Survey of the tax increase In a survey of 1,000 people, 420 are opposed to the tax increase. a) Construct a 95% confidence interval for the proportion of those people opposed to the tax increase. b) Interpret the CI in terms of the question. c) Is the estimate in part (a) valid? Explain. My work: I was able to solve a) and get the answer of [0.4205,0.4195]. For b) I am confused as to how to interpret the CI in terms of the question... could it be that its the percentage range of the possible number of people opposed to the tax increase? For c) I think that it is valid because the number of people opposed to the tax increase is 420, which lies in the range of the confidence interval.

spremani0r

spremani0r

Answered question

2022-09-06

Survey of the tax increase
In a survey of 1,000 people, 420 are opposed to the tax increase.
a) Construct a 95% confidence interval for the proportion of those people opposed to the tax increase.
b) Interpret the CI in terms of the question.
c) Is the estimate in part (a) valid? Explain.
My work:
I was able to solve a) and get the answer of [0.4205,0.4195].
For b) I am confused as to how to interpret the CI in terms of the question... could it be that its the percentage range of the possible number of people opposed to the tax increase?
For c) I think that it is valid because the number of people opposed to the tax increase is 420, which lies in the range of the confidence interval.
EDIT:
To solve for a) I did p ± 1.96 x [ 0.42 x 0 × 0.58 1000 , which equalled 0.42 ± 0.0005...
I now see that I forgot to apply the sqrt before multiplying by 1.96. The correct answer should then be: 0.42 ± 1.96x0.0156 = 0.42 ± 0.0306, which results in the range of [0.4506, 0.3894]

Answer & Explanation

Maggie Tanner

Maggie Tanner

Beginner2022-09-07Added 18 answers

Confidence intervals need to be written with the lesser endpoint preceding the greater endpoint; e.g., [0.3894,0.4506].
The classical, frequentist view holds that a confidence interval is a kind of interval estimate of a population parameter, in which if we were to conduct many such surveys with the same sample size n=1000, and were to compute for each survey its associated confidence interval, we would find that on average, 95% of the resulting confidence intervals would contain the true value of the population parameter.
In other words, the result we were given, 420 out of 1000 votes in opposition to a tax increase, comprises a single random realization whose outcome may be different if we sampled another 1000 people. We don't know what the true proportion of voters opposed to a tax increase really is. In order to find out, we'd have to ask every single person who is eligible to vote. If there are millions of voters, that could be extremely resource-intensive. So when we sample, we sacrifice precision for expediency. A confidence interval, like a point estimate, is also an estimator of the true value of the population proportion. But unlike a point estimate, a confidence interval lets us express our uncertainty in our knowledge of the true value of the parameter. The wider the interval, the less certain we are. And in the frequentist view, the confidence interval does not mean that the parameter falls within it 95% of the time, because the parameter is the quantity that is fixed, and the interval, being an estimate derived from randomly sampled data, is itself random--as I explained above, each time you conduct a survey, you might get a different outcome and a different confidence interval. So, what the CI means is that the coverage probability is 95%; i.e., the probability that the survey would collect data that results in a CI that contains the parameter is 95%.
For the last part of the question, the meaning is not whether the CI contains the parameter. Rather, what it is really asking is, "Is the CI you constructed a reasonable one in light of the data observed?" To help you understand what this means, suppose I conducted a survey that asked, "do you think every child under the age of 10 should be given handguns," and I found that in a sample of n=1000 people, 2 people said "Yes" and everyone else said "No." The resulting 95% CI under the same procedure as you used to calculate your question would be:
2 1000 ± 1.96 0.002 ( 1 0.002 ) 1000 [ 0.000769 , 0.004769 ] .
Is this a reasonable interval? No, because the lower bound is negative. It's also not reasonable because, despite the rather large sample size, the number of "Yes" responses is too small to assume the normal approximation to the binomial distribution. This question is so skewed that you would need a different method to compute a confidence interval. But for your question, 420 out of 1000 is reasonably balanced, and the sample size is large, so a normal approximation (and the Wald-type CI you are using) is reasonable.

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