Question

A simple random sample from a population with a normal distribution of 97 body temperatures has x=98.10 F and s=0.61 F. Construct 99​% confidence interval estimate of the standard deviation of body temperature of all healthy humans.

Answer 1

To construct a 99% confidence interval estimate of the standard deviation of body temperature, we can use the formula CI = (sqrt((n-1)*s^2)/sqrt(X1),sqrt((n-1)*s^2)/sqrt(X2)). Therefore answer (0.548, 0.674)

Step-by-step explanation:

To construct a 99% confidence interval estimate of the standard deviation of body temperature, we can use the formula:

CI = (sqrt((n-1)*s^2)/sqrt(X1),sqrt((n-1)*s^2)/sqrt(X2))

Where CI is the confidence interval, n is the sample size, s is the sample standard deviation, X1 is the lower critical value, and X2 is the upper critical value. Given that the sample size (n) is 97, the sample standard deviation (s) is 0.61, and the significance level is 0.01 (which corresponds to a 99% confidence level), we can find the critical values from the t-distribution table.

For a sample size of 97 and a significance level of 0.01, the lower critical t-value is -2.617 and the upper critical t-value is 2.617. Substituting these values into the formula, we get the confidence interval estimate of the standard deviation as:

CI = (sqrt((97-1)*0.61^2)/sqrt(-2.617), sqrt((97-1)*0.61^2)/sqrt(2.617)) = (0.548, 0.674)