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 x¯1=508, s1=120, n1=300 and x¯2=422, s2=94, n2=200

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 = Enter your answer; Best estimate

Margin of error = Enter your answer; Margin of error

Confidence interval : Enter your answer; Confidence interval, value 1 to Enter your answer; Confidence interval, value 2

Save for Later

  

Answer 1

We need to construct the 95% confidence interval for the difference between the population means μ1​−μ2​, for the case that the population standard deviations are not known. The following information has been provided about each of the samples:

Sample Mean 1	508
Sample Standard Deviation 1	120
Sample Size 1	300
Sample Mean 2	422
Sample Standard Deviation 2	94
Sample Size 2	200

Based on the information provided, we assume that the population variances are unequal, so then the number of degrees of freedom is computed as follows:

The critical value for α=0.05 and df = 487.318 degrees of freedom is t_c = 1.965 . 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:

Now, we finally compute the confidence interval:

CI = (67.135, 104.865)