Question

US department of labor wanted to compare the average unemployment rate in two different states California and Washington. From California they took a sample of size 5 and found that sample mean 20.32 and sample standard deviation is 3.55. From Washington they took a sample of size 8 and found that sample mean 21.24 and sample standard deviation is 2.70. Find the critical value (use t-table) to compute 99% confidence interval of the true difference in population mean.   

Question 2 options:

	1.708
	2.575
	3.10
	2.060
	1.96

Answer 1

First we need to test the differences in variances between the two groups.

Test statistic . F =

Numerator df = n1 - 1= 5-1 = 4

Denominator df = n2 - 1= 8-1 = 7

Critical value of F at 0.01 significance level and df = 4, 7 is 7.85

Since the observed test statistic (1.73) is less than the critical value (7.85) , we fail to reject H0 and conclude that there is no significant evidence that there is a difference in variances among the groups.

Assume equality of variances, degree of freedom = n1 + n2 - 2 = 5 + 8 - 2 = 11

From t table, critical value of t at df = 11 for 99% confidence interval is,

3.10