Question

You are the operations manager for an airline and you are considering a higher fare level for passengers in aisle seats. How many randomly selected air passengers must you​ survey? Assume that you want to be 95​% confident that the sample percentage is within 4.5 percentage points of the true population percentage.
a. Assume that nothing is known about the percentage of passengers who prefer aisle seats.
b. Assume that a prior survey suggests that about 36?% of air passengers prefer an aisle seat.

Answer 1

Answer:

$a) n\approx 475 \ passengers$

$b)\ n\approx 438\ passengers$

Step-by-step explanation:

a. #Assume that nothing has changed.

The confidence interval is 95%. the level of significance is $1-\alpha=95\%, \alpha=0.05$

Margin of error, ME=4.5%=0.045

-Denote the sample proportion of passengers who prefer aisle seats as$\hat p$.

- When $\hat p$ is unknown, assume its prior value as 0.:

$1-\alpha=95\%, \alpha=0.05\\\alpha/2=0.025\\\\z_{0.025}=1.96$

The sample size required is thus calculated as:

$n=\hat p(1-\hat p)\times (z_{0.5\alpha}/ME)^2, ME=0.045\\\\=0.5(1-0.5)\times (1.96/0.045)^2\\\\=474.27\approx 475$

Hence, the sample size required is approximately 475 passengers

b. Given the confidence level of 95%, ME=0.045 and $\hat p=0.36$

The sample size is calculated as(when $\hat p$ is known):

$n=\frac{[z_{0.5\alpha}]^2\hat p(1-\hat p)}{(ME)^2}\\\\\\z_{0.5\alpha}=1.96, \\\\n={1.96^2\times 0.36(1-0.36)}{0.045^2}\\\\=437.08\\\\n\approx 438\ passengers$

Hence, the sample required given 36% prefer an aisle seat is approximately 438 passengers.