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 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 \(n = 105\) (large, \(n>30\)), we can use the standard normal distribution. For a \(99\%\) confidence interval, the significance level \(\alpha=1 - 0.99=0.01\), and \(\alpha/2=0.005\).
From the standard - normal distribution table, \(z_{\alpha/2}=z_{0.005} = 2.576\)

Step2: Calculate the margin of error \(E\)

The formula for the margin of error \(E\) is \(E = z_{\alpha/2}\frac{s}{\sqrt{n}}\), where \(s = 0.64\), \(n = 105\)

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Step3: Calculate the confidence interval

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

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Answer:

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

Since \(98.6^{\circ}F\) is within the confidence interval \(98.539^{\circ}F<\mu<98.861^{\circ}F\), the sample does not provide sufficient evidence to reject the use of \(98.6^{\circ}F\) as the mean body temperature.