Solution: We are given:
Margin of error [tex]E=0.055[/tex]
Confidence level[tex]=90\%[/tex]
[tex]\therefore \alpha =100\%-90\%=10\%[/tex] or [tex]0.1[/tex]
Since we are not given the proportion of passengers who prefer aisle seats, therefore we assume it to be [tex]0.5[/tex]
[tex]\therefore p=0.5[/tex]
Now, the formula for finding the sample size is:
[tex]n=p(1-p)\left( \frac{z_{\frac{0.1} {2}} }{E} \right)^{2}[/tex]
Where:
[tex]z_{\frac{0.1}{2}}=1.645[/tex] is the critical value at 0.1 significance level
[tex]\therefore n=0.5(1-0.5) \left(\frac{1.645}{0.055}\right)^{2}[/tex]
[tex]=224[/tex] rounded to nearest integer
