Question

A statistics teacher wants to see if there is any difference in the abilities of students enrolled in statistics today and those enrolled five years ago. A sample of final examination scores from students enrolled today and from students enrolled five years ago was taken. You are given the following information. Today (Population 1) Five Years Ago (Population 2) Sample Mean 85 88 Sample Variance 112.5 54 Sample Size 45 36 The 92.50% confidence interval for the difference between the two population means is (Round your answers to 2 decimal places.)

Answer 1

answer)

As the population variance is unknown, we will use t table to estimate the interval.

First we need to estimate the degrees of freedom

Degrees of freedom is equal to smaller of n1-1,n2-1

Here smaller is 36

So degrees of freedom is 36

For df 36 and 92.5 % confidence level, critical value t is 1.835

Margin of error (MOE) = t*standard error

Standard error = √{(v1/n1)+(v2/n2)}

V1 = 112.5

V2 = 54

N1 = 45, n2 = 36

t = 1.835

MOE = 3.67

confidence interval is given by

(Mean 1 - Mean 2) - MOE < u < (Mean 1 - Mean 2) + MOE

(-3-3.67) < u < (-3+3.67)

(-6.67<u<0.67)