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Use the $t$-distribution to find a confidence interval for a difference in means $\mu_{1}-\mu_{2}$ given the relevant sample results. Give the best estimate for $\mu_{1}-\mu_{2}$, the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed.
A $90 \%$ confidence interval for $\mu_{1}-\mu_{2}$ using the sample results $\bar{x}_{1}=76.8, s_{1}=11.3, n_{1}=25$ and $\bar{x}_{2}=62.2, s_{2}=8.0, n_{2}=20$
Enter the exact answer for the best estimate and round your answers for the margin of error and the confidence interval to two decimal places.
Best estimate $=$
Margin of error $=$
Confidence interval : to
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Step 1 ：Given the sample results are \(\bar{x}_{1}=76.8, s_{1}=11.3, n_{1}=25\) and \(\bar{x}_{2}=62.2, s_{2}=8.0, n_{2}=20\).

Step 2 ：The best estimate for the difference in means \(\mu_{1}-\mu_{2}\) is the difference in sample means \(\bar{x}_{1}-\bar{x}_{2}\), which is \(76.8 - 62.2 = 14.6\).

Step 3 ：The margin of error can be calculated using the formula for the confidence interval of the difference in means for a t-distribution, which is \(t_{\alpha/2} \cdot \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}\). Here, \(t_{\alpha/2}\) is the t-score for a 90% confidence interval, which we can find using a t-distribution table or a statistical function. In this case, the t-score is approximately 1.729.

Step 4 ：Substituting the given values into the formula, we get the margin of error as \(1.729 \cdot \sqrt{\frac{11.3^{2}}{25} + \frac{8.0^{2}}{20}} \approx 4.98\).

Step 5 ：The confidence interval is then the best estimate plus or minus the margin of error. So, the 90% confidence interval for \(\mu_{1}-\mu_{2}\) is approximately from \(14.6 - 4.98 = 9.62\) to \(14.6 + 4.98 = 19.58\).

Step 6 ：Final Answer: The best estimate for \(\mu_{1}-\mu_{2}\) is \(\boxed{14.60}\). The margin of error is approximately \(\boxed{4.98}\). The 90% confidence interval for \(\mu_{1}-\mu_{2}\) is approximately from \(\boxed{9.62}\) to \(\boxed{19.58}\).