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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 9595​% confident that the sample percentage is within 2.52.5 percentage points of the true population percentage. complete parts​ (a) and​ (b) below.

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nobillionaireNobley
Recall that margin of error is given by:

[tex]M=z_{\alpha/2}\sqrt{ \frac{p(1-p)}{n}} \\ \\ \Rightarrow\frac{M^2}{\left(z_{\alpha/2}\right)^2}= \frac{p(1-p)}{n} \\ \\ \Rightarrow n= \frac{\left(z_{\alpha/2}\right)^2p(1-p)}{M^2} [/tex]

Given that you want to be 95​% confident that the sample percentage is within 2.5 percentage points of the true population percentage.

This means that [tex]z_{\alpha/2}=1.96[/tex] and M = 2.5% = 0.025.

Part A:

When nothing is known about the percentage of persengers who prefer aisle seat, we make use of p = 50% = 0.5.

Thus,

[tex]n=\frac{1.96^2\times0.5\times0.5}{(0.025)^2}= \frac{0.9604}{0.000625} =1,537[/tex]

Therefore, 1,537 randomly selected air passengers must be​ surveyed to be 95​% confident that the sample percentage is within 2.5 percentage points of the true population percentage.



Part B:

Given that recent surveys surgest that about 38% of all air passengers prefer an aisle seat, thus p = 38% = 0.38

Thus,

[tex]n=\frac{1.96^2\times0.38\times0.62}{(0.025)^2}= \frac{0.9051}{0.000625} =1,449[/tex]

Therefore, 1,449 randomly selected air passengers must be​ surveyed to be 95​% confident that the sample percentage is within 2.5 percentage points of the true population percentage.
ankitprmr2
  • 1,537  randomly selected air passengers must be surveyed to be 95 percent confident within 2.5 percentage points of the sample percentage by using p=50 percent.
  • 1,449  randomly selected air passengers must be surveyed to be 95 percent confident within 2.5 percentage points of the sample percentage by using p=38 percent.

Step by Step Explanation:

You want 95% confidence within 2.5% points of the actual population.

The margin of error can be formulated as given below:

[tex]\begin{aligned}M&=Z_{\alpha/2}} \times \sqrt{\dfrac{p(p-1)}{n}}\\\dfrac{M^2}{[Z_{\alpha/2}]^2}}&=\dfrac{p(p-1)}{n}\\n&=\dfrac{[Z_{\alpha/2}]^2 \times p(p-1)}{M^2} \end{aligned}[/tex]

Part-(A)

Using p=50%

Substitute the values in the above formula and solve for the n.

Thus,

n=1537.

Part-(B)

Substitute the values in the above formula and solve for the n.

Thus,

n=1449.

To know more, please refer to the link:

https://brainly.com/question/19435930?referrer=searchResults

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