Solution to Two groups of children were given visual acity tests, 1 was composed of 11 children … - Sikademy
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Archangel Macsika

Two groups of children were given visual acity tests, 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 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

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


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:




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



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)


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).

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