Two samples are taken with the following sample means, sizes, and standard deviations ¯x1x¯1 = 39 ¯x2x¯2 = 35 n1n1 = 74 n2n2 = 63 s1s1 = 4 s2s2 = 5 Find a 86% confidence interval, round answers to the nearest hundredth. ____________ < μ1−μ2μ1-μ2 <________________

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Two samples are taken with the following sample means, sizes, and standard deviations ¯x1x¯1 = 39 ¯x2x¯2 = 35 n1n1 = 74 n2n2 = 63 s1s1 = 4 s2s2 = 5 Find a 86% confidence interval, round answers to the nearest hundredth. ____________ < μ1−μ2μ1-μ2 <________________

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We need to construct the confidence interval for the difference between the population means , 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 (\bar X_1) = 39
Sample Standard Deviation 1 (s_1) = 4
Sample Size 1 (N_1) = 74
Sample Mean 2 (\bar X_2) = 35
Sample Standard Deviation 2 (s_2) = 5
Sample Size 2 (N_2) = 63

Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are df = n_1 + n_2 -2 = 74 + 63 - 2 = 135

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The critical value for and  degrees of freedom is .

\begin{array}{ccl} s_p &amp; = &amp; \displaystyle \sqrt{ \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2} } \\ \\ &amp; = &amp; \displaystyle \sqrt{ \frac{(74-1)\times 4^2 + (63-1) \times 5^2}{74 + 63 - 2} } \\ \\ &amp; = &amp; 4.487 \end{array}

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