## Related questions with answers

Because of safety considerations, in May 2003 the Federal Aviation Administration (FAA) changed its guidelines for how small commuter airlines must estimate passenger weights. Under the old rule, airlines used 180 pounds as a typical passenger weight (including carry-on luggage) in warm months and 185 pounds as a typical weight in cold months. The Alaska Journal of Commerce (May 25, 2003) reported that Frontier Airlines conducted a study to estimate average passenger plus carry-on weights. They found an average summer weight of 183 pounds and a winter average of 190 pounds. Suppose that each of these estimates was based on a random sample of 100 passengers and that the sample standard deviations were 20 pounds for the summer weights and 23 pounds for the winter weights. a. Construct and interpret a $95\%$ confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers. b. Construct and interpret a $95\%$ confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers. c. The new FAA recommendations are 190 pounds for summer and 195 pounds for winter. Comment on these recommendations in light of the confidence interval estimates from Parts (a) and (b).

Solution

Verifieda. To construct a $95\%$ confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers, we should take note of the following. Recall that there are two formulae for the confidence interval. One is when the population standard deviation is known and one is when it is unknown. Since the population standard deviation is unknown, our formula for the confidence interval is the $t$ confidence interval for $\mu$:

$\left(\bar x - \text{critical value} \cdot \frac{s}{\sqrt{n}}, \bar x + \text{critical value} \cdot \frac{s}{\sqrt{n}} \right)$

where,

- $\bar x$ is the sample mean
- critical value is $t_{\frac{\alpha}{2}}$
- $s$ is the sample standard deviation
- $n$ is the sample size

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