equal and that the two populations are normally distributed. Step 1 of 3 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

Formula for Confidence Interval for Difference in two Population means when population Standard deviation are not known and

the population variances are not equal is assumed

(11 – 12) = (a/2,4) XV

Degrees of Freedom: A- [(si/ni) + (sỉ/n2) + 1 ni-1 ny-1

$\alpha$ for 90% confidence level = (100-90)/100 =0.10

$\alpha$/2 = 0.10/2 =0.05

Sample 1 : Male

Number of males in the sample : n₁ = 12

Sample  SAT scoring mean for male : $\overline{x}_{1}$ = 1053

Sample standard deviation for male : s₁ = 30

Sample 2 : Female

Number of females in the sample : n₂ = 18

Sample  SAT scoring mean for female : $\overline{x}_{2}$ = 1155

Sample standard deviation for male : s₂ = 42

Step 1 of 3 : Find the critical value that should be used in constructing the confidence interval

Degrees of Freedom: A = [(sínı) + (s/n2)) [(302/12) + (422/18) (302/12)2 (422/18) 2 12- 1 7 18-1 n-1 + n2-1 [(75) + (98) 511.3636 + 564.9412 29929 1076.3048 m = 27.807227

Critical value :

ta/2,0 = 70.05.27 = 1.7033

critical value that should be used in constructing the confidence interval = 1.703.

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90% confidence interval for the true mean difference between the scoring mean for male applicants and female applicants

+ (11-12) Eta/2.01 V ni 302 422 = (1053 – 1155) +1.703317 +10 = -102 +1.7033 x 13.1529 = -102 – 22.4034= (-124.4034, -79.5966)

90% confidence interval for the true mean difference between the scoring mean for male applicants and female applicants

= (-124.4034, -79.5966)