Question

question 2
a random sample of 1,200 college students is surveyed about video - streaming habits, and 870 of these students indicate that they are subscribed to at least one paid streaming service. when a 95% confidence interval is constructed based on this sample of data, the interval is found to be from 0.701 to 0.759. a colleague argues that if the researchers had instead used a 99% level of confidence, the 99% confidence interval would reduce uncertainty and make the estimate more exact. is the colleague correct?
yes, the colleague is correct because a higher confidence level results in a smaller margin of error.
no, the colleague is incorrect because the confidence level only affects how likely it is that the true population proportion falls outside the interval.
yes, the colleague is correct because a 99% confidence interval must be narrower than a 95% confidence interval in order to indicate stronger evidence.
no, the colleague is incorrect because a higher confidence level leads to a wider interval and therefore less precision.
yes, the colleague is correct because both intervals will have the same center and only slightly different widths.

Explanation:

The formula for a confidence interval for a proportion is \(\hat{p}\pm z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\), where \(z\) is the critical value. A higher confidence level (e.g., 99% instead of 95%) corresponds to a larger \(z\) - value. A larger \(z\) - value increases the margin of error (\(z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\)), making the confidence interval wider. A wider interval means less precision.

Answer:

No, the colleague is incorrect because a higher confidence level leads to a wider interval and therefore less precision.