Problem

In order to compare the means of two populations, independent random samples of 395 observations are selected from each population, with the results found in the table to the right. Complete parts a through e below. \[ \begin{array}{ll} \hline \text { Sample 1 } & \text { Sample 2 } \\ \hline \bar{x}_{1}=5,318 & \bar{x}_{2}=5,284 \\ s_{1}=142 & s_{2}=190 \\ \hline \end{array} \] a. Use a $95 \%$ confidence interval to estimate the difference between the population means $\left(\mu_{1}-\mu_{2}\right)$. Interpret the confidence interval. The confidence interval is (Round to one decimal place as needed)

Solution

Step 1 :Given values are sample means \(\bar{x}_{1}=5318\) and \(\bar{x}_{2}=5284\), standard deviations \(s_{1}=142\) and \(s_{2}=190\), and sample sizes \(n_{1}=n_{2}=395\). The z-score for a 95% confidence level is 1.96.

Step 2 :Calculate the standard error (SE) using the formula \(SE = \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}\). Substituting the given values, we get \(SE = \sqrt{\frac{142^{2}}{395} + \frac{190^{2}}{395}} = 11.934844210508738\).

Step 3 :Calculate the confidence interval (CI) using the formula \(CI = (\bar{x}_{1} - \bar{x}_{2}) \pm z*SE\). Substituting the given values, we get \(CI = (5318 - 5284) \pm 1.96*11.934844210508738\).

Step 4 :Calculate the lower and upper bounds of the confidence interval. The lower bound is \(CI_{lower} = (5318 - 5284) - 1.96*11.934844210508738 = 10.607705347402874\) and the upper bound is \(CI_{upper} = (5318 - 5284) + 1.96*11.934844210508738 = 57.39229465259713\).

Step 5 :Final Answer: The 95% confidence interval for the difference between the population means \(\mu_{1} - \mu_{2}\) is \(\boxed{(10.6, 57.4)}\). This means we are 95% confident that the true difference between the population means falls within this interval.

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Source: https://solvelyapp.com/problems/17650/

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