In constructing a confidence interval for a mean with unknown variance with a sample of 25 items, Bob used z instead of t. “Well, at least my interval will be wider than necessary, so it was a conservative error,” said he. Is Bob’s statement correct?
A. Yes.
B. No.
C. It depends on µ.
The correct answer and explanation is :
The correct answer is B. No.
Explanation:
In constructing a confidence interval for a population mean when the population variance is unknown, the correct approach is to use the t-distribution, not the z-distribution. This is because when the population variance is unknown, we have to estimate it from the sample, and the sample mean follows a t-distribution rather than a normal distribution.
The general formula for a confidence interval when using the t-distribution is:
[
\bar{x} \pm t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}}
]
Where:
- (\bar{x}) is the sample mean.
- (t_{\alpha/2, n-1}) is the critical value from the t-distribution with (n-1) degrees of freedom.
- (s) is the sample standard deviation.
- (n) is the sample size.
- (\alpha) is the significance level (e.g., for a 95% confidence interval, (\alpha = 0.05)).
The z-distribution, on the other hand, is used when the population variance is known or the sample size is large (typically (n > 30), where the central limit theorem applies and the sample mean is approximately normally distributed even if the population variance is unknown).
Bob’s mistake was using the z-distribution instead of the t-distribution. The z-distribution assumes that the population variance is known, which is not the case here. The critical value for the t-distribution is larger than the z-value for smaller sample sizes, so using the z-distribution instead of the t-distribution results in an incorrectly narrower confidence interval.
Bob incorrectly assumed that using the z-distribution would lead to a wider interval and thus a more conservative estimate. In fact, the z-distribution would result in a narrower confidence interval because the critical value associated with the z-distribution is smaller than that of the t-distribution (especially for smaller sample sizes like 25). Therefore, his statement is incorrect.
In summary:
- The correct distribution to use is the t-distribution when the population variance is unknown and the sample size is small.
- Using the z-distribution would lead to an incorrectly narrower confidence interval, not a wider one, as Bob assumed.