Question

In a random sample of 50 college seniors, 18 indicated that they were planning to pursue a graduate degree. Find a 98% confidence interval for the true proportion of all college seniors planning to pursue a graduate degree, and interpret the result, and state any assumptions you have made.

Answer 1

p = x/n = 18/50 = .36

So, 98% CI has Z = 2.33, hence, 

We have the CI as:

p +/- Z*sqrt(p*p'/n)

= .36 +/- 2.33*sqrt(.36*.64/50)

= 0.202 to 0.518

So, 98% CI is (0.202,0.518)

98% confidence interval is a range of values that you can be 98% certain contains the true mean of the population

i.e. (.202, .518) is the range of values that we are 98% sure that it contains the true population.

We make the following assumptions for drawing the Confidence interval:

• The data must be sampled randomly.
• The sample values must be independent of each other
• When the sample is drawn without replacement (usually the case), the sample size, n, should be no more than 10% of the population
• The sample size must be sufficiently large