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 $\alpha =2\mathrm{\%}$ 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 $\mu$. (c) Would the result change if the sample variance were 0.24 from the data rather than knowing ${\sigma}^{2}=0.24$?

My Answer:

(a) I calculated the positive square root of the variance (0.24) to get the standard deviation of the population - $\sigma =0.49$. 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) $17.87<\mu <18.05$

(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).