A large city newspaper periodically reports the mean cost of dinner for two people at restaurants in the city. The newspaper staff will collect data from a random sample of restaurants in the city and estimate the mean price using a 90 percent confidence interval. In past years, the standard deviation has always been very close to $35. Assuming that the population standard deviation is $35, which of the following is the minimum sample size needed to obtain a margin of error of no more than $5 ?

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We can use the formula for margin of error given in the formula page. The margin of error is the product of the critical value and the standard error of the statistic.

Confidence interval: statistic \(\pm\) (critical value)(standared error of the statistic)

The critical value that corresponds to a 90 percent confidence interval is 1.645. This can be obtained by the corresponding \(p\)-value of 0.05 for a 90 percent confidence interval.

The standard error is given by this formula:

$$ s_{\bar{X}}=\frac{s}{\sqrt{n}} $$ $$ = \frac{35}{\sqrt{n}} $$

We need the margin of error to be less than 5.

$$ \text{(critical value)(standard error of the statistic)} \lt 5 $$ $$ (1.645)\left(\frac{35}{\sqrt{n}}\right) \lt 5 $$ $$ \sqrt{n} \gt \frac{(1.645)(35)}{5} $$ $$ n \gt \left(\frac{(1.645)(35)}{5}\right)^2 $$ $$ n \gt 132.59 $$