$\newcommand{\szdp}[1]{\!\left(#1\right)}$ Background Question: A manufacturer claimed that at least $20\%$ of the public preferred his product. A sample of $100$ persons is taken to check his claim. With $\alpha=0.05,$ how small would the sample percentage have to be before the claim could legitimately be refuted? (Notice that this would involve a one-tailed test of the hypothesis.) Note: This is Exercise 10.18 in Mathematical Statistics with Applications, 5th Ed., by Wackerly, Mendenhall, and Sheaffer.

Answer to Background Question: We would have $H_0: p=0.2, H_a: p<0.2.$ At the $\alpha=0.05$ level, the rejection region would be $z<-1.645.$ This would require: \begin{align*} z&<-1.645\\ \frac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}&<-1.645\\ \hat{p}-p_0&<-1.645\sqrt{\dfrac{p_0(1-p_0)}{n}}\\ \hat{p}&<p_0-1.645\sqrt{\dfrac{p_0(1-p_0)}{n}}\\ \hat{p}&<0.2-1.645\sqrt{\dfrac{0.2(0.8)}{100}}\\ \hat{p}&<0.1342. \end{align*}

Problem Statement: Refer to the Background Question. Calculate the value of $\beta$ for the alternative $p_a=0.15.$ Note: This is Exercise 10.27 in Mathematical Statistics with Applications, 5th. Ed., by Wackerly, Mendenhall, and Sheaffer.

My Work So Far: In the background question, we had $p_0=0.2,\; n=100.$ We found that the rejection region was $z<-2.575,$ corresponding to $\hat{p}<0.097.$ We have that \begin{align*} \beta &=P\szdp{\frac{\hat{p}-p_a}{\sqrt{p_0(1-p_0)/n}}\ge \frac{p_0-p_a}{\sqrt{p_0(1-p_0)/n}}}\\ &=P(z\ge 1.25)\\ &=0.1056. \end{align*}

My Question: The book does not have an example of calculating $\beta$ for a proportion, so I'm trying to go by analogy, and it's not working. The book's answer is $0.67.$ Where am I going wrong?

  • 2
    $\begingroup$ How did you obtain the criterion $Z \lt -2.575$ for the rejection region?? $\endgroup$
    – whuber
    Jul 28, 2021 at 19:31
  • $\begingroup$ @whuber: For $\alpha=0.05,$ a standard normal distribution in a one-sided test would have $1.645,$ not $2.575.$ Must have looked that up in the table wrong. Let's see: fixing that error in the background question would mean we would have to have $\hat{p}<0.1342.$ However, that doesn't affect the key calculation in the main problem, at least not the way I've written it. $\endgroup$ Jul 28, 2021 at 19:36
  • $\begingroup$ @Dave The required power is not specified. I have copied down both problem statements word-for-word. $\endgroup$ Jul 28, 2021 at 19:36
  • $\begingroup$ The critical region is unchanged, so it was crucial first to get it right. Now, given that the true probability is $0.15,$ what is the chance the test statistic will lie in that critical region? $\endgroup$
    – whuber
    Jul 28, 2021 at 19:37
  • $\begingroup$ @whuber I think I'm confused about terminology. What do you mean by "true probability"? We are supposed to calculate the probability that the test statistic will not lie in the rejection region, correct? $\endgroup$ Jul 28, 2021 at 19:42

1 Answer 1


I think I was plugging numbers in wrong, even before I corrected the $0.097$ to $0.1342.$ It should be \begin{align*} \beta &=P\!\left(\frac{\hat{p}-p_a}{\sqrt{p_a(1-p_a)/n}}\ge \frac{0.1342-0.15}{\sqrt{0.15(0.85)/100}}\right)\\ &=P(z\ge -0.448)\\ &=0.67 \end{align*}


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.