Question

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all​ students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. The 9898​% confidence interval for p is 59 plus or minus 0.08 .59±0.08. Interpret this interval. A. We are 9898​% confident that between 5151​% and 6767​% of the sampled students receive some sort of financial aid. B. We are 9898​% confident that the true proportion of all students receiving financial aid is between 0. 51.51 and 0. 67..67. C. 9898​% of the students receive between 5151​% and 6767​% of their tuition in financial aid. D. We are 9898​% confident that​ 59% of the students are on some sort of financial aid.

Answer 1

Explanation:

Confidence interval is written as

Sample proportion ± margin of error

Margin of error = z × √pq/n

Where

z represents the z score corresponding to the confidence level

p = sample proportion. It also means probability of success

q = probability of failure

q = 1 - p

p = x/n

Where

n represents the number of samples

x represents the number of success

From the information given,

n = 200

x = 118

p = 118/200 = 0.59

q = 1 - 0.59 = 0.41

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.98 = 0.02

α/2 = 0.02/2 = 0.01

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.01 = 0.9

The z score corresponding to the area on the z table is 2.326. Thus, confidence level of 98% is 2.326

Therefore, the 98% confidence interval is

0.59 ± 2.326√(0.59)(0.41)/200

= 0.59 ± 0.08

The lower boundary is

0.59 - 0.08 = 0.51

The upper boundary is

0.59 + 0.08 = 0.67

The correct option is

B. We are 98​% confident that the true proportion of all students receiving financial aid is between 0. 51 and 0. 67