Two groups of children were given visual acity tests,group.group 1 was composed of 11 children who receive their health care from private physician.the mean score for this group was 26 with standard deviation of 5.group 2 was composed of 14 children who receive their health care from the health department and had an average score of 21 with a standard deviation of assume normally distributed population with equal variance calculate the 95%confidence interval

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

$F=\dfrac{s_1^2}{s_2^2}=\dfrac{5^2}{6^2}=0.694444$

The critical values for $df_1=11-1=10,df_2=14-1=13, \alpha=0.05$ are $F_L = 0.2791,$ $F_U=3.2497,$ and since $F = 0.694444,$ then the null hypothesis of equal variances is not rejected.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the degrees of freedom are $df=n_1-1+n_2-1 = 11-1+14-1=23,$ assuming that the population variances are equal.

Hence, it is found that the critical value for this two-tailed test for $\alpha=0.05,df=23$ degrees of freedom is $t_c=2.068658.$

Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows:

$s_p=\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}$

$=\sqrt{\dfrac{(11-1)5^2+(14-1)6^2}{11+14-2}}$

$=5.587253$

Since we assume that the population variances are equal, the standard error is computed as follows:

$se=s_p\sqrt{\dfrac{1}{n_1}+\dfrac{1}{n_2}}=5.587253\sqrt{\dfrac{1}{11}+\dfrac{1}{14}}$

$=2.251168$

Now, we finally compute the confidence interval:

$CI=(\bar{x_1}-\bar{x_2}-t_c\times se, \bar{x_1}-\bar{x_2}+t_c\times se)$

$=(26-21-2.068658\times 2.251168,$

$26-21+2.068658\times 2.251168)$

$=(0.3431,9.6569)$

Therefore, based on the data provided, the 95% confidence interval for the difference between the population means $\mu_1 - \mu_2$ is $0.343 1< \mu_1 - \mu_2 < 9.6569,$ which indicates that we are 95% confident that the true difference between population means is contained by the interval $(0.3431, 9.6569).$