Question

Given two independent random samples with the following results: ni = 15 n2 = 13 Xi = 153 X2 = 114 $i = 19 S2 = 21 Use this data to find the 95 % confidence interval for the true difference between the population means. Assume that the population variances are equal and that the two populations are normally distributed. Copy Data Step 1 of 3: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

1.

Since , the population variances are equals.

Now , df=degrees of freedom=n1+n2-2=15+13-2=26

The pooled estimate is ,

$S_{p}=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}=\sqrt{\frac{(15-1)19^2+(13-1)21^2}{15+13-2}}=19.95$

Step 1 of 3 : The critical value is ,

$t_{df,\alpha/2}=t_{23,0.05/2}=2.056$ ; From t-table

Step 2 of 3 : The standard error is ,

$SE=S_{p}\sqrt{\frac{/1}{n_1}+\frac{1}{n_2}}=19.95*\sqrt{\frac{1}{15}+\frac{1}{13}}=7.56\approx 8$

Step 3 of 3 : The 95% confidence interval is ,

$(\bar{X_1}-\bar{X_2})\pm t_{df,\alpha/2}*S_{p}\sqrt{\frac{/1}{n_1}+\frac{1}{n_2}}$

1 (153 - 114) 12.056 * 19.95 * + الا 13

$39\pm 15.54$