Question

A data set includes

103

body temperatures of healthy adult humans having a mean of

98.3°F

and a standard deviation of

0.73°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.

LOADING...

Click here to view page 1 of the standard normal distribution table.

LOADING...

Click here to view page 2 of the standard normal distribution table.

LOADING...

What is the confidence interval estimate of the population mean

μ​?

nothing°F<μ<nothing°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 :

degrees of freedom = n - 1 = 103 - 102

t/2,df = t_0.005,102 = 2.625

Margin of error = E = t_/2,df * (s /n)

= 2.625 * (0.73 / 103)

Margin of error = E = 0.189

The 99% confidence interval estimate of the population mean is,

- E < < + E

98.3 - 0.189 < < 98.3 + 0.189

( 98.111°F < < 98.489°F )

A.This suggests that the mean body temperature is significantly lower than 98.6°F.