Answer:
[tex]a) n\approx 475 \ passengers[/tex]
[tex]b)\ n\approx 438\ passengers[/tex]
Step-by-step explanation:
a. #Assume that nothing has changed.
The confidence interval is 95%. the level of significance is [tex]1-\alpha=95\%, \alpha=0.05[/tex]
Margin of error, ME=4.5%=0.045
-Denote the sample proportion of passengers who prefer aisle seats as[tex]\hat p[/tex].
- When [tex]\hat p[/tex] is unknown, assume its prior value as 0.:
[tex]1-\alpha=95\%, \alpha=0.05\\\alpha/2=0.025\\\\z_{0.025}=1.96[/tex]
The sample size required is thus calculated as:
[tex]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[/tex]
Hence, the sample size required is approximately 475 passengers
b. Given the confidence level of 95%, ME=0.045 and [tex]\hat p=0.36[/tex]
The sample size is calculated as(when [tex]\hat p[/tex] is known):
[tex]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[/tex]
Hence, the sample required given 36% prefer an aisle seat is approximately 438 passengers.
