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Okay so I have an exam coming up pretty soon and I really can't get the hang of how to calculate sample size when all you are given is standard deviation.

For instance:

A normal population has a standard deviation of 15. How large a sample should be drawn to estimate with 95% confidence the population mean to within 1.5?

Or:

A statistician wants to estimate the mean weekly family expenditure on clothes. He believes that the largest weekly expenditure is $650 and the lowest is \$150. a. Determine with 99% confidence the number of families that must be sampled to estimate the mean weekly family expenditure on clothes to within \$15.

Or:

A social scientist claims that the average adult watches less than 26 hours of television per week. He collects data on 25 individuals’ television viewing habits and finds that the mean number of hours that the 25 people spent watching television was 22.4 hours. If the population standard deviation is known to be eight hours, can we conclude at the 1% significance level that he is right?

Or:

Domino’s Pizza in Big Rapids, Michigan, advertises that they deliver your pizza within 15 minutes of placing an order or it is free. A sample of 25 customers is selected at random. The average delivery time in the sample was 13 minutes with a sample standard deviation of 4 minutes. a) Test to determine if we can infer at the 5% significance level that the population mean is less than 15 minutes.

A solution to any of these would probably help me figure out the rest of them, but solutions to all would be kindly appreciated.

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    $\begingroup$ The usual strategy is to begin with a formula for the confidence interval's width, which will depend on the standard deviation, sample size, and desired confidence level. Given any two of those three numbers, solve for the third. At what stage of this process would you like help? $\endgroup$ – whuber Jan 17 at 22:21
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I'll help you get started and answer the first, and then maybe you can try the rest based on that and then come back for more help if needed. So the first question asks:

A normal population has a standard deviation of 15. How large a sample should be drawn to estimate with 95% confidence the population mean to within 1.5?

Instead of just giving you a formula, I'll try to walk you through how you could get to the formula.

The first point I'll make is that you actually know a lot more than just the standard deviation as you first wrote, so let's gather what you know and work from there. First, you know that the distribution is normal, and so you know that if your sample mean is $\hat{\mu}$, it must be that

$$\hat{\mu} \sim N(\mu,(\frac{\sigma}{\sqrt{n}})^2)$$

and in particular, we have $$\frac{\sqrt{n}(\hat{\mu} - \mu)}{\sigma} \sim N(0,1)$$

And we know the value $\sigma = 15$. Okay, next, what else do we know? Well the question asks for a number $n$ such that the probability of $|\hat{\mu} - \mu| \leq 1.5$ is 95%. So let's write that correctly as wanting to find an $n$ such that $$P(|\hat{\mu} - \mu| \leq 1.5) = .95$$

We can't work much with that, but what if we multiplied both sides inside the probability by $\frac{\sqrt{n}}{\sigma}$? Then we have

$$P(|\frac{\sqrt{n}(\hat{\mu} - \mu)}{\sigma}| \leq \frac{\sqrt{n}1.5}{\sigma}) = .95$$

which is now equivalent to

$$P(|N(0,1)| \leq \frac{\sqrt{n}1.5}{\sigma}) = .95$$

and using zscores or whatever method, you know that a standard normal distribution is within 95% of the distribution (two tails) when it is within $[-1.96,1.96]$. So now we know that $\frac{\sqrt{n}1.5}{\sigma}$ must be equal to 1.96 so that the above holds. So now we simply solve for $n$ and get that

$$n = (\frac{1.96*\sigma}{1.5})^2$$

Plugging in for $\sigma$, you should get 485 (always round up!).

More generally, we essentially derived the formula for normal distribution more generally: if we have a desired confidence range $1-\alpha$ (in this problem, $\alpha = .05$) and a desired width $w$, then we find the $z_{\alpha/2}$ z-score and have the following equation that links them all together: $$\alpha_{z/2} = \frac{\sqrt{n}w}{\sigma}$$

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