Question

a data set includes 103 body temperatures of healthy adult humans having a mean of 98.9°f and a standard deviation of 0.67°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: Calculate the degrees of freedom

Degrees of freedom \(df=n - 1\), where \(n = 103\). So \(df=103-1 = 102\).

Step2: Find the critical value \(t_{\alpha/2}\)

For a \(99\%\) confidence interval, \(\alpha=1 - 0.99=0.01\), and \(\alpha/2=0.005\). Using the t - distribution table (or a calculator with t - distribution functions), for \(df = 102\approx100\) (since 102 is close to 100 in the t - table), \(t_{\alpha/2}= 2.626\).

Step3: Calculate the margin of error \(E\)

The formula for the margin of error \(E=t_{\alpha/2}\times\frac{s}{\sqrt{n}}\), where \(s = 0.67\), \(n = 103\).

$$E=2.626\times\frac{0.67}{\sqrt{103}}$$

$$E=2.626\times\frac{0.67}{10.149}$$

$$E=2.626\times0.066$$

$$E\approx 0.173$$

Step4: Calculate the confidence interval

The confidence interval is \(\bar{x}-E<\mu<\bar{x}+E\), where \(\bar{x}=98.9\)

$$98.9 - 0.173<\mu<98.9+0.173$$

Answer:

\(98.727^{\circ}F<\mu<99.073^{\circ}F\)

Since \(98.6^{\circ}F\) is not in the confidence interval \( (98.727,99.073)\), the sample suggests that \(98.6^{\circ}F\) is not a reasonable value for the mean body temperature of all healthy humans.