Question

A random sample of Miami University students finds that 100 of the 500 sampled participate in intramural sports. Find a 95% confidence interval for the true proportion of all students that participate in intramural sports.

Group of answer choices

[0.171, 0.229]

[0.190, 0.210]

[99.965, 100.035]

[0.165, 0.235]

Answer 1

Here, 100 of the 500 sampled participate in intramural sports.

Therefore, the sample proportion , = 100/500

= 0.2

z-value for 95% confidence interval = 1.96

The 95% confidence interval is = z*[(1-)/n]

= 0.2 0.035

= (0.165 , 0.235)

Answer : [0.165, 0.235]