Question

A data set includes 106 body temperatures of healthy adult humans having a mean of 98.9 degrees F and a standard deviation of 0.62 degrees 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 degrees F as the mean body temperature? What is the confidence interval estimate of the population mean? Round to three decimal places as needed.

Answer 1

Answer:

Step-by-step explanation:

Sample mean x bar = 98.9

Std dev s = 0.62

Sample size n = 106

Std error = $\frac{s}{\sqrt{n} } =0.0602$

Since population std deviation not known we can use t critical value.

t critical for 99% with df = 105 is 1.98

Margin of error = ±1.98(0.0602) =±0.1192

Confidence interval for sample mean = (98.9±0.1192)

=(98.781, 99.019)

2) Sample mean confidence interval suggests that sample mean is different  from the population since 98.6 is not contained in the confidence interval.

3) Population interval confidence mean would be

98.6±0.1192

=(98.4808, 98.7192)