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 99​% confident that the sample percentage is within 5.5 percentage points of the true population percentage. Complete parts​ (a) and​ (b) below. a. Assume that nothing is known about the percentage of passengers who prefer aisle seats. nequals 549 ​(Round up to the nearest​ integer.) b. Assume that a prior survey suggests that about 34​% of air passengers prefer an aisle seat. nequals nothing ​(Round up to the nearest​ integer.)

Answer 1

Answer:

a) $n=\frac{0.5(1-0.5)}{(\frac{0.055}{2.58})^2}=550.116$  

And rounded up we have that n=551

b) $n=\frac{0.34(1-0.34)}{(\frac{0.055}{2.58})^2}=493.78$  

And rounded up we have that n=494

Step-by-step explanation:

Previous concept

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.05$. And the critical value would be given by:

$z_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58$

Part a

The margin of error for the proportion interval is given by this formula:  

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$    (a)  

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$   (b)  

We can assume that $\hat p =0.5$ since we don't know prior info. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.055}{2.58})^2}=550.116$  

And rounded up we have that n=551

Part b

$n=\frac{0.34(1-0.34)}{(\frac{0.055}{2.58})^2}=493.78$  

And rounded up we have that n=494

Answer 2

Answer:

(a) n=551

(b) n=494

Step-by-step explanation:

Given information:

Confidence level $c=0.99$

Significance level $\alpha=1-c=1-0.99=0.01$

$\frac{\alpha}{2}=0.05$

Then critical value,

$z_{\frac{\alpha}{2}=-2.58$

Sample size =n

Marginal error $ME=5.5$%

$ME=\pm 0.05$

By use of formula ,

$ME=z_{\frac{\alpha}{2}}\sqrt\frac{\widehat{p}(1-\widehat{p})}{n}$

$n=\frac{\widehat{p}(1-\widehat{p})}{(\frac{ME}{z_{\frac{\alpha}{2}}})^2}$

(a)If nothing is known we take $\widehat{p}=0.5$

On substitution,

$n=\frac{{0.5}(1-{0.5})}{(\frac{0.05}{2.58})^2}=550.116 \approx551$

(b) Given 34%,

$\widehat{p}=0.34$

On substitution,

$n=\frac{{0.34}(1-{0.34})}{(\frac{0.05}{2.58})^2}=493.78 \approx494$

Hence,answer is

(a) n=551

(b) n=494

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