Question

Suppose that 37% of ONU students have taken an online course.               

A random sample of 200 ONU students is taken, of which 78 have taken an online course.

               

1. What is the probability of obtaining a sample proportion this high or higher?

2. There is some thought that the proportion taking online courses is no longer 37%.

Based on the random sample, construct a 95% confidence interval for the proportion of ONU students who have taken an online course.

                                                                                                               

      c) Compute the margin of error for this interval.            d) Does the interval confirm that

the interval online course rate is no longer 37%? __________

Explain in at most 2 sentences.

Answer 1

SOLUTION-

POPULATION PROPORTION(p) = 37%

SAMPLE SIZE(n) = 200

SUCCESS(x) = 78

1.) PROBABILITY OF SAMPLE PROPORTION THIS HIGH () = 78/200 = 0.39

2.) THE HYPOTHESIS THAT WE FRAME IS,

THE TEST STATISTIC IS,

  

  

NOW A 95% CONFIDENCE INTERVAL FOR POPULATION PROPORTION IS ,

MARGIN OF ERROR(E) =

SO, THE 95% CONFIDENCE INTERVAL IS,

YES, THE INTERVAL SUGGESTS THAT THE ONLINE COURSE RATE IS NO LONGER THAN 37% AS THE SAMPLE PROPORTION IS CONTAINED WITHIN THE INTERVAL.

****IN CASE OF DOUBT, COMMENT BELOW. ALSO LIKE THE SOLUTION, IF POSSIBLE.