Question

A survey of an urban university showed that 750 of 1,100 students sampled attended a home football game during the season. Using the 90% level of confidence, what is the confidence interval for the proportion of students attending a football game?

a. 0.7510 and 0.8290
b. 0.6592 and 0.7044
c. 0.6659 and 0.6941
d. 0.6795 and 0.6805

Answer 1

Answer:

b. 0.6592 and 0.7044

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence interval $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

A survey of an urban university showed that 750 of 1,100 students sampled attended a home football game during the season. This means that $n = 1100, p = \frac{750}{1100} = 0.6818$

90% confidence interval

So $\alpha = 0.1$, z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6818 - 1.645\sqrt{\frac{0.6818*0.3182}{1100}} = 0.6592$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6818 + 1.645\sqrt{\frac{0.6818*0.3182}{1100}} = 0.7044$

So the correct answer is:

b. 0.6592 and 0.7044

Answer 2

I think they are correct ^^