A sporting goods manufacturer claims that the variance of string tensions for any decent tennis racquet should be about 9 pounds. The string tensions of 18 randomly selected tennis racquets produced a variance of 8.13 pounds. Find the p-value to test the manufacturer's claim (assume the population is normally distributed).

I tried attempting the question as follows:
${H_o}: {\sigma ^2} = 9$
${H_1}: {\sigma ^2} \ne 9$
We are also given the following:
$s^2 = 8.13$
Test Statistics:
$\chi^2 = {{(n-1)S^2}\over{\sigma^2}}$
Calculating Test Statistics under $H_o$ gives us:
$\chi^2 = {{(18-1)8.13}\over 9} = 15.35667$

For calculating the p-value I proceeded as follows:
$p-value = 2\min\{P[\chi^2>15.35667],\ P[\chi^2<15.35667]\}\ =2\times 0.430192 = 0.860384$

I am not certain if my method for calculating the p-value is correct or not. Can someone please tell if the above process is valid?

Thanks in advance!

  • 1
    $\begingroup$ You can't have a variance of 8.13 pounds. Either you have a standard deviation of 8.13 pounds, or a variance of 8.13 pounds^2 . $\endgroup$
    – Gloomy
    Commented Jan 5, 2018 at 9:07

2 Answers 2


What you are dealing with in this question is a two-sided variance test, which is a specific case of a two-sided test with an asymmetric null distribution. The p-value is the total area under the null density for all values in the lower and upper tails of that density that are at least as "extreme" (i.e., at least as conducive to the alternative hypothesis) as the observed test statistic. Because this test has an asymmetric null distribution, we need to specify exactly what we mean by "extreme".

Lowest-density p-value calculation: The most sensible thing method of two-sided hypothesis testing is to interpret "more extreme" as meaning a lower value of the null density. This is the interpretation used in a standard likelihood-ratio (LR) test. Under this method , the p-value is the probability of falling in the "lowest density region", where the density cut-off is the density at the observed test statistic. With an asymmetric null distribution, this leads you to a p-value calculated with unequal tails.

To implement this method for the two-sided variance test, let $\text{Chi-Sq}(\cdot | n-1)$ denote the chi-squared density function with $n-1$ degrees-of-freedom. For a given test statistic $\chi_\text{obs}^2$, the p-value for the two-sided test is given by:

$$p(\chi_\text{obs}^2) = \mathbb{P} \Big( \text{Chi-Sq}( \chi^2 | n-1) \leqslant \text{Chi-Sq}( \chi_\text{obs}^2|n-1) \Big| H_0 \Big).$$

In your particular problem you have $n=18$ and $\chi_\text{obs}^2 = 15.35667$. At this observed value of the test statistic you have density $\text{Chi-Sq}( \chi_\text{obs}^2|n-1) = \text{Chi-Sq}( 15.35667|17) = 0.07188203$, and this density cut-off also occurs at the lower point $\chi^2=14.64890$. This means that your two-sided p-value can be calculated as:

$$\begin{equation} \begin{aligned} p(\chi_\text{obs}^2) &= \mathbb{P} \Big( \text{Chi-Sq}( \chi^2 | 17) \leqslant 0.07188203 \Big| H_0 \Big) \\[6pt] &= \mathbb{P} \Big( \chi^2 \leqslant 14.64890 \Big| H_0 \Big) + \mathbb{P} \Big( \chi^2 \geqslant 15.35667 \Big| H_0 \Big) \\[6pt] &= 0.3792454 + 0.5698078 \\[6pt] &= 0.9490532. \end{aligned} \end{equation}$$

This is a large p-value and so we would not reject the null hypothesis in this case. Hence, there is no significant evidence to falsify the null hypothesis that $\sigma^2 = 9$.

A common alternative to the above test is to use a simpler calculation that takes the smallest of the two tail areas from above and below the observed test statistic, and then doubles this value. This is often used as a "quick-and-nasty" approximation to the above method, since it is relatively simple and does not require the calculation of a density cut-off value. (This is the method you are applying in your question.) Under this method, the p-value is approximated as:

$$\hat{p}(\chi_\text{obs}^2) = 2 \cdot \min \Big( \mathbb{P}(\chi^2 \leqslant \chi_\text{obs}^2 | H_0), \mathbb{P}(\chi^2 \geqslant \chi_\text{obs}^2 | H_0) \Big).$$

In your particular case this is an odd calculation, since the observed test statistic is above the mode of the null density, but its left-tail is the smaller probability. Your calculation of this approximate p-value is correct, but it is notable that it is not a very good approximation in this case (and in general, this is not a very good way to get p-values for a two-sided test with an asymmetric null distribution).


  • $\begingroup$ Problem with this equal-density criterion is that it is not invariant under nonlinear transformation of the test statistic, so is essentially arbitrary $\endgroup$ Commented Oct 27, 2018 at 7:26
  • 1
    $\begingroup$ @Kjetil Since the test statistic has a meaning and interpretation, one might instead suppose that nonlinear transformations of it are arbitrary, rather than the equal-density criterion! $\endgroup$
    – whuber
    Commented May 10, 2019 at 15:25
  • $\begingroup$ Thank you for clarifying this, however finding both density cutoff does not looks like something trivial. Do you know if there is a numerical gold standard to do so ? and if by any chance it had already been implemented in a python library ? $\endgroup$
    – Tobbey
    Commented Sep 9, 2019 at 14:54

The formula you are using for p-value calculation is incorrect.

The right way to do it would be:

p-value = probability ($χ2$ <= value) with (n-2) degrees of freedom (chi-square distribution)

here value is 15.35667 here and (n-2) =16. Now identify the corresponding p-value using any tool that is available online or a table from a standard statistical text. Based on that, you can reach your conclusion.

  • 1
    $\begingroup$ Why did you take the degree of freedom as $(n-2)$? Doesn't ${{(n-1)S^2}\over {\sigma ^2}} {~} \chi ^2_{n-1}$ $\endgroup$ Commented Feb 14, 2016 at 4:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.