Question 2 Background: Facebook: Top Source for News A poll based on a random sa
ID: 3124917 • Letter: Q
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Question 2 Background: Facebook: Top Source for News A poll based on a random sample of 1,820 American adults (aged 18 and older, living in all 50 U.S. states and the District of Columbia) was conducted to study the source of news for Americans. The study reports that the proportion of all Americans who use Facebook as their source of news is estimated to be 61 %. Question 2 Subquestions 2.a Give a 99% conservative confidence interval for the proportion of all such Americans who would say Facebook is their source of news. Show all work. 2 point(s) No answer entered. Click above to enter an answer. 2.b The study went on to report the results by age group. If the results were used to make a 99% conservative confidence interval for the proportion of all Americans who are in their 30s (using only the subset of the 1820 Americans surveyed who are in their 30s), the corresponding margin of (sampling) error would have been... 1 point(s) points O smaller O largerExplanation / Answer
a)
Note that
p^ = point estimate of the population proportion = x / n = 0.61
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.011433035
Now, for the critical z,
alpha/2 = 0.005
Thus, z(alpha/2) = 2.575829304
Thus,
Margin of error = z(alpha/2)*sp = 0.029449546
lower bound = p^ - z(alpha/2) * sp = 0.580550454
upper bound = p^ + z(alpha/2) * sp = 0.639449546
Thus, the confidence interval is
( 0.580550454 , 0.639449546 ) [ANSWER]
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b)
LARGER. [ANSWER]
Note that smaller sample sizes yield larger margin of error.
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c)
RESPONSE BIAS. [ANSWER]
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d)
Note that
n = z(alpha/2)^2 p (1 - p) / E^2
where
alpha/2 = 0.05
Using a table/technology,
z(alpha/2) = 1.644853627
Also,
E = 0.03
p = 0.61
Thus,
n = 715.1653197
Rounding up,
n = 716 [ANSWER]
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