A poll of 10,000 Americans found that 4% were vegetarians. a) What has to be ass
ID: 3023135 • Letter: A
Question
A poll of 10,000 Americans found that 4% were vegetarians. a) What has to be assumed about this sample in order to construct a confidence interval for the population proportion of vegetarians?
b) Construct a 99% confidence interval for the population proportion. Explain why the interval is so narrow even though the confidence level is high.
99% Confidence Interval: ___________________________________________________
Explanation for why CI is narrow:
c) Can you conclude that fewer than 10% of Americans are vegetarians? Explain your reasoning.
Explanation / Answer
a)
That this sample is less than 5% of the population.
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b)
Note that
p^ = point estimate of the population proportion = x / n = 0.04
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.001959592
Now, for the critical z,
alpha/2 = 0.005
Thus, z(alpha/2) = 2.575829304
Thus,
Margin of error = z(alpha/2)*sp = 0.005047574
lower bound = p^ - z(alpha/2) * sp = 0.034952426
upper bound = p^ + z(alpha/2) * sp = 0.045047574
Thus, the confidence interval is
( 0.034952426 , 0.045047574 ) [ANSWER]
It is narrow because of the large sample size, which mininized the standard error of the sample proportion. [ANSWER]
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c)
YES, because the whole interval is less than 10%.