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 at 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? F<< °F (Round to three decimal places as needed.)

Answer 1

given data are,

sample mean ($\bar{x}$) = 98.9

sample sd (s) = 0.63

sample size (n) = 106

degrees of freedom = (n-1) = (106-1) = 105

t critical value for df = 105, 99 % confidence level , both tailed test be:-

[ from t distribution table ]

=2.6235

the 99 % confidence interval for population mean be:-

tt>

0.63 98.9 2.6235 106

(98.739, 99.061

the temperature 98.6 suggests that,

the mean body temperature could be higher than 98.6 degree F.

[ because we are 99% confident that population mean body temperature will be in the interval (98.739 , 99.061) ]

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Answer 2

98.742 degrees fahrenheit < u< 99.058 degrees fahrenheit