A data set includes 106106 body temperatures of healthy adult humans having a mean of 98.798.7degrees°F and a standard deviation of 0.660.66degrees°F. Construct a 9999% 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?

The 99% confidence interval estimate of the mean body temperature of all healthy humans is between 96.97 ºF and 100.43 ºF. This suggests that the use of 98.6 °F as the mean body temperature is consistent, because it is a part of the interval

Step-by-step explanation:

We have the standard deviation of the sample, so we use the students t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 106 - 1 = 105

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 105 degrees of freedom(y-axis) and a confidence level of ). So we have T = 2.6235

The margin of error is:

M = T*s = 2.6235*0.66 = 1.73.

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 98.7 - 1.73 = 96.97 ºF.

The upper end of the interval is the sample mean added to M. So it is 98.7 + 1.73 = 100.43 ºF.

The 99% confidence interval estimate of the mean body temperature of all healthy humans is between 96.97 ºF and 100.43 ºF. This suggests that the use of 98.6 °F as the mean body temperature is consistent, because it is a part of the interval

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