1
$\begingroup$

In a random sample of 86 athletes participating in the 2012 Summer Olympics, 63 were found to have sponsorship by private companies. Based on this sample, conduct a hypothesis test to see if there evidence to support the claim that the true proportion (among all athletes participating in the 2012 Summer Olympics) of those with private company sponsorship is greater than 70%. Which distribution is used to compute the p‐value, and what range does the p‐value lie in?

Among the answers I had confusion between the following options

a. Z distribution, p < 0.5 b. t distribution, p < 0.5

I believe that the statistic that needs to be used is the t stat, since we have a sample of the overall population and we are deriving the population sd from the sample sd. But the official answer stated is that a Z distribution is supposed to be used.

Can anyone please weigh in?

$\endgroup$
  • 1
    $\begingroup$ See stats.stackexchange.com/a/411707/919 for a discussion of the z test, t test (both of which are approximations) and tests based on the exact Binomial distribution. It also explains why the official answer is (a) even though (b) is perfectly correct. $\endgroup$ – whuber Jul 30 at 16:51
  • 1
    $\begingroup$ @whuber Thank you so much. That cleared my doubt. $\endgroup$ – Arindam Bose Jul 30 at 18:27
1
$\begingroup$

Sporty question. Why not use a binomial test for this purpose? You want to test whether your

$H_1: P > 0.70$

as opposed to the opposite

$H_0: P \leq 0.70$

The binomial test reasons on a fraction in a distribution, and it is exact.

Reference

S. Siegel, N.J. Castellan. Nonparametric statistics for the behavirioral sciences, McGraw-Hill, 1988, chapter 4.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for the response. It was indeed quite insightful. $\endgroup$ – Arindam Bose Jul 30 at 18:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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