Tom Hanks, Human Resources Manager with ANZ bank, is reviewing the employee training programs of ANZ branches. His staff randomly selected personnel files for 100 tellers in the Southern Region and determined that their mean training time was 25 hours. Assume that the population standard deviation is 5 hours. The 95% confidence interval for the population mean of training times is ________.

Question

Tom Hanks, Human Resources Manager with ANZ bank, is reviewing the employee training programs of ANZ branches. His staff randomly selected personnel files for 100 tellers in the Southern Region and determined that their mean training time was 25 hours. Assume that the population standard deviation is 5 hours. The 95% confidence interval for the population mean of training times is ________.

BlackTom AI · Accepted Answer

We’re given a sample mean of 25 hours, a population standard deviation of 5 hours, and a sample size of 100. The task is to form a 95% confidence interval for the population mean.

First, recall the formula for a z-based confidence interval when the population standard deviation is known: mean ± z*(sigma/√n).

Compute the standard error: sigma/√n = 5/√100 = 5/10 = 0.5.
Determine the z-value for 95% confidence: z ≈ 1.96 (the critical value from the standard normal distribution for 0.975 in each tail).
Calculate the margin of error: 1.96 × 0.5 = 0.98.
Construct the interval by adding and subtracting the margin from the sample mean: 25 − 0.98 = 24.02, and 25 + 0.98 = 25.98.

Now evaluate each option:
- Option "13 to 23": This interval is far from the calculated center and width; the margins would be about 12 on the low side and 2 on the high side, which doesn’t match the data. It is not plausible given a mean of 25 and the computed 0.98 margin.
- Option "0": An interval of length zero would imply no uncertainty about the mean, which contradicts the presence of a finite standard deviation, sample size, and the standard theory of confidence intervals. This is incorrect.
- Option "24.02 to 25.98": This interval centers at 25 and spans 0.98 on each side, matching the computed margin of error (0.98) with a standard error of 0.5 and z of 1.96. It is consistent with the given data and the 95% confidence level.
- Option "24.18 to 25.82": This interval has a half-width of 0.64, which would correspond to a margin of error of 0.64. That would require a z*(sigma/√n) of 0.64, i.e., a z-value of 0.64/0.5 = 1.28, not the 1.96 used for 95% confidence, so this option does not align with the 95% level.

In summary, the correct construction uses the sample mean 25, standard error 0.5, and 1.96 as the critical value, yielding a 95% CI of 24.02 to 25.98. The other options either misstate the interval bounds or imply an incorrect margin of error that does not correspond to a 95% confidence level.

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