Exercise 2 Last year, ballots in Champaign-Urbana contained the following questi
ID: 3362866 • Letter: E
Question
Exercise 2 Last year, ballots in Champaign-Urbana contained the following question to assess public opinion on an issue: Should the State of Illinois legalize and regulate the sale and use of marijuana in a similar fashion as the State of Colorado?" Suppose that we would like to understand Champaign-Urbana's 2017 opinion on marijuana legalization. To satisfy our curiosity, we obtain a random sample of 120 Champaign-Urbanians and find that 87 support marijuana legalization. (a Construct a 99% confidence interval for p, the true proportion of Champaign-Urbania s that support marijuana legalization. (b) Suppose that a pollster wants to estimate the true proportion of Champaign-Urbanians that support marijuana legalization to within 0.04, with 95% confidence. How many Champaign-Urbanians should this pollster poll? Assume the pollster has no prior knowledge about the proportion c) Now assume the pollster believes that support for legalization is somewhere between 65% and 85% and they would like to estimate the true proportion of Champaign-Urbanians that support marijuana legalization to within 0.04, with 90% confidence. How many Champaign-Urbanians should this pollster poll?Explanation / Answer
Back-up Theory
100(1 - )% Confidence Interval for population proportion, p is: pcap ± Z/2{pcap(1 - pcap)/n},………… (1)
where pcap = sample proportion, Z/2 = upper (/2) percent point of N(0, 1) and n = sample size.
Sample size, n = {p.(1 - p).Z2/2}/E2 ………………………….........................................…………………(2)
where p is the initial estimate or a pre-determined/notional value , E is the error margin and Z/2 = upper (/2) percent point of N(0, 1).
Preparatory Work
Given, in a sample of 120 respondents, 87 support marijuana legalization, proportion of Champaign-Urbanians supporting marijuana legalization is: pcap = 87/120 = 0.725
Upper 0.5% point of N(0, 1) = Z0.005 = 2.57583; .upper 2.5% point of N(0, 1) = Z0.025 = 1.96;
upper 5% point of N(0, 1) = Z0.05 = 1.645
Part (a)
Given
n =
120
x =
87
pcap =
0.725
1-pcap =
0.275
=
0.01
1 - /2
0.995
Z/2 =
2.57583
99% CI for p:
0.725
±
0.10499
Lower Bound =
0.62001
Upper Bound =
0.82999
ANSWER
Part (b)
Substituting p = 0.725, Z/2 = 1.96, and E = 0.04, in (2) under Back-up Theory,
Sample size, n = {0.725.(0.275).1.962}/0.042
= 479 ANSWER
Part (c)
Substituting p = 0.65, Z/2 = 1.645, and E = 0.04, in (2) under Back-up Theory,
Sample size, n = {0.65.(0.35).1.6452}/0.042
= 385 ANSWER 1
Substituting p = 0.85, Z/2 = 1.645, and E = 0.04, in (2) under Back-up Theory,
Sample size, n = {0.85.(0.15).1.6452}/0.042
= 216 ANSWER 2
Combining Answer 1 and Answer 2, if the expected p is between 65% and 85%, sample size should be more than 216 but less than 385. ANSWER 3
Given
n =
120
x =
87
pcap =
0.725
1-pcap =
0.275
=
0.01
1 - /2
0.995
Z/2 =
2.57583
99% CI for p:
0.725
±
0.10499
Lower Bound =
0.62001
Upper Bound =
0.82999
ANSWER
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