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Len [333]
3 years ago
6

Liam wants to estimate the percentage of people who lease a car. He surveys 240 individuals and finds that 54 lease a car. Find

the margin of error for the confidence interval for the population proportion with a 95% confidence level. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576
Mathematics
1 answer:
scoundrel [369]3 years ago
6 0

Answer:

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

The confidence interval is 95% and the significance level is \alpha=1-0.95=0.05 and \alpha/2 =0.025 and the critical value would be:

z_{\alpha/2}= 1.96

And the margin of error would be:

ME = 1.96 \sqrt{\frac{0.225 (1-0.225)}{240}}= 0.0528

Step-by-step explanation:

We have the following info given:

n =240 the sample size selected

X=54 the number of people who lease a car

\hat p =\frac{54}{240}= 0.225 the estimated proportion of people who lease a car

The margin of error is given by:

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

The confidence interval is 95% and the significance level is \alpha=1-0.95=0.05 and \alpha/2 =0.025 and the critical value would be:

z_{\alpha/2}= 1.96

And the margin of error would be:

ME = 1.96 \sqrt{\frac{0.225 (1-0.225)}{240}}= 0.0528

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