The manager of a car company will select a random sample of its customers to create a 90 percent confidence interval to estimate the proportion of its customers who have children. Of the following, which is the smallest sample size that will result in a margin of error of no more than 6 percentage points?

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$$ \text{Confidence Interval} : \text{statistic} \pm (\text{critical value})(\text{standard error of statistic}) $$

To be within 6 percentage points, we need to find the critical value and standard error. The critical value can be found using the z table. A 90% confidence interval corresponds to a 0.05 tail probability, which correspoinds to a z-score of 1.645.

The standard error of the the sample statistic for population proportion can be found in the formula sheet. Putting it all together:

$$ 0.06 \geq (1.645)\left(\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right) $$

We assume a probability of 0.5 to anticipate the maximum margin of error.

$$ 0.06 \geq (1.645)\left(\sqrt{\frac{(0.5)(0.5)}{n}}\right) $$ $$ \frac{0.06}{1.645}\geq \frac{0.5}{\sqrt{n}} $$ $$ \sqrt{n}\geq \frac{(0.5)(1.645)}{0.06} $$ $$ n \geq \left(\frac{(0.5)(1.645)}{0.06}\right) $$ $$ n \geq 187.918 $$ $$ n= \boxed{200} $$