Random samples of students at 128 four-year colleges were interviewed several ti
ID: 3155296 • Letter: R
Question
Random samples of students at 128 four-year colleges were interviewed several times since 1991. Of the students who reported using intravenous drugs, the percentage who reported injecting daily was 43.8% of 13,249 students in 1991 and 48.5% of 9600 students in 2000.
a. Estimate the difference between the proportions in 2000 and 1991, and interpret.
b. Find the standard error for this difference.
c. Construct and interpret a 90% confidence interval to estimate the true change.
d. State the assumptions for the confidence interval in (c) to be valid.
Explanation / Answer
a)
p1^ - p2^ = 0.438-0.485 = -0.047
Hence, the best estimate that we have for the difference between the 2000 and 1991 proportions of students who use intravenous drugs daily is 0.047 or 4.7% more for the 2000 proportion.
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b)
Getting p1^ and p2^,
p1^ = x1/n1 = 0.438
p2 = x2/n2 = 0.485
Also, the standard error of the difference is
sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] = 0.006678132 [ANSWER]
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c)
For the 90% confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.05
z(alpha/2) = 1.644853627
Margin of error = z(alpha/2)*sd = 0.010984549
lower bound = p1^ - p2^ - z(alpha/2) * sd = -0.057984549
upper bound = p1^ - p2^ + z(alpha/2) * sd = -0.036015451
Thus, the confidence interval is
( -0.057984549 , -0.036015451 ) [ANSWER]
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d)
We assume that all observations in the two groups are independent, and that the sample size is large enough for the sampling distrbutions of proportion to be approximately normal.
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