Question

A data set includes 103 body temperatures of healthy adult humans having a mean of 98.9degreesf and a standard deviation of 0.67degreesf. 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 degreesf as the mean body​ temperature?

Answer 1

Answer:

Step-by-step explanation:

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × s/√n

Where

s = sample standard deviation = 0.67

n = number of samples = 103

From the information given, the population standard deviation is unknown hence, we would use the t distribution to find the z score

In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 103 - 1 = 102

Since confidence level = 99% = 0.99, α = 1 - CL = 1 – 0.99 = 0.01

α/2 = 0.01/2 = 0.005

the area to the right of z0.005 is 0.005 and the area to the left of z0.005 is 1 - 0.005 = 0.995

Looking at the t distribution table,

z = 2.6249

Margin of error = 2.6249 × 0.67/√103

= 0.173

Confidence interval = 98.6 ± 0.173

This suggests that the mean body temperature could very possibly be

98.6degrees°F.