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Working on some math for school I came across the following exercise:

John is examining how large a proportion of the population want to buy a 
newly released phone model. The answers are given as the confidence interval
(0.3, 0.42) which has the confidence level 0.95. How many persons (n) were
asked?

According to the book, for a confidence level of 0.95 I should use the following formula to calculate the sample size:

$\bar{z}+1.96{\sigma \over \sqrt{n}}-(\bar{z}-1.96{\sigma \over \sqrt{n}})$

$\bar{z}+1.96{\sigma \over \sqrt{n}}-\bar{z}+1.96{\sigma \over \sqrt{n}} = 3.92{\sigma \over \sqrt{n}}$

This makes sense. As I am not given the standard deviation, I calculate it as suggested in this answer on roughly the same topic. I get

$\sigma = \sqrt{0.3(1-0.3)}\approx 0.46$

So I enter this into the expression and get

${{3.92*0.46} \over \sqrt{n}} = {1.56 \over \sqrt{n}}$

Now this is where it all falls apart for me. According to the book I should now set the above expression to be equal to the maximum length of the interval, and then calculate n from there, which makes sense. But I was never given an interval length. So how do I procede?

(Looking at the answer tells me that $n=246$, which gives the length $l\approx 0.01$, but as I can't use this in my calculations it doesn't really help me.)

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    $\begingroup$ $3.92\times 0.46=1.8032 \not = 1.56$ $\endgroup$
    – Henry
    Oct 22, 2015 at 16:13
  • $\begingroup$ The confidence interval length is $0.42-0.3=0.12$ $\endgroup$
    – Henry
    Oct 22, 2015 at 16:14
  • $\begingroup$ @Henry Of course, thanks! I accidentally entered $3.42*0.46$ instead of $3.92*0.46$.. Thanks for pointing that out! $\endgroup$ Oct 22, 2015 at 16:38

2 Answers 2

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You are probably expected to use the mid-point of the confidence interval $\frac{0.3+0.42}{2} = 0.36$. This would make $\sigma=\sqrt{0.36(1-0.36)}=0.48$.

The length of the confidence interval should then be $2 \times 1.96 \times \frac{0.48}{\sqrt{n}}$. But the length of the confidence interval is in fact $0.42-0.3=0.12$.

Solving this gives $n \approx 245.86$ and rounding to an integer gives $246$.

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  • $\begingroup$ This works wonders! Thanks man, I really appreciate it! $\endgroup$ Oct 23, 2015 at 12:24
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You have the confidence interval $$\left[\bar{z} - 1.96 \frac{\sigma}{\sqrt{n}}, \bar{z} + 1.96 \frac{\sigma}{\sqrt{n}}\right]$$ The length from this interval is the upper end minus the lower end $$\bar{z} + 1.96 \frac{\sigma}{\sqrt{n}} - \left( \bar{z} - 1.96 \frac{\sigma}{\sqrt{n}} \right) = 2 \cdot 1.96 \frac{\sigma}{\sqrt{n}}$$ Now you know the length of the given confidence interval is 0.12. With this you can compute $$2\cdot 1.96 \frac{\sigma}{\sqrt{n}} = 0.12$$ and this is the case when $$\left(\frac{3.92 \cdot \sigma}{0.12}\right)^2 \geq n$$

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