Question

a data set includes 109 body temperatures of healthy adult humans having a mean of 98.2°f and a standard deviation of 0.62°f. construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. what does the sample suggest about the use of 98.6°f as the mean body temperature? click here to view a t distribution table click here to view page 1 of the standard normal distribution table click here to view page 2 of the standard normal distribution table what is the confidence interval estimate of the population mean μ? °f < μ < °f (round to three decimal places as needed.)

Explanation:

Step1: Determine the critical value

Since the sample size \(n = 109\) (which is large, \(n>30\)), we can use the standard normal distribution (\(z -\)distribution) for a \(99\%\) confidence interval. The significance level \(\alpha=1 - 0.99=0.01\), and \(\alpha/2=0.005\). Looking up in the standard - normal distribution table, the \(z -\)score \(z_{\alpha/2}=z_{0.005}\approx2.576\)

Step2: Calculate the margin of error

The formula for the margin of error \(E\) is \(E = z_{\alpha/2}\frac{s}{\sqrt{n}}\), where the sample mean \(\bar{x}=98.2\), the sample standard deviation \(s = 0.62\), and \(n = 109\).

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Step3: Calculate the confidence interval

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

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Answer:

\(98.047^{\circ}F<\mu<98.353^{\circ}F\)