Question

Question 1

An airline company wishes to know the proportion of business class travelers flying the Kuala Lumpur-to-Hong Kong route. In a random sample of 350 passengers, 190 are business class passengers.

 

a. Determine the point estimate of the true proportion of business class passengers.

b. Construct a 95% confidence interval estimate of the average waiting time for all customers.

c. Construct a 99% confidence interval estimate of the average waiting time for all customers.

d. Referring to b) and c), what will happen to the width of confidence interval?

Answer 1

a. The point estimate of the true proportion of business class passengers

$$\hat{p}=\dfrac{X}{N}=\dfrac{190}{350}=\dfrac{19}{35}\approx0.542857$$

b. The critical value for $\alpha=0.05$ is $z_c=z_{1-\alpha/2}=1.96.$

The corresponding confidence interval is computed as shown below:

$$CI(proportion)$$

$$=(\hat{p}-z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}},\hat{p}+z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})$$

$$=(\dfrac{19}{35}-1.96\sqrt{\dfrac{\dfrac{19}{35}(1-\dfrac{19}{35})}{350}},$$

$$\dfrac{19}{35}+1.96\sqrt{\dfrac{\dfrac{19}{35}(1+\dfrac{19}{35})}{350}})$$

$$=(0.491,0.595)$$

Therefore, based on the data provided, the 95% confidence interval for the population proportion is $0.491<x<0.595,$ which indicates that we are 95% confident that the true population proportion $p$ is contained by the interval $(0.491,0.595).$

c. The critical value for $\alpha=0.01$ is $z_c=z_{1-\alpha/2}=2.576.$

The corresponding confidence interval is computed as shown below:

$$CI(proportion)$$

$$=(\hat{p}-z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}},\hat{p}+z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})$$

$$=(\dfrac{19}{35}-2.576\sqrt{\dfrac{\dfrac{19}{35}(1-\dfrac{19}{35})}{350}},$$

$$\dfrac{19}{35}+2.576\sqrt{\dfrac{\dfrac{19}{35}(1+\dfrac{19}{35})}{350}})$$

$$=(0.474,0.611)$$

Therefore, based on the data provided, the 99% confidence interval for the population proportion is $0.474<x<0.611,$ which indicates that we are 99% confident that the true population proportion $p$ is contained by the interval $(0.474,0.611).$

d. If you want a higher level of confidence, that confidence interval will be wider.

Increasing the confidence will increase the margin of error resulting in a wider interval.