In a random sample of 50 college seniors, 18 indicated that they were planning to pursue a graduate degree.

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.

Expert Solution

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

SUN BUNRA answered 2 years ago

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