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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).

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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.

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  • $\begingroup$ So, technically, my answer to part c is correct? $\endgroup