Question

Given two independent random samples with the following results: ni = 18 12 = 10 y = 164 12 = 136 Si = 18 $2 = 32 Use this data to find the 99 % confidence interval for the true difference between the population means. Assume that the population variances are not equal and that the two populations are normally distributed. Copy Data Step 1 of 3: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

Solution :

=> The 99% confidence interval of the true difference between the population means is (−5.078,61.078)

Explanation :-

Given that n1 = 18, x1-bar = 164 , s1 = 18

n2 = 10 , x2-bar = 136 , s2 = 32

Based on the informationa provided, we assume that the population variances are unequal, so then the number of degrees of freedom is computed as follows: df = 11121131 - 2 = 12.94 _ (si/n +33/72) En + / + TB+1 The critical value for a = 0.01 and df = 12.94 degrees of freedom is to = ti-a/-1 =3.015. The corresponding confidence interval is computed as shown below: Since we assume that the population variances are unequal, the standard error is computed as follows: 332 1 + 2 V 721 722 = 192 322 1 19 + 10 10.973 = Now, we finally compute the confidence interval: CI = (1-- te x se, ž1 - 2 + te x se) = (164 - 136 - 3.015 x 10.973, 164 - 136 +3.015 X 10.973) = (-5.079, 61.079) Therefore, based on the data provided, the 99% confidence interval for the difference between the population means Hi-Ha is -5.078 <H1 -1 < 61.078, which indicates that we are 99% confident that the true difference between population means is contained by the interval (-5.078, 61.078).