Question

a data set includes 103 body temperatures of healthy adult humans having a mean of 98.9°f and a standard deviation of 0.65°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 the t distribution table. click here to view critical t values 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

The sample size \(n = 103\), so the degrees of freedom \(df=n - 1=103-1 = 102\). For a \(99\%\) confidence interval, \(\alpha=1 - 0.99=0.01\), and \(\frac{\alpha}{2}=0.005\). Using the t - distribution table (or a calculator with t - distribution functions), the critical value \(t_{\frac{\alpha}{2}}\approx 2.626\) (approximate value for \(df = 102\), when \(df\) is large, we can also use the standard normal approximation for critical values, but for a more accurate result with t - distribution).

Step2: Calculate the margin of error

The formula for the margin of error \(E=t_{\frac{\alpha}{2}}\times\frac{s}{\sqrt{n}}\), where \(s = 0.65\), \(n = 103\), and \(t_{\frac{\alpha}{2}}\approx2.626\).

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Step3: Calculate the confidence interval

The sample mean \(\bar{x}=98.9\). The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x} + E\).

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Answer:

\(98.732^{\circ}F<\mu<99.068^{\circ}F\)