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.

Step-by-step 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

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