Question

A data set includes 104 body temperatures of healthy adult humans having a mean of 98.7degrees°F and a standard deviation of 0.66degrees°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 98.6degrees°F as the mean body​ temperature?

Answer 1

Solution:

Note that, Population standard deviation() is unknown..So we use t distribution.

Our aim is to construct 99% confidence interval.   

c = 0.99

= 1 - c = 1 - 0.99 = 0.01

  /2 = 0.01 2 = 0.005

Also, d.f = n - 1 = 104 - 1 = 103   

    =    =  _0.005,103 = 2.624

( use t table or t calculator to find this value..)

The margin of error is given by

E =  /2,d.f. * ( / n )

= 2.624 * (0.66 / 104)

= 0.1698

Now , confidence interval for mean() is given by:

( - E ) <   <  ( + E)

(98.7 - 0.1698)   <   <  (98.7 + 0.1698)

98.5302 <   < 98.8698

Required.99% confidence interval is (98.5302 , 98.8698)

Now , the value 98.6 is inside the interval.

So ,

there is no evidence to reject the claim that the  mean body temperature is 98.6 degrees°F