Professor York randomly surveyed 240 students at Oxnard University and found that 150 of the students surveyed watch more than 10 hours of television weekly.Develop a 95 percent confidence interval to estimate the true proportion of students who watch more than 10 hours of television each week.The confidence interval is
A)0.533 to 0.717
B)0.564 to 0.686
C)0.552 to 0.698
D)0.551 to 0.739

Question

Quizplus · Accepted Answer

To find the confidence interval, first calculate the point estimate:
point estimate = number of students who watch more than 10 hours of TV / total number of students surveyed
point estimate = 150/240
point estimate = 0.625

Next, calculate the margin of error:
margin of error = z* * standard error
* z-value for 95% confidence level = 1.96
standard error = sqrt( (point estimate * (1 - point estimate)) / sample size )
standard error = sqrt( (0.625 * 0.375) / 240 )
standard error = 0.042

margin of error = 1.96 * 0.042
margin of error = 0.082

Finally, construct the confidence interval:
confidence interval = point estimate ± margin of error
confidence interval = 0.625 ± 0.082
confidence interval = (0.543, 0.707)

Therefore, the answer is B.

Professor York Randomly Surveyed 240 Students at Oxnard University and Found