Question

A survey of an urban university (with a finite population of 25,450) showed that 750 of 1,100 students sampled attended a home football game during the season. Using the 99% level of confidence, what is the confidence interval for the proportion of students attending a football game?

Answer 1

Answer: $(0.644,\ 0.716)$

Step-by-step explanation:

Given : Sample size = 1100

The number of students attended a home football game during the season = 750

Then, the proportion of students attending a football game =$\dfrac{750}{1100}=0.681818181818\approx0.68$

Critical value = $z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.576$

Now, the confidence interval for the population proportion is given by :-

$p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}\\\\=0.68\pm (2.576)\sqrt{\dfrac{0.68(0.32)}{1100}}\\\\\approx0.68\pm0.036\\\\=(0.644,0.716)$

Hence, the 99% confidence interval for the proportion of students attending a football game = $(0.644,\ 0.716)$