Question

State legislators propose cutting funding to public universities. College administrators oppose this, claiming that across the state, freshmen enrollment has increased dramatically over the last decade. To substantiate their claim, they randomly sample 10 institutions. The administrators find that the average freshmen enrollment 10 years ago was 934.1 with a standard deviation of 120.52 (sample 1). They independently sample another IO institutions and find that the average freshmen enrollment is currently 1302.2 witn a standard deviation of 178.6 (sample 2). Assuming both populations are normal and have equal variance, construct a 98% confidence interval on tne mean difference in freshmen enrollment at state public institutions over the last decade.

Answer 1

Explanation:

The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

Where

x1 = sample mean of sample 1

x2 = sample mean of sample 2

s1 = sample standard deviation for sample 1

s2 = sample standard deviation for sample 2

n1 = number of samples in sample 1

n2 = number of samples in sample 2

From the information given,

x1 = 934.1

s1 = 120.52

x2 = 1302.2

s2 = 178.6

For a 98% confidence interval, we would determine the z score from the t distribution table because the number of samples are small

Degree of freedom =

(n1 - 1) + (n2 - 1) = (10 - 1) + (10 - 1) = 18

z = 2.552

x1 - x2 = 934.1 - 1302.2 = -368.1

√(s1²/n1 + s2²/n2) = √(120.52²/10 + 178.6²/10) = √(1452.50704 + 3189.796)

= 68.13

Margin of error = 2.552 × 68.13 = 173.87

Therefore, confidence interval =

-368.1 ± 173.87