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 86 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 answer to 2 decimal places).

Answer 1

n1 = 45

= 86

s1^2 = 112.5

n2 = 36

= 88

s2^2 = 54

Claim: There is difference in the abilities of students enrolled in statistics today and those enrolled five years ago.

The null and alternative hypothesis is

For doing this test first we have to check the two groups have population variances are equal or not.

Null and alternative hypothesis is

Test statistic is

F = largest sample variance / Smallest sample variances

F = 112.5 / 54 = 2.083

Degrees of freedom => n1 - 1 , n2 - 1 => 45 - 1 , 36 - 1 => 44 , 35

Critical value = 1.607     ( Using f table)

Critical value < test statistic so we reject null hypothesis.

Conclusion: The population variances are not equal.

So we have to use here unpooled variance.

Test statistic is

Degrees of freedom = Min( n1 - 1 , n2 - 1 ) = Min(45 - 1 , 36 - 1 ) = Min(44,35) = 35

P-value = 2*P(T < -1) = 2*0.1621 = 0.3242

P-value > 0.075 we fail to reject null hypothesis.

Conclusion:

There is no difference in the abilities of students enrolled in statistics today and those enrolled five years ago.