i think this question calls for a confidence interval calculation but im not sure. what formula do i use and how do i go about using it. using calculators i came up with an interval of 3.1. is this right?

Jones and Smith are currently locked in an intense race for Congress. A local talking head recently claimed that 55% of the likely voters support Jones while only 45% support Smith. In order to test that proposition, you commission a poll that randomly sampled 1000 likely voters. In reviewing the results of the poll, you observe that the sampled population proportion was actually only 51.5% for Jones. Assuming that the sampling was correctly obtained and that voter preferences are stable over the period in question, is there suficient evidence to conclude that the population proportion di ffers from the 55% that expert claimed?

  • $\begingroup$ What is your confidence level? $\endgroup$ – user99680 Dec 1 '13 at 1:49
  • $\begingroup$ it doesnt say. im assuming 95% $\endgroup$ – student Dec 1 '13 at 1:50
  • $\begingroup$ Let me write an answer below, since it is hard to write details in the explanation. $\endgroup$ – user99680 Dec 1 '13 at 1:51
  • $\begingroup$ I assumed $95$%, and did get an interval of width $0.31$, as you said. Now you just need to see if $.515$ is in the acceptance region or not. $\endgroup$ – user99680 Dec 1 '13 at 2:12

The confidence interval for population proportions at the $\alpha$ level of confidence is given by: $$p' +/- z_{\alpha} \sqrt {\frac {{[p'(1-p')}]}{n}}$$

Where $p'$ is the sample proportion, and $z_\alpha$ is the $z-$-value associated with the confidence level $\alpha$. In this case, $p'=0.515$ , and $n=1,000$, so that we get: $0.515 +/- z_{\alpha}\sqrt{(0.158)} $. If $\alpha= 95$% , then $z_{\alpha}=1.96$, and we get the interval $$(0.515-0.31, 0.515+0.31)=(0.484, 0.546)$$.

Now we see that the proportion of $55$% falls outside of the acceptance region.

The idea is this: you assume a certain distribution of your test statistic (here normal, with proportion 0.515), and use the distribution to decide how like your sample values are, given the assumptions on the distribution. In our case, we conclude that a value of sample $0.55$-or-more has probability less than $95$% of coming from this (normally-distributed, centered at 0.515) population of proportions, so we reject, or do not accept.

  • $\begingroup$ thanks! so this means that the experts 55% estimate is a good estimate as its clearly within the approx 67% window, correct? $\endgroup$ – student Dec 1 '13 at 2:18
  • $\begingroup$ Sorry, I don't see what you mean with the $67$%. The issue, I think is to see whether you can accept the proportion of $55$% at this confidence level of $95$% , and the acceptance depends on whether $.55$ falls in the confidence interval. $\endgroup$ – user99680 Dec 1 '13 at 2:20
  • $\begingroup$ Please see the last paragraph I just included. $\endgroup$ – user99680 Dec 1 '13 at 2:31

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