Question

A data set includes 107 body temperatures of healthy adult humans having a mean of 98.7°F and a standard deviation of 0.72°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 u? °F<</°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 significantly lower than 98.6°F. B. This suggests that the mean body temperature could very possibly be 98.6°F. C. This suggests that the mean body temperature is significantly higher than 98.6°F.

Answer 1

solution:

given information as follows:

sample size = n = 107

sample mean = =98.7

X

sample standard deviation = S = 0.72

confidence level = 99% = 0.99

$\alpha$ = 1 - 0.99 = 0.01

since population standard deviation is not known so we use t distribution

degree of freedom = df = n-1 = 107 - 1 = 106

critical value of t =               (using t table)

to.012.106df 2.623

margin of error

$E=t_c*\frac{S}{\sqrt{n}}=2.623*\frac{0.72}{\sqrt{107}}=0.1826$

confidence interval

XIE

upper boundary of interval = 98.7 + 0.183 = 98.883

lower boundary = 98.7 - 0.183 = 98.517

so, confidence interval is

98.517°F <μ 98.883°F

b)

since mean temperature 98.6 degree F is contained in the interval so

B: this suggest that the mean body temperature could very possibly be 98.6 degree farenhite.