Assume that both populations are normally distributed. a) Test whether mu 1 not equals mu 2 at the alpha equals 0.05 level of significance for the given sample data. b) Construct a 95% confidence interval about mu 1 minus mu 2. Sample 1 Sample 2 n 19 19 x overbar 16.2 14.1 s 4.5 3.1

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

sample size 1

sample size 2

sample mean for group 1

sample mean for group 2

sample deviation for group 1

sample deviation for group 2

Solution to the problem

For this case the confidence interval is given by:

And the degrees of freedom are given by:

We want a 95% of confidence o then the significance level is 1-0.95 =0.05 and if we find a critical value in the t distribution with 36 degrees of freedom we got:

And replacing we got:

16

step-by-step explanation:

2 x 40 = 80

80/5

= 16

3

step-by-step explanation: