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For (a), $H_0 = 0.75$ and $H_a \ne 0.75$.

For (b), how can I get an answer with a $z$-value? Why can't I use a $t$-value to answer this question? Is the idea: $p \pm Z (p(1-p)/N)^{0.5}$?

gung - Reinstate Monica
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sss
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  • How would you calculate the sample standard deviation for a t-statistic (and while that's possible, if non-obvious -- why would the result have a t-distribution)? Note that "infected" and "not-infected" are two states, so the sample observations are not drawn from a normal distribution. Please show your actual reasoning or an actual attempt at the question. Why do you think it's a Z -- are you working from some answers? (If so I suggest you put the answers away, and try to solve these without ever looking at the answers, by using the information you have been supplied in the subject) – Glen_b May 17 '16 at 00:46

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T-values are used for inference with means because the sample standard deviation has to be used as an estimate for a population standard deviation, rather than using the population standard deviation directly (this is generally unknown).

Proportions are a little different in that the standard deviation of the hypothetical curve with mean $ p $ can be calculated by $ \sqrt{\frac{p(1-p)}{n}} $. (This has to do with the Central Limit Theorem.) The z-score of your hypothesis test is calculated by $ \frac{\hat p - p}{\sqrt{\frac{p(1-p)}{n}}} = z$. You can use the $ z $ value you calculate to find a P-value and reject or fail to reject $ H_0 $

ZachTheRiah
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  • n/m. I see what you mean now. – ZachTheRiah May 17 '16 at 00:56
  • I calculated based on the equation that you gave to me. It was 1.686 while the answer is 1.756. I think it is not simply calculation mistake. Is there another equation for getting z value? anway, thank you for answering. – sss May 17 '16 at 01:04
  • @sss please show your working. I get $z=-1.756$ using Zach's formula. My guess is you used the sample proportion in the denominator; in Zach's formula $\hat{p}$ is the sample proportion and $p$ is the hypothesized proportion. Note there are no $\hat{\ }$'s over the $p$'s in the denominator – Glen_b May 17 '16 at 01:07
  • Please show any working in your question. Note the rest of my comment above (I was editing it as you replied, sorry). My guess was correct. The approach you took would also converge to a standard normal distribution as $n\to\infty$, but relies on an additional theorem over the CLT and would have a flipped sign compared to the direction of the difference you'd want to look at, which may be confusing to beginners. The usual version of the proportions test -- and the one you've apparently been taught -- uses the null hypothesis values for the standard error of the proportion. – Glen_b May 17 '16 at 01:11
  • I did like 0.75-0.716 / (0.716*0.284/500)^0.5 – sss May 17 '16 at 01:11
  • oh then I should put 0.716 for ^p! – sss May 17 '16 at 01:12
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    Remember, $ p $ is the population parameter (the thing you're hypothesizing; 0.75.) $ \hat p $ is the sample you took of your population (0.716) – ZachTheRiah May 17 '16 at 01:13
  • Oh, now I got -1.756. Then, according to p with ^ +- 1.756(0.750.25/500)^0.5, 0.682 < p < 0.75, 75% is out of range that H0 can be rejected? – sss May 17 '16 at 01:19
  • I am confused that how to calculate range of P from Zach's formula. Do I need to find the range of P from the table? – sss May 17 '16 at 01:21
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    If you're talking about a confidence interval (a range for $ p $), that is different than a hypothesis test. The OP asked if there is evidence, so all you need to do is a hypothesis test. – ZachTheRiah May 17 '16 at 01:55