Question

A random sample with 150 students has 45 female students. Estimate the population proportion of female students at the 99% level of confidence.

a. Find the right boundary of the estimation?

b. Find the margin of error.

Answer 1

Here from a random sample of 150 students, 45 are female students which is a sample proportion of which means 30% of the students in the sample are female.

Firstly we will have to calculate the margin of error:

margin of error = where z is the z-score, n=population size, = sample proportion.

Here = 30%, n = 150 and z-score for 99% confidence interval is 2.575 (see the table below)

Confidence Level	Critical Value (Z-score)
0.90	1.645
0.91	1.70
0.92	1.75
0.93	1.81
0.94	1.88
0.95	1.96
0.96	2.05
0.97	2.17
0.98	2.33
0.99	2.575

Now putting these values in the formula we get:

   = 9.635%

Hence margin of error is 9.635%.

We can interpret the margin of error by saying we are 99% confident that the proportion of all students who are female is within 0.9635 of our sample proportion of 0.30. We can then write the interval in the following form:

where MOE = Margin Of Error

When we add and subtract the margin of error from the sample proportion, the confidence interval is -0.6635 to 1.2635

Hence the right boundary of the estimation = 1.2635

We are 99% confident that the proportion of all students who are female is between -0.6635 to 1.2635.