A popular restaurant takes a random sample n = 25 customers and records how long each occupied a table. They found a sample mean of 1.2 hours and a sample standard deviation of 0.3 hours. Find the 95 percent confidence interval for the mean.

Question

BlackTom AI · Accepted Answer

We start by identifying the task: construct a 95% confidence interval for the population mean based on a sample.

- Given: n = 25, sample mean x̄ = 1.2 hours, sample standard deviation s = 0.3 hours.
- Since the population standard deviation is unknown and the sample size is small (n < 30), we should use the t-distribution with df = n − 1 = 24, rather than the normal z-distribution.
- The standard error of the mean is SE = s / sqrt(n) = 0.3 / sqrt(25) = 0.3 / 5 = 0.06 hours.
- For a 95% confidence level and 24 degrees of freedom, the critical t-value (two-tailed) is approximately t* ≈ 2.064.
- The margin of error (ME) is ME = t* × SE ≈ 2.064 × 0.06 ≈ 0.12384, which rounds to about 0.124.
- Therefore, the 95% confidence interval for the mean is x̄ ± ME ≈ 1.2 ± 0.124, corresponding to the interval from roughly 1.076 to 1.324 hours.

Now evaluate the answer choices:
- Option A: "1.2 ± .124" matches the computed margin of error and the central estimate, yielding the interval 1.076 to 1.324. This aligns with the t-based 95% CI for n = 25.
- Option B: "1.2 ± .118" uses a margin of error of 0.118, which corresponds to using a z-value (1.96) rather than the t-value, or rounding differences that are not appropriate here with s unknown and n small; thus this is not the correct t-based 95% CI.
- Option C: "1.2 ± .609" is far too wide; a margin of 0.609 would imply an enormous standard error or an incorrect calculation, not plausible from s = 0.3 and n = 25.
- Option D: "1.2 ± .588" is also far too wide and inconsistent with the data; it would imply a much larger standard error than provided.

In summary, the approach uses the proper t-distribution for a small sample with unknown population standard deviation, yields a margin of error about 0.124, and the option that matches this result is the one labeled 1.2 ± .124.

OSC 2030-007 Fall 2025 Exam 2

A popular restaurant takes a random sample n = 25 customers and records how long each occupied a table. They found a sample mean of 1.2 hours and a sample standard deviation of 0.3 hours. Find the 95 percent confidence interval for the mean.

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