Question

a data set includes 105 body temperatures of healthy adult humans having a mean of 98.7°f and a standard deviation of 0.64°f. construct a 99% confidence interval estimate of the mean body temperature of all healthy humans.
what is the confidence interval estimate of the population mean μ?
□°f < μ < □°f
(round to three decimal places as needed.)

Explanation:

Step1: Find the critical value

Since the sample size \(n = 105\) (\(n>30\)), we can use the standard normal distribution. For a \(99\%\) confidence interval, the significance level \(\alpha=1 - 0.99=0.01\), and \(\alpha/2=0.005\). The critical value \(z_{\alpha/2}\) is the \(z\) - score such that the area to the right of \(z_{\alpha/2}\) is \(0.005\). From the standard normal table, \(z_{\alpha/2}=2.576\).

Step2: Calculate the margin of error

The formula for the margin of error \(E\) is \(E = z_{\alpha/2}\times\frac{\sigma}{\sqrt{n}}\). Given \(\sigma = 0.64\), \(n = 105\), and \(z_{\alpha/2}=2.576\).

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Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu <\bar{x} + E\). Given \(\bar{x}=98.7\).

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Answer:

\(98.539^{\circ}F<\mu<98.861^{\circ}F\)