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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: $p \pm Z (p(1-p)/N)^{0.5}$?

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  • $\begingroup$ 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) $\endgroup$
    – Glen_b
    May 17, 2016 at 0:46

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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 $

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  • $\begingroup$ n/m. I see what you mean now. $\endgroup$ May 17, 2016 at 0:56
  • $\begingroup$ 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. $\endgroup$
    – sss
    May 17, 2016 at 1:04
  • $\begingroup$ @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 $\endgroup$
    – Glen_b
    May 17, 2016 at 1:07
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    $\begingroup$ 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) $\endgroup$ May 17, 2016 at 1:13
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    $\begingroup$ 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. $\endgroup$ May 17, 2016 at 1:55

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