Answer to Question #192586 in Statistics and Probability for Rick Stanbeck

Question

Explain that all conditions for the sampling distribution for sample proportions have been met.

A local travel company believes that 64% of the U.S have never taken a vacation where you must fly on a plane to get there. The company is interested in a new marketing plan targeting these people and they would like to determine if they should launch a campaign in your town with a population of 1576 and it is found that out of 89 selected adults 63 have never taken such a vacation.

Part 2. Assuming you already checked the conditions, find the 90% confidence interval for the true proportion of all adults in your hometown who have never taken a vacation where a flight is required to get there

Answer 1

condition for the sampling distribution are-

Some sample proportions will be on the low side — say, 0.55 or 0.58 — while others will be on the high side — say, 0.61 or 0.66. It is reasonable to expect all the sample proportions in repeated random samples to average out to the underlying population proportion, 0.6. In other words, the mean of the distribution of p-hat should be p.

Sample proportions closest to 0.6 would be most common, and sample proportions far from 0.6 in either direction would be progressively less likely. In other words, the shape of the distribution of sample proportion should bulge in the middle and taper at the ends: it should be somewhat normal.

$p=0.64,$

Sample size n=89

Probability of adults take vacation $\hat{p}=\dfrac{63}{89}=0.707$

Part 2.

$\alpha=1-0.9=0.1$

$\dfrac{\alpha}{2}=0.05,Z_{0.05}=1.64$

90% confidence interval is-

$=\hat{p}\pm z_{0.05}\sqrt{\dfrac{p(1-p)}{n}}$

$=0.707\pm 1.64\sqrt{\dfrac{0.64(1-0.64)}{89}}\\[9pt]=0.707\pm 0.0834\\[9pt]=(0.6235,0.7904)$