Question

A recent study from the Department of Education shows that approximately 11% of students are in private schools. A random sample of 450 students from a wide geographic area indicated that 55 attended private schools.

a) Estimate the true proportion of students attending private schools with 95% confidence.

b) Are the study results by the Department of Education consistent with the sample? Why or why not?

Answer 1

Part a

Here, we have to find the confidence interval for the population proportion of students attending private schools.

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

x = 55

n = 450

P = 55/450 = 0.122222222

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.122222222 ± 1.96* sqrt(0.122222222 *(1 – 0.122222222)/450)

Confidence Interval = 0.122222222 ± 1.96* 0.0154

Confidence Interval = 0.122222222 ± 0.0303

Lower limit = 0.122222222 - 0.0303 = 0.0920

Upper limit = 0.122222222 + 0.0303 = 0.1525

Confidence interval = (0.0920, 0.1525)

We are 95% confident that the true population proportion of students attending private schools will lies between 9.2% and 15.3%.

b) Are the study results by the Department of Education consistent with the sample? Why or why not?

Yes, the study results by the Department of Education are consistent with the sample, because the 11% lies within the above interval 9.2% and 15.3%.