Answer to Question #279120 in Statistics and Probability for Syaa

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

$H_0: \sigma = 7.20 \\ H_1: \sigma < 7.20$

The chi-square variance test statistic

$χ^2 = \frac{(n-1) \times s^2}{\sigma^2} \\ = \frac{(25-1) \times 3.50^2}{7.20^2} \\ = 5.671 \\ d.f. = n-1 \\ = 25-1 \\ = 24$

From the chi-square distribution table, for the left-tailed test, the critical value is 13.848 at a 0.95 significance level.

The chi-square variance test statistics is fall in the rejection region, so the null hypothesis is rejected at a 5% level of significance. There is sufficient evidence to conclude that the standard deviation for normally distributed waiting times for customers on Friday afternoon is less than 7.20 minutes. The result is statistically significant.