Question

a data set includes 106 body temperatures of healthy adult humans having a mean of 98.9°f and a standard deviation of 0.63°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.)
what does this suggest about the use of 98.6°f as the mean body temperature?
a. this suggests that the mean body temperature is higher than 98.6°f.
b. this suggests that the mean body temperature is lower than 98.6°f.
c. this suggests that the mean body temperature could very possibly be 98.6°f.

Explanation:

Step1: Determine the critical value

Since the sample size \(n = 106\) (large, \(n>30\)), we can use the standard normal distribution. For a \(99\%\) confidence interval, \(\alpha=1 - 0.99=0.01\), and \(\alpha/2=0.005\). From the standard - normal distribution table, \(z_{\alpha/2}=z_{0.005} = 2.576\).

Step2: Calculate the margin of error

The formula for the margin of error \(E\) is \(E = z_{\alpha/2}\frac{s}{\sqrt{n}}\). Given \(s = 0.63\), \(n = 106\), and \(z_{\alpha/2}=2.576\).

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Step3: Calculate the confidence interval

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

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Step4: Analyze the value \(98.6^{\circ}F\)

Since \(98.6<98.742\), the value \(98.6^{\circ}F\) is not in the confidence interval \((98.742,99.058)\)

Answer:

\(98.742^{\circ}F<\mu<99.058^{\circ}F\)

A. This suggests that the mean body temperature is higher than \(98.6^{\circ}F\)