Question

A consumer group surveyed 146 airplane travelers after a flight and found that 132 of them would fly that airline again. Find the standard error for the sample proportion of airline travelers who would fly on that airline again. Enter your answer as a decimal rounded to three decimal places.

Answer 1

Answer:

$\hat p =\frac{X}{n}$

And replacing we got:

$\hat p =\frac{132}{146}= 0.904$

And for this case the standard error assuming normality would be given by:

$SE= \sqrt{\frac{\hat p (1-\hat p)}{n}}$

And replacing we got:

$SE= \sqrt{\frac{0.904*(1-0.904)}{146}}= 0.024$

Step-by-step explanation:

For this problem we know the following notation:

$n= 146$ represent the sample size selected

$X= 132$ represent the number of airplane travelers who after a flight  would fly that airline again

The estimated proportion for this case would be:

$\hat p =\frac{X}{n}$

And replacing we got:

$\hat p =\frac{132}{146}= 0.904$

And for this case the standard error assuming normality would be given by:

$SE= \sqrt{\frac{\hat p (1-\hat p)}{n}}$

And replacing we got:

$SE= \sqrt{\frac{0.904*(1-0.904)}{146}}= 0.024$