Question

Given two independent random samples with the following results: n1=18x‾1=141s1=13   n2=12x‾2=161s2=12

Use this data to find the 98% 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.

Step 2 of 3 : Find the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places.

Answer 1

Given ,

Sample 1 :

n1 = 18 ,   = 141 , s₁ = 13

Sample 2 :

n2 = 12 ,   = 161 , s₁ = 12

We have to find 98% confidence interval for the true difference between the population means.

Assume that the population variances are not equal.

We have to use t distribution because population standard deviations are not known.

Formula :

Where E is margin of error.

Where , is two tailed t critical value at given confidence level

Given confidence level = 98% = 0.98

Significance level = = 1- 0.98 = 0.02 , / 2 = 0.01

Now let's find degrees of freedom ( df ).

Since population variances are not equal and unknown , we need to use following formula to find df.

So critical value is ,

{ Using Excel ,   =TINV( , df ) , the function returns two tailed inverse of t distribution

=TINV(0.02 , 25.03160776 ) = 2.485107175 }

So margin of error is ,

Now , let's find confidence interval.

The 98% confidence interval for the true difference between the population means is ,