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A political action committee is interested in finding out what proportion of voters will

support an environmental initiative. Obtain a sample size that will ensure a margin of

error of at most 0.09 for a 95% confidence interval. Similar initiatives in the past have

gotten 93% support.

1 Answer

7 votes

Answer:


n=(0.93(1-0.93))/(((0.09)/(1.96))^2)=30.875

And rounded up we have that n=31

Explanation:

First we need to find the critical value for the margin of error desired. The confidence level is 0.95 so then the significance is
\alpha=1-0.95=0.05 and the critical value for this case would be:


z_(\alpha/2) =\pm 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 0.09 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)

The estimated proportion for this case is
\hat p =0.93. And replacing into equation (b) the values from part a we got:


n=(0.93(1-0.93))/(((0.09)/(1.96))^2)=30.875

And rounded up we have that n=31

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