First we need to adjust the standard deviation to compensate for the sample size n=25, so we need:
s=σ/√n=2/5=0.4g/100mL, because in a large population the spread of data will be greater than in a sample, so the standard deviation of the comparatively small sample needs to be smaller. s is the standard error of the mean, and is a special kind of standard deviation.
(a) 95% corresponds to a Z-score of 1.96, that is, 1.96 standard deviations above or below the mean. Because we are dealing with an interval where the mean sits in the middle of the interval, we have a lower and upper limit, which is why we refer to a 2-tailed value for the Z-score. The corresponding 1-tailed value would be 97.5%, because 2.5% of the distribution lies in each tail, the combined tails make 5%, hence the 95% CI.
s=0.4g/100mL, 1.96s=0.784g/100mL, CI=(15±0.784)g/100mL or [14.22,15.78] approx. If the mean of the sample is used instead of the population mean the CI for the sample=(13±0.784)g/100mL or [12.22,13.784].
(b) s=2/6=⅓, 2/7, 2/8=¼ respectively for the three sample sizes. The corresponding sample CIs are:
13±0.65 or [12.35,13.65], 13±0.56 or [12.46,13.56]. 13±0.49 or [12.51,13.49].
(c) As the sample size increases the standard error decreases so the CI decreases. The reason is that there should be less deviation from the mean the more data can be sampled.
