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Question:

A leading optometrist claims that $40\%$ of people wear contact lenses. You decide to test the claim by surveying a sample of 12 students.

How many students in your sample would need to reply that they wear contact lenses for there to be significant evidence at the $5\%$ level that the true percentage is less than $40\%$?


$H_0: p = 0.4$

$f = \frac{X}{n} = \frac{X}{12}$

I need to find an $X$ such that $\hat{p}$ would be in the blue area indicated

enter image description here

From the $z$ test table I have a value of $z = 1.645$ would mean significance at a $5\%$ level

So

$$ 1.645 = \frac{\frac{X}{12} - 0.4}{\frac{\sqrt{0.4 (0.6)}}{\sqrt{12}}} $$

Solving this for $X$ gives

$$ 12\left(\frac{\sqrt{0.4 (0.6)}}{\sqrt{12}}( 1.645) + 0.4 \right) = X $$

Then $X \approx 7.59$, meaning that $8$ people would have to say that they wear contact lenses for there to be a significant difference to $H_0$.


This can't be correct - because $\frac{8}{12} > 0.4$ and we were expected to find how many would mean that the true percentage is lower than $40\%$. Having 8 people wear contacts would imply that the percentage is greater, not less.

I'm not sure how to approach this

Edit

As in the drawing a $z$ value of less than the mean is required, so $-1.645$ is required (rather than $1.645$)

Which gives

$$ 12\left(\frac{\sqrt{0.4 (0.6)}}{\sqrt{12}}( -1.645) + 0.4 \right) = X $$

Which gives a value of $X \approx 2.0083$

Which would mean that if only $2$ students stated they wore contacts, this would be significant evidence at a $5\%$ level that $H_0$ is false.

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A z-value of 1.645 is above the mean (5% of values lie above it and 95% of values lie below it).

You need (as shown in the drawing) a z-value that is below the mean, not one that is above the mean.

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  • $\begingroup$ ah yes... I've edited the post - does the reasoning make sense now? thanks $\endgroup$ – baxx Apr 27 '17 at 11:06
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    $\begingroup$ Well, yes the calculations themselves seem okay. However, the normal approximation is actually quite poor when $n$ is only $12$, but if you're meant to use that normal approximation down that far (rather than use an exact calculation) then you seem to be doing those calculations correctly. (A continuity correction to the normal approximation gets quite a bit closer to the exact probability, but the exact calculations aren't really any more work). $\endgroup$ – Glen_b Apr 27 '17 at 11:42

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