Answer to Question #292198 in Statistics and Probability for Emu

Question

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

Answer 1

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 $(0.3431, 9.6569).$