Question

Carl conducted an experiment to determine if the there is a difference in mean body temperature between men and women. He found that the mean body temperature for men in a sample of 100 men was 91.1 with a population standard deviation of 0.52 and the mean body temperature for women in a sample of 100 women was 97.6 with a population standard deviation of 0.45.
Assuming the population of body temperatures for men and women is normally distributed, calculate the 98% confidence interval and the margin of error for the mean body temperature for both men and women. Using complete sentences, expla

Answer 1

The \((1-\alpha)\times100\%\) confidence interval for a difference in means will have the form
\[\left((\bar{x}_1-\bar{x}_2)-Z_{\alpha/2}\sqrt{\frac{{\sigma_1}^2}{n_1}+\frac{{\sigma_2}^2}{n_2}},~(\bar{x}_1-\bar{x}_2)+Z_{\alpha/2}\sqrt{\frac{{\sigma_1}^2}{n_1}+\frac{{\sigma_2}^2}{n_2}}\right)\]
You want a 98% CI, so \(1-\alpha=.98~\iff~~\alpha=0.02~~\iff~~\dfrac{\alpha}{2}=0.01\).

The cutoff/critical \(Z\) value is \(Z_{\alpha/2}=2.33\). Everything else is given to you, so just plug in the relevant info.

The margin of error is the critical value times the standard error - that is, the \(Z_{\alpha/2}\sqrt{\cdots}\) expression.