Respuesta :
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:
[tex]CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
The information provided is:
n = 200
X = 24
Confidence level = 90%
Compute the value of sample proportion as follows:
[tex]\hat p=\frac{X}{n}=\frac{24}{200}=0.12[/tex]
Compute the critical value of z for 90% confidence level as follows:
[tex]z_{\alpha/2}=z_{0.10/2}=z_{0.05}=1.645[/tex]
Compute the 90% confidence interval for the population proportion of all train travelers who do not buy a ticket as follows:
[tex]CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
[tex]=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)[/tex]
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.
