Here is the full question:
The fill of a cereal box machine is required to 18 ounces, with the variance already established at 0.24. The past 150 boxes revealed an average of 17.96 ounces. A testing error of α = 2% is considered acceptable. If the box overfills, profits are lost; if the box underfills, the consumer is cheated. (a) Can we conclude that machine is set at 18 ounces? (b) Find a 98% confidence interval for μ. (c) Would the result change if the sample variance were 0.24 from the data rather than knowing $σ^2$ = 0.24?
My Answer:
(a) I calculated the positive square root of the variance (0.24) to get the standard deviation of the population - σ = 0.49. I found the acceptable region to be between -2.33 and 2.33, so, for a value of z = -0.9998, I concluded that the machine is set at 18 ounces.
(b) 17.87 < μ < 18.05
(c) I understood the question as: "will there be a change in the result of part (b) if 0.24 changed from being a population variance to a sample variance". My answer is that when only the standard deviation of a sample is known and n>30, the same formula will be used to calculate the 98% confidence interval, so there will be no change in the result of part (b).
Parts (a) and (b) are correct. (Although for part (a) one might quibble that we are not accepting that the machine pours an average of $18$ ounces, rather failing to reject that it does. In any event this quibble is for whoever wrote the problem, not you.)
Part (c) is right in the sense that for a sample size as large as $150,$ using a normal distribution rather than a t distribution doesn't change the answer significantly. There is a theoretical difference though.