Question

In a survey, the planning value for the population proportion is p* = 0.35. How large a sample should be taken to provide a 95% confidence interval with a margin of error of 0.05?

Answer 1

`n=(0.35(1-0.35))/(((0.05)/(1.96))^2)=349.59`

And rounded up we have that n=350

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`p` represent the real population proportion of interest

`\aht p` represent the estimated proportion for the sample

n is the sample size required (variable of interest)

`z` represent the critical value for the margin of error

Solution to the problem

The population proportion have the following distribution

`p \sim N(p,\sqrt{(\hat p(1-\hat p))/(n)})`

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by
`\alpha=1-0.95=0.05` and
`\alpha/2 =0.025`. And the critical value would be given by:

`z_(\alpha/2)=-1.96, z_(1-\alpha/2)=1.96`

The margin of error for the proportion interval is given by this formula:

`ME=z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}` (a)

And on this case we have that
`ME =\pm 0.05` and we are interested in order to find the value of n, if we solve n from equation (a) we got:

`n=(\hat p (1-\hat p))/(((ME)/(z))^2)` (b)

And replacing into equation (b) the values from part a we got:

`n=(0.35(1-0.35))/(((0.05)/(1.96))^2)=349.59`

And rounded up we have that n=350