Question

The GO transportation system of buses and commuter trains operates on the honor system. Train travelers are expected to buy their tickets before boarding the train. Only a small number of people will be checked on the train to see whether they bought a ticket. Suppose that a random sample of 200 train travelers was sampled and 24 of them had failed to buy a ticket. Estimate with 90% confidence the proportion of all train travelers who do not buy a ticket. Interpret the confidence interval you found.

Answer 1

Answer:

The 90% confidence interval for the population proportion of all train travelers who do not buy a ticket is (0.08, 0.16).

Step-by-step explanation:

The (1 - α)% confidence interval for the population proportion is:

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

The information provided is:

n = 200

X = 24

Confidence level = 90%

Compute the value of sample proportion as follows:

$\hat p=\frac{X}{n}=\frac{24}{200}=0.12$

Compute the critical value of z for 90% confidence level as follows:

$z_{\alpha/2}=z_{0.10/2}=z_{0.05}=1.645$

Compute the 90% confidence interval for the population proportion of all train travelers who do not buy a ticket as follows:

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

     $=0.12\pm 1.645\times\sqrt{\frac{0.12(1-0.12)}{200}}\\\\=0.12\pm0.038\\=(0.082, 0.158)\\\approx (0.08, 0.16)$

The 90% confidence interval for the population proportion of all train travelers who do not buy a ticket is (0.08, 0.16).

The 90% confidence interval (0.08, 0.16) for the population proportion of all train travelers who do not buy a ticket implies that there is a 0.90 probability that the true proportion lies in this interval.

Or if 100 such intervals are computed then 90 of those intervals will consist of the true proportion.