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 13 male applicants results in a SAT scoring mean of 1061 with a standard deviation of 27. A random sample of 9 female applicants results in a SAT scoring mean of 1026 with a standard deviation of 56 . Using this data, find the 98% 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.

Step 2 of 3: Find the standard error of the sampling distribution to be used in constructing the confidence interval. Round your answer to the nearest whole number.

Step 3 of 3: Construct the 98% confidence interval. Round your answers to the nearest whole number.

.....lower endpoint...upper endpoint...

Answer 1

Answer:

1.

Given,

s1 = 27 , n1 = 13

s2 = 56 , n2 = 9

let us consider,

significance level = n1 - 1 or n2 - 1[in unequal population variances]

significance level = 1 - 0.98

= 0.02

degree of freedom = n1 - 1

= 13 - 1

= 12

Critical value(two tailed) = t 0.02,12 = 2.681

2.

Now consider,

Standard error = sqrt(s1^2/n1 + s2^2/n2)

substitute values

= sqrt(27^2/13 + 56^2/9)

= sqrt(404.5213675)

= 20.113

SE = 20

3.

98% confidence interval is given as follows

= (x1bar - x2bar) +/- t*SE

substitute values

= (1061 - 1026) +/- 2.681*20

= 35 +/- 53.62

= (35 - 53.62 , 35 + 53.62)

= (-18.62 , 88.62)

= (- 19 , 89)

Upper limit = 89

Lower limit = - 19