Question

Questions 15-18 are related to the following: In a sample of n = 600 IUPUI undergraduate students polled x = 216 stated that they are planning to go to graduate school. 15 The point estimate of the population proportion of students planning to go to graduate school is,

0.38
0.37
0.36
0.35
The standard error of the sample proportion is,
0.0196
0.0213
0.0243
0.0277
The margin or error for a 95% confidence interval for the population proportion is,
0.024
0.028
0.034
0.038
In the previous question to build an interval estimate such that 19 of every 20 such intervals would capture the population proportion within ±3 percentage points margin of error, the minimum sample size is ______. Use the sample proportion in the previous question as the planning value.
								956
								964
								972
984

Answer 1

15)

p = 216/600 = 0.36

16)

std.error = sqrt(p *(1-p)/n)
=sqrt(0.36 *(1-0.36)/600)
= 0.0196

17)

sample proportion, = 0.36
sample size, n = 600
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.36 * (1 - 0.36)/600) = 0.0196

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0196
ME = 0.038

18)

The following information is provided,
Significance Level, α = 0.05, Margin of Error, E = 0.03

The provided estimate of proportion p is, p = 0.36
The critical value for significance level, α = 0.05 is 1.96.

The following formula is used to compute the minimum sample size required to estimate the population proportion p within the required margin of error:
n >= p*(1-p)*(zc/E)^2
n = 0.36*(1 - 0.36)*(1.96/0.03)^2
n = 983.45

Therefore, the sample size needed to satisfy the condition n >= 983.45 and it must be an integer number, we conclude that the minimum required sample size is n = 984
Ans : Sample size, n = 984