can i do this?

**rclements3** · August 6th 2009, 03:16 AM

3. A researcher is interested in estimating the proportion of voters who favor a tax on e-commerce. How large a sample would she need to collect to estimate this proportion to within 5% with 95% accuracy? (4.5 points)

Can this be answered using minitab or do I have to do it all pen and paper? Anyone know how to figure this out?

**CaptainBlack** · August 6th 2009, 05:21 AM

Originally Posted by rclements3

3. A researcher is interested in estimating the proportion of voters who favor a tax on e-commerce. How large a sample would she need to collect to estimate this proportion to within 5% with 95% accuracy? (4.5 points)

Can this be answered using minitab or do I have to do it all pen and paper? Anyone know how to figure this out?

Assume the sample size will have to be large, so we can use the normal approximation where we need. Let the sample size be $N$ and the proportion who favour the tax be $p$ . Then the expected number in the sample will be $pN$ , and we will estimate this by $\tilde p= n/N$ where $n$ is the number in the sample favouring the tax. Then the variance of $\tilde p$ is $p(1-p)/N$ and this has a maximum when $p=0.5$ , so we assume $p=0.5$ for the purposes of calculating the variance.

Now a 95% inteval for $\tilde p$ is of width $\le 2 \times 1.96 /[2\times \sqrt{N}]$ (this is assuming the normal approximation) and we want this to be less than or equal $0.1$ ( that is $2\times 0.05$ )

So we put:

$1.96 /(2\sqrt{N})=0.05$

and solve for $N$ to get the required sample size.

CB

**rclements3** · August 6th 2009, 12:29 PM

I'm coming up with N = 385, does this sound about right?

**CaptainBlack** · August 6th 2009, 02:04 PM

Originally Posted by rclements3

I'm coming up with N = 385, does this sound about right?

More or less, I've only done the calculation approximately and have 400, which is consistent with simulation

CB

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