Question

a survey of an urban university (population of 25,450) 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? multiple choice [0.7510, 0.8290] [0.6592, 0.7044] [0.6659, 0.6941] [0.6795, 0.6805]

Answer 1

The confidence interval for the proportion of students attending a football game is [0.6592, 0.7044].

Confidence interval can be calculated by below formula

$Pcap - Z_{\alpha /2}\sqrt{\frac{Pcap(1 - Pcap)}{n} } < p < Pcap + Z_{\alpha /2}\sqrt{\frac{Pcap(1 - Pcap)}{n} }$

where the left part is the lower interval and right side is the higher interval.

Pcap = 750/1100

Pcap = 0.6818

Again to calculate $Z_{\alpha /2}$ , as we know that the level of confidence is 90 percent so its value will be 1.645

now putting the values on the both side we can get ,

0.6818 - 1.645 x 0.014 < p  <  0.6818 + 1.645 x 0.014

0.6818 - 0.023  < p < 0.6818 + 0.023

0.6588 < p < 0.7048

the nearest value is [0.6592, 0.7044]

Hence the confidence interval for the proportion of students attending a football game is [0.6592, 0.7044]

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