Answer to Question #146639 in Statistics and Probability for Azlan

Question

With individual lines at its various windows, a post office founds that the standard deviation for normally distributed waiting times for customers on Friday afternoon is 7.2 minutes. The post office experiments with a single main waiting line and founds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes. With a
significance level of 5%, test the claim that a single line causes lower variation among waiting times (shorter waiting times) for customers.

Answer 1

$H_0:\sigma^2=7.2^2$

$H_a:\sigma^2<7.2^2$

This is a left-tailed test.

Calculate the test-statistic:

$$\chi^2=\frac{(n-1)s^2}{\sigma^2}=\frac{24\cdot3.5^2}{7.2^2}=5.67$$

where s = 3.5, n = 25.

df = 25 – 1 = 24

Probability statement: p-value = $P(\chi^2<5.67)=0.000042$

$\alpha=0.05>0.000042$

Since $\alpha$ > p-value we reject the null hypothesis.

At the 5% significance level the data do provide sufficient evidence to conclude that a single line causes a lower variation among the waiting times.

We are 95% confident to conclude that with a single line, the customer waiting times vary less than 7.2 minutes.