Question

(05.01, 05.04 MC) A polling agency showed the following two statements to a random sample of 1,500 18-year-olds who are registered to vote in the United States. Healthcare statement: Improving the healthcare system should be given priority over improving the education system. Education statement: Improving the education system should be given priority over improving the healthcare system. The order in which the statements were shown was selected randomly for each person in the sample. After reading the statements, each person was asked to choose the statement that was most consistent with his or her opinion. The results are shown in the table. Percent of Sample Healthcare Statement Education Statement No Preference 39% 54% 7% Part A: Assume the conditions for inference have been met. Construct and interpret a 98% confidence interval for the proportion of all 18-year-old registered voters in the United States who would have chosen the education statement. (3 points) Part B: One of the conditions for inference that is met is that the number of those who chose the education statement and the number of those who did not choose the education statement are both greater than 10. Explain why it is necessary to satisfy this condition. (3 points) Part C: A suggestion is made to use a two-sample z-interval for a difference between proportions to investigate whether the difference in proportions between 18-year-old registered voters in the United States who would have chosen the healthcare statement and 18-year-old registered voters in the United States who would have chosen the education statement is statistically significant. Is the two-sample z-interval for a difference between proportions an appropriate procedure to investigate the difference? Justify your answer. (4 points) (10 points)

Answer 1

Solution

Let p be the proportion of all 18-year-old registered voters in the United States who would have chosen the education statement. [i.e., p is the population proportion]

Back-up Theory

100(1 - α) % Confidence Interval for the population proportion, p is:

p_cap ± MoE, …………………………………………………………………. (1)

where

MoE = Z_α/2[√{p_cap(1 –p_cap)/n}] ……………………………………………..(2)

with

Z_α/2 is the upper (α/2)% point of N(0, 1),

pcap = sample proportion, and

n = sample size.

Now, to work out the solution,

Part (A)

98% confidence interval for p (in percentage form) is: 51.0% < p < 57.0% i.e., 0.510 < p < 0.570 Answer 1

Calculation Details

n	1500
X	810
p' = pcap	0.54
F = p'(1-p')/n	0.000166
sqrtF	0.012869
α	0.02
1 - (α/2)	0.99
Zα/2	2.326348
LB	0.510063
UB	0.569937

Part (B)

The above CI which has Z_α/2, is derived assuming that Binomial probabilities can be approximated by Normal probabilities, for which the pre-requisite is the given condition. Answer 2

Part (C)

The two-sample interval for (p₁ – p₂) is based on the assumption that the corresponding estimates, p_1cap and p_2cap are derived from two independent samples. In the present case, the two estimates 54% and 39% are derived from the same sample of 1500 subjects. So, this data cannot be used to form CI for (p₁ – p₂). Answer 3

[If 1500 subjects were divided randomly into two groups of say size n1 and n2 and the survey proceeding in identical way, the data can be used to form CI for (p₁ – p₂) .]   

DONE