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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?

1 Answer

3 votes

Answer:


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

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