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lutik1710 [3]
3 years ago
9

We are interested in conducting a study to determine the percentage of voters of a state would vote for the incumbent governor.

What is the minimum sample size needed to estimate the population proportion with a margin of error of .05 or less at 95% confidenc
Mathematics
1 answer:
Vladimir [108]3 years ago
4 0

Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16  

And rounded up we have that n=385

Step-by-step explanation:

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{\frac{\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=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

The estimated proportion for this case is \hat p=0.5. And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16  

And rounded up we have that n=385

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