Question

Give a 90% confidence interval, for Hi - p2 given the following information. ni = 25, 71 = 2.15, $1 = 0.78 n2 = 55, 12 = 2.28, 82 = 0.75 + Use Technology. Rounded to 2 decimal places.

Answer 1

We need to construct the 90% 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	2.15
Sample Standard Deviation 1	0.78
Sample Size 1	25
Sample Mean 2	2.28
Sample Standard Deviation 2	0.75
Sample Size 2	55

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:

$df = \frac{ (s_1^2/n_1 +s_2^2/n_2)^2 }{ \frac{(s_1^2/n_1)^2}{n_1+1} + \frac{(s_2^2/n_2)^2}{n_2+1}} - 2 = 46.471$

The critical value for α=0.1 and df = 46.471 degrees of freedom is t_c = 1.678. 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:

$se= \displaystyle \sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} } =\displaystyle \sqrt{ \frac{0.78^2}{25} + \frac{0.75^2}{55}} = 0.186$

Now, we finally compute the confidence interval:

CI = (X1 - X, – te x se, Xi - X2 + te x se)

$CI= \displaystyle \left( 2.15 - 2.28 - 1.678 \times 0.186, 2.15 - 2.28 + 1.678 \times 0.186 \right)$

$CI= \displaystyle \left(-0.13 - 0.31 , -0.13 + 0.31 \right)$

$CI= \displaystyle \left(-0.13 +/- 0.31 \right)$