An environmental group wanted to estimate the proportion of fresh produce sales identified as organic in a local grocery store. In the winter, the group obtained a random sample of sales from the store and used the data to construct the 95 percent \(z\)-interval for a proportion (0.087, 0.133). Six months later in the summer, the group obtained a second random sample of sales from the store. The second sample was the same size as the first, and the proportion of sales identified as organic was 0.4. How does the 95 percent \(z\)-interval for a proportion constructed from the summer sample compare to the winter interval?

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For a population proportion, the confidence interval is:

$$ \text{statistic} \pm (\text{critical value})(\text{standard error of statistic}) $$ $$ = \hat{p}\pm (z^*)\left(\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right) $$

The point estimate for the winter sample is:

$$ \frac{0.087+0.133}{2} = 0.11$$

Thus the interval for the winer sample is:

$$ 0.11 \pm .023 $$

Because the population size and critical value are the same, the only thing that influences the interval is the sample proportion \(\hat{p}\). Since the summer sample has a greater value (0.40), the interval will be larger.

Compare the two: