With individual lines at its various windows, a post office finds that the standard deviation for waiting times for customers on Friday afternoon is 7.2 minutes. The post office experiments with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes on a Friday afternoon.

With a significance level of 5%, test the claim that a single line causes lower variation among waiting times for customers.

Id like to test the hypothesis that the waiting times for a single line are longer. For this problem, I get Chi critical value of 36.41502850. Which is much larger than noted in the solution. If Chi squared test statistic value is larger than the critical value, then the null hypothesis ( that the waiting times for singe line has a more variation could be rejected? ) What if one would like to determine if one has to wait longer than 7.2 minutes? Then I guess the appropriate F score would be 1-$/alpha$ quantile?


2 Answers 2


First you have to assume Normality...

Then your Hypothesis' System to be verified is the following

$$\begin{cases} \mathcal{H}_0: & \sigma^2=51.84 \\ \mathcal{H}_1: & \sigma^2<51.84 \end{cases}$$

thus your critical chi value at 5% is $13.8484$

As your Observed Statistic is $\frac{24\times3.5^2}{7.2^2}\approx5.67$ this means that you are at the left of your critical value: you reject $\mathcal{H}_0$ that is: there are enough evidences against the null hypothesis to agree with the claim that "a single line causes lower variation among waiting times for customers".

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  • $\begingroup$ How did you found the critical value? At 0.05 and df=24, I obtained much larger number. $\endgroup$
    – user
    Dec 2, 2020 at 16:26
  • $\begingroup$ @user : perhaps you calculated the right quantile..$\approx 36.41503$ $\endgroup$
    – tommik
    Dec 2, 2020 at 16:29
  • $\begingroup$ Perhaps you got confused by the labeling of the table and cut 5% of the area from the upper tail of the distribution. Remember that the mean of CHISQ(24) is 24. $\endgroup$
    – BruceET
    Dec 3, 2020 at 0:36

@Tomik (+1) has already indicated that $\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(\nu = 25-1 = 24).$ Notice that your problem gives standard deviations, not variances. You should already have enough hints to do do the test. So I just show output from a statistical software package that does this test without requiring the data instead of just the SD.

Output from a recent release of Minitab.

Test and CI for One Variance 


Null hypothesis         σ = 7.2
Alternative hypothesis  σ < 7.2

The chi-square method is only for the normal distribution.


 N  StDev  Variance
25   3.50      12.3

95% One-Sided Confidence Intervals

              for   Upper Bound
Method      StDev  for Variance
Chi-Square   4.61          21.2


Method      Statistic  DF  P-Value
Chi-Square       5.67  24    0.000

The value 13.85 cuts probability 0.05 from the lower tail of $\mathsf{Chisq}(\nu = 24).$ [You may be able to get that value from a printed chi-squared CDF table; or with software.] The P-value of your test is the area under the density curve of this distribution to the left of 5.67; it's 0.000042. (about 0). [This computation requires software.]


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