Question

Use the t-distribution to find a confidence interval for a difference in means μ1-μ2 given the relevant sample results. Give the best estimate for μ1-μ2, the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed. A 95% confidence interval for μ1-μ2 using the sample results x̅1 = 75.5, s1=11.5, n1=35 and x̅2=69.3, s2=7.0, n2=20 Enter the exact answer for the best estimate and round your answers for the margin of error and the confidence interval to two decimal places Best estimate =__6.2_________ Margin of error =______________ Confidence interval :_____ to _______

Answer 1

Solution :

The difference in sample means is the unbiased point estimate of the difference between two population means.

We have,  x̅1 = 75.5 and  x̅2 = 69.3

(x̅1 - x̅2) = (75.5 - 69.3) = 6.2

Hence, best estimate of μ1 - μ2 is 6.2.

The margin of error to estimate μ1 - μ2 at 95% confidence level is given as follows :

Where,

and t_{(0.05/2, n1 + n2 - 2)} is critical t value to construct 95% confidence interval for μ1 - μ2.

We have,  s1 = 11.5, n1 = 35, s2 = 7.0, and n2 = 20

Using t-table we get, t_{(0.05/2, n1 + n2 - 2)} = 2.0057

The margin of error is 5.69.

The 95% confidence interval for μ1-μ2 is given as follows :

Where, E is margin of error.

Confidence interval : 0.51 to 11.89