Question

The admissions officer at a small college compares the scores on the Scholastic Aptitude Test (SAT) for the school's in-state and out-of-state applicants. A random sample of 8 in-state applicants results in a SAT scoring mean of 1048 with a standard deviation of 44. A random sample of 1616 out-of-state applicants results in a SAT scoring mean of 1147 with a standard deviation of 43. Using this data, find the 98% confidence interval for the true mean difference between the scoring mean for in-state applicants and out-of-state applicants. Assume that the population variances are not equal and that the two populations are normally distributed.

Step 2 of 3 : Find the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places.

Answer 1

Here we have given that

n1=number of in-state applicants  = 8

=Mean SAT score of in-state applicants = 1147

S1=sample standard deviation of in-state applicants =43

n2=number of in-state applicants  = 1616

=Mean SAT score of in-state applicants = 1147

S2=sample standard deviation of in-state applicants =43

Here, We assume that the population variances are not equal and that the two populations are normally distributed.

Now, We want to find he 98 % confidence interval for difference in the two population means

Formula is as follows

first we find the t-critical value,

Degrees of freedom =min(n1-1, n2-1)= min(7,1615)= 7

C=confidence level =0.98

=level of significance= 1-c = 1-0.98 =0.02

We get

T-critical=2.998   (using EXCEL=TINV(Probability=0.02, D.F=7)

Now we get the 95% confidence interval for the difference in the two population mean

i.e. here we get E=margin of error= =46.747302

we get 98% CI is as follows,

Lower limit= -145.747

Upper limit= -52.253

Interpretation:

This interval shows that we are 98% confident that the difference in the population mean will falls within that interval.