
Mary Ann S. answered 07/19/22
Ph.D. Educational Measurement, Doctoral Minor in Statistics.
First, figure out what the problem is asking:
- Confidence interval of a mean.
Next examine what information you have been given:
- sample mean = 53 ounces
- population standard deviation assumed known. 12.4 ounces.
- N = 29.
- population standard deviation is known, but, technically, we should use a t-test because the sample size is < 30. Let's run both a Z and a single sample t-test and compare the results!
- calculate the standard error: sigma/sqrt(N) = 12.4/sqrt(29) = 2.228
- The critical value for a 95% confidence interval for Z = 1.96
- The critical value for a 95% confidence interval for t on N -1 = 28 degrees of freedom = 2.048
CI95 using z scores = 53 +- 1.96*2.228 = 53 +- 4.3669
CI95 using t scores = 53 +- 2.048*2.228 = 53 +- 4.5629
The t-score-based interval is only very slightly wider than that based on the z score.