Question

A state’s Department of education reports that 43 percent of all college students in the state are foreign students. The state university wonders if the state’s claim is valid for diversity purposes. Admission officers from the university want to ensure that a significant portion of the applicants accepted were in fact foreign students. They sampled over 6,000 recently admitted students to gain an estimate for their university.

A) The admissions officers want to estimate the true percentage of foreign students on campus to within ±5%, with 95% confidence. How many applications should they sample?

B) They actually select a random sample of 550 applications, and found that 45%were foreigners. Create the 90 percent confidence interval. Be sure to verify the conditions.

C) Interpret what 90% confidence means in this context.

D) Should the admissions officers conclude that the percentage of foreign students in the college is lower than statewide enrollment rate of 43%? Explain.

Answer 1

Ans A:

Given the margin of error should be within 5%

i.e., Critical value*   

For 95% confidence level, Critical value = 1.96

=> 1.96*

To find the least possible value of n, the numerator should be considered at maximum population proportion which is 0.5

=> 1.96*

squaring on both sides gives

3.841*(0.25/n)   => n

i.e., they should sample 385 applications

Ans B:

n = 550, mean = 0.45 i.e., 247.5 students are foreigners

Estimator of true variance = =

= 0.248

=> s = 0.498 => 0.5 approx

= = 0.02

Therefore, the 90% confidence interval = (0.45 - 1.645*0.02, 0.45 + 1.645*0.02 ) = (0.417, 0.483)

i.e., the 90% confidence interval is between 41.7% to 48.3%

Ans 3:

In this context, 90% confidence interval means that, there is a 90% chance that the percentage of foreign students is in between 41.7% to 48.3% when a sample of 550 students is considered