Worked Interval for a Campus Poll

From Campus Poll Counts to an Interval

Suppose a student government wants to know whether a majority of students support a proposed change to campus dining hours. A poll can estimate the fraction who support the proposal, but the poll's sample proportion will generally differ from the population proportion. A confidence interval uses the observed count to give a range of plausible values for that population proportion, while the sampling design and approximation conditions determine whether the range is defensible.

The calculation starts with two counts: the number of sampled students who support the proposal and the number who do not. The interval is useful only if the poll represents the population of interest reasonably well. A precise calculation cannot repair a poll that systematically misses some students.

Definition: Let \(p\) be the proportion of all students in the target population who support the proposal. If \(x\) of the \(n\) sampled students say they support it, the sample proportion is \(\hat{p}=x/n\). The remaining \(n-x\) sampled students are counted as not supporting it, provided each sampled student gives one usable yes-or-no response.

The Interval and Its Conditions

For a two-sided confidence interval based on the usual normal approximation, the calculation is

\hat{p}\ \pm\ z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

Here \(z^*\) is the positive standard-normal critical value for the chosen confidence level. For a 95% confidence level, \(z^*\) is approximately 1.96. The square-root term is the estimated standard error, and multiplying it by \(z^*\) gives the margin of error. Subtracting and adding that margin to \(\hat{p}\) gives the lower and upper endpoints.

Before using the formula, check the design and the normal approximation. For a campus poll, the checks are:

Random sampling: The students should be selected using a random mechanism from the population the poll is meant to describe. A poll that anyone can choose to answer is not a random sample.
Independence: Responses should be independent, or approximately so. When sampling without replacement from a finite student population, the usual 10% condition is that the sample size is no more than 10% of the population size.
Success-failure counts: The sample should include at least 10 supporters and at least 10 non-supporters. In counts, check \(x\geq 10\) and \(n-x\geq 10\). These observed counts are a practical check that a normal approximation is reasonable; they do not prove the model assumptions.

The interval also depends on the poll measuring the intended response consistently. For example, the question should make clear what the proposal is, and the response options should allow students to report support or non-support. Missing answers and nonresponse need attention: if those who do not respond differ systematically from those who do, the completed responses may not represent all students.

Why the Formula Uses This Standard Error

A short derivation explains the role of the estimated standard error. For each sampled student, define \(Y_i=1\) if that student supports the proposal and \(Y_i=0\) otherwise. Under random sampling, each \(Y_i\) has mean \(p\) and variance \(p(1-p)\). With independent observations, the sample proportion is their average, so its variance is

\operatorname{Var}(\hat{p}) =\operatorname{Var}\left(\frac{Y_1+\cdots+Y_n}{n}\right) =\frac{1}{n^2}\,n p(1-p) =\frac{p(1-p)}{n}.

Thus the standard deviation of \(\hat{p}\) is \(\sqrt{p(1-p)/n}\). The population proportion \(p\) is unknown, so the interval estimates this standard deviation by replacing \(p\) with \(\hat{p}\). When the success-failure counts are large enough for the normal approximation, the standardized sample proportion is approximately standard normal. Using \(z^*\) to capture the central confidence level and solving for \(p\) gives the interval formula above. This is an approximate interval: its coverage depends on the design and approximation conditions being adequate.

Worked Example: A 95% Interval for Campus Support

Suppose, hypothetically, that a university has 8,000 enrolled students. A random sample of 400 students is asked whether they support extending campus dining hours, and all 400 give a usable response. Of these, 236 say yes. Calculate and interpret a 95% confidence interval for the proportion of all enrolled students who support the proposal.

Check the sampling design. The sample is described as random. Sampling without replacement, 400 is less than 10% of 8,000, since \(0.10(8{,}000)=800\). The independence condition is therefore reasonable under the 10% rule. The question is about all enrolled students, matching the stated sampling population.

Check the approximation. There are 236 supporters and \(400-236=164\) non-supporters. Both counts exceed 10, so the success-failure check is satisfied.

Calculate the sample proportion and standard error.

\hat{p}=\frac{x}{n}=\frac{236}{400}=0.590

SE=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} =\sqrt{\frac{0.590(0.410)}{400}} =\sqrt{0.00060475} \approx 0.02459

For 95% confidence, \(z^*=1.96\). The margin of error is \(1.96(0.02459)\approx 0.04820\), or about 4.82 percentage points. Therefore,

0.590\pm 1.96(0.02459) =0.590\pm 0.04820 =(0.5418,\ 0.6382)

Interpret in context. The 95% confidence interval is approximately 54.2% to 63.8%. Under the stated sampling and approximation conditions, the method gives a range of plausible values for the proportion of all enrolled students who support extending dining hours. The interval's endpoints are estimates of a population proportion, not a range containing 95% of individual students.

