A data set includes 109 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.54degreesF. Construct a 99​% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6degreesF as the mean body​ temperature?

What is the confidence interval estimate of the population mean u​?

Respuesta :

Answer:

The 99​% confidence interval estimate of the mean body temperature of all healthy humans is between 96.88ºF and 99.72ºF.

98.6ºF is part of the interval, so the sample suggests that the use of 98.6ºF as the mean body​ temperature is correct.

Step-by-step explanation:

We have the standard deviation for the sample. So we use the 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 = 36 - 1 = 35

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 108 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.99}{2} = 0.995[/tex]. So we have T = 2.63

The margin of error is:

M = T*s = 2.63*0.54 = 1.42

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 98.3 - 1.42 = 96.88ºF.

The upper end of the interval is the sample mean added to M. So it is 98.3 + 1.42 = 99.72ºF.

The 99​% confidence interval estimate of the mean body temperature of all healthy humans is between 96.88ºF and 99.72ºF.

98.6ºF is part of the interval, so the sample suggests that the use of 98.6ºF as the mean body​ temperature is correct.