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)

The following null and alternative hypotheses need to be tested:

$H_0:\sigma^2\leq 3.4^2$

$H_1:\sigma^2>3.4^2$

This corresponds to a right-tailed test, for which a Chi-Square test for a single population variance will be used.

$$\chi^2=\dfrac{(n-1)s^2 }{\sigma^2}$$

Test of a single variance statistic where:

$n=30$ is sample size

$s=4.1$ minutes is sample standard deviation

$\sigma=3.4$ minutes is population standard deviation

$df=30-1$ degrees of freedom

$$\chi^2=\dfrac{(30-1)(4.1)^2 }{(3.4)^2}\approx42.1704$$

b) The p-value for right-tailed test, $df=30-1=29$ degrees of freedom,

$\chi^2=42.1704$ is $p= 0.054209.$

c) The significance level is $\alpha = 0.05.$

Since p-value is $p=0.054209>0.05=\alpha,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population variance $\sigma^2$ is greater than $3.4^2,$ at the $\alpha=0.05$ significance level.