Checking the Design Before Trusting the Calculation

The same arithmetic can produce a misleading result if the way students entered the poll does not support inference to the campus population. A voluntary online poll, for example, may attract students with especially strong opinions. The number of responses can be large and both response counts can exceed 10, yet the sample may still be systematically unrepresentative. Random selection and adequate response are separate from the success-failure check.

The next example makes the finite-population check explicit. It also illustrates that reporting the sample size and the population size helps readers assess the independence condition.

Worked Example: A Random Poll With a 10% Check

In a hypothetical campus with 12,000 students, a student association randomly selects 350 students to ask whether they support a new evening shuttle route. All selected students respond; 203 support it and 147 do not. Find a 95% confidence interval for the campus support proportion and check the conditions.

Design check. The selection is random. The sample is no more than 10% of the population because \(0.10(12{,}000)=1{,}200\), and \(350\leq1{,}200\). The independence condition is reasonable under this check.

Approximation check. There are 203 supporters and 147 non-supporters, both at least 10. The normal approximation check is satisfied.

Calculate. The observed support proportion is \(203/350=0.580\). Its estimated standard error is

SE=\sqrt{\frac{0.580(1-0.580)}{350}} =\sqrt{\frac{0.2436}{350}} =\sqrt{0.000696} \approx 0.02638.

Using \(z^*=1.96\), the margin of error is \(1.96(0.02638)\approx0.05171\). The interval is

0.580\pm0.05171=(0.5283,\ 0.6317).

The 95% confidence interval is about 52.8% to 63.2%. In context, it gives a range of plausible values for the proportion of all 12,000 students who support the evening shuttle route, provided the random-sampling and response assumptions are credible.

Confidence Level and the Width of the Interval

A higher confidence level uses a larger critical value and therefore produces a wider interval when the sample data are unchanged. This is not a change in the observed support proportion; it is a change in how much uncertainty the interval procedure is designed to cover over repeated samples. The next calculation shows how that choice enters a campus-poll result.

Worked Example: A 99% Interval From Poll Counts

Suppose, hypothetically, that a random poll of 500 students at a campus of 8,000 students asks whether they support a proposed study-space renovation. All students sampled respond, and 310 say yes. Calculate a 99% confidence interval.

The sample is less than 10% of the population because \(500\leq0.10(8{,}000)=800\). Random selection is stated, and the counts are 310 supporters and \(500-310=190\) non-supporters, so both design and success-failure checks are satisfactory.

The sample proportion is \(310/500=0.620\). The estimated standard error is

SE=\sqrt{\frac{0.620(1-0.620)}{500}} =\sqrt{\frac{0.2356}{500}} =\sqrt{0.0004712} \approx0.02171.

For a 99% confidence level, \(z^*\approx2.576\). The margin of error is \(2.576(0.02171)\approx0.05592\), giving

0.620\pm0.05592=(0.5641,\ 0.6759).

The 99% confidence interval is approximately 56.4% to 67.6%. It is wider than a 95% interval from the same sample would be because the larger critical value increases the margin of error. In context, the interval estimates the proportion of all students at this campus who support the renovation.

Common Mistake: Treating a Large Response Count as a Good Sample

A common mistake is to reason, “There were hundreds of responses and more than 10 yes and no answers, so the interval must represent the whole campus.” This mixes up two different issues. The success-failure counts help justify a normal approximation for the calculation; they do not establish that the respondents represent the target population. A voluntary poll can have plenty of responses and still overrepresent students more motivated to answer.

The correction is to assess the sampling design and the approximation separately. Ask whether the sample was randomly selected from the intended population, whether nonresponse could distort who is represented, and whether the independence and success-failure checks are reasonable. If the sample is not representative, the calculated interval describes the uncertainty from the model under its assumptions; it does not account for selection bias or other systematic errors.

Key takeaway: For a campus support poll, calculate \(\hat{p}=x/n\), check random sampling, independence, and the observed counts of supporters and non-supporters, then use \(\hat{p}\pm z^*\sqrt{\hat{p}(1-\hat{p})/n}\). Interpret the resulting range for the campus population only when the design and approximation assumptions are credible.

Check Your Understanding

Use the poll setting and interval method from this tutorial to answer the following questions.

A random sample of 250 students includes 145 supporters of a proposal. What are the sample proportion, the number of non-supporters, and the two observed success-failure counts?
A campus has 5,000 students, and 600 are sampled without replacement. Does the sample satisfy the 10% condition? Show the comparison.
Why do 200 responses to an opt-in campus poll not, by themselves, establish that a confidence interval represents all students?
In the first worked example, what campus proportion does the interval estimate, and why is it incorrect to say that 95% of individual students fall inside the interval?
If the confidence level is raised while the sample proportion and sample size stay the same, what happens to the margin of error and why?