A coin is tossed $40$ times. Define $T$ as the number of tails.

i) Define the region of rejection by $|T-20|\geq5.$ Calculate $\alpha,$ the significance level ---

$\displaystyle\alpha = P(y\leq15 \lor y\geq25)= 1- P(16 \leq y\leq24) = 1-\sum_{T=16}^{24}{{40\choose{T}}\left(\frac{1}{2}\right)^T\left(\frac{1}{2}\right)^{40-T}} \approx 0.154.$

ii) Let $T=26.$ Compute the p-value. ---

We have $\displaystyle \frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} = \frac{\frac{26}{40}-0.5}{\sqrt{\frac{0.5(1-0.5)}{40}}} \approx 1.897.$ We have $H_0: p = 0.5$ and $H_a: p \neq 0.5,$ since we are testing whether it is fair or not (the side it favors doesn't matter). Hence, the p-value is $2(\phi(1.897))=0.9426.$ This also entails, using the former significance level, that we fail to reject the null.

iii) Calculate $\beta,$ the probability of a type II error for $H_0: p=0.5$ and $H_a: p = 0.6.$ ---

We have $\displaystyle z = \frac{\phi^{-1}(0.5,\sqrt{\frac{0.5(1-0.5)}{40}}, 0.846) - 0.6}{\sqrt{\frac{0.5(1-0.5)}{40}}} \approx \frac{0.581-0.6}{0.079057}\approx 0.240333,$ where $\phi^{-1}$ is the inverse normal function, with parameters $\mu, s, $ and $p$ (probability), respectively. The area to the right of this is $0.405,$ which is $\beta.$

This is a homework problem that I'm a little unsure on. Specifically, I don't fully understand how to calculate the p-value for proportions in this two-tailed scenario. Also, I'm not sure if I correctly calculated $\beta.$ I don't know whether one must take into account the fact that it's (presumably) a two-tailed test. Any pointers/guidance would be greatly appreciated.

  • $\begingroup$ Since this is a homework question, you should add the [self-study] tag and make sure you read its wiki. $\endgroup$
    – Danica
    Apr 9, 2015 at 3:39

1 Answer 1


your method is correct but you need some correction

i) $\alpha = P(T\leq15 \lor T\geq25)= 1- P(16 \leq T\leq24) = 1-\sum_{T=16}^{24}{{40\choose{T}}\left(\frac{1}{2}\right)^T\left(\frac{1}{2}\right)^{40-T}} \approx 0.154.$

ii) $p~~value=2*P(T\ge26|p=0.5)\approx0.0807$ So,likely fail to reject.

iii) $\beta(type~II~error)=P(16\le T \le 24 | p=0.6)=\sum_{T=16}^{24}{{40\choose{T}}\left(0.6\right)^T\left(0.4\right)^{40-T}}\approx0.846$

  • $\begingroup$ The "correction" is wrong because it fails to include the possibilities $Y=15$ or $Y=25$, both of which are explicitly in the critical region. Normally that would be a minor issue, but (a) it changes the answer substantially and (b) it seems to be one of the key issues in this question. The computations of the p-value and power implicitly use a different statistical test (based on a Normal approximation rather than the exact Binomial distribution), which furthers the confusion. $\endgroup$
    – whuber
    Nov 9, 2017 at 13:45
  • $\begingroup$ @whuber thanks for your input, I have edited my answer. $\endgroup$ Nov 18, 2017 at 9:58

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.