Question

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

Answer 1

Answer:

$(16.2-14.1) -2.03 \sqrt{\frac{4.5^2}{19} +\frac{3.1^2}{19}} = -0.445$

$(16.2-14.1) +2.03 \sqrt{\frac{4.5^2}{19} +\frac{3.1^2}{19}} = 4.645$

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".

$n_1 = 19$ sample size 1

$n_2 = 19$ sample size 2

$\\bar X_1 = 16.2$ sample mean for group 1

$\\bar X_2 = 14.2$ sample mean for group 2

$s_1 = 4.5$ sample deviation for group 1

$s_2 = 3.1$ sample deviation for group 2

Solution to the problem

For this case the confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

And the degrees of freedom are given by:

$df = n_1 +n_2 -2 = 19+19 -2= 36$

We want a 95% of confidence o then the significance level is 1-0.95 =0.05 and $\alpha/2 = 0.025$ if we find a critical value in the t distribution with 36 degrees of freedom we got:

$t_{cric} =2.03$

And replacing we got:

$(16.2-14.1) -2.03 \sqrt{\frac{4.5^2}{19} +\frac{3.1^2}{19}} = -0.445$

$(16.2-14.1) +2.03 \sqrt{\frac{4.5^2}{19} +\frac{3.1^2}{19}} = 4.645$