Question #146639
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.
1
Expert's answer
2020-12-01T02:18:37-0500

H0:σ2=7.22H_0:\sigma^2=7.2^2

Ha:σ2<7.22H_a:\sigma^2<7.2^2

This is a left-tailed test.

Calculate the test-statistic:


χ2=(n1)s2σ2=243.527.22=5.67\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(χ2<5.67)=0.000042P(\chi^2<5.67)=0.000042

α=0.05>0.000042\alpha=0.05>0.000042

Since α\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.


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