Answer to Question #281128 in Statistics and Probability for Nik

Question

The average waiting time in a doctor’s office varies. The standard deviation of waiting times in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s office has a standard deviation of waiting times of 4.1 minutes. One doctor believes the variance of waiting times is greater than originally thought. a) What is the test statistic? b) What is the p-value? c) What can you conclude at the 5% significance level?

Answer 1

a) Test statistic: $\chi^2=\frac{(n-1)s^2}{\sigma^2}=\frac{(30-1)*4.1^2}{3.4^2}=42.17.$

b) P-value: $p=P(\chi^2>42..17)=0.0542.$

c) Since the p-value is greater than 0.05, fail to reject the null hypothesis.

There is no sufficient evidence that the variance of waiting times is greater than originally thought.