Answer to Question #280725 in Statistics and Probability for Emir

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

Solution:

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 $p= 0.054209.$(Using chi-square table at df=29, p=0.054209).

Also, by using MINITAB, find chi-square variance test statistics with the help of following steps is:

1) Import the data.

2) Select the Stat and choose the Basic Statistics option.

3) Select the 1 Variance and choose variable option and put Sample in different columns.

4) Select Confidence level and Alternative hypothesis from Options.

5) Enter Hypothesized Value.

6) Click Ok.

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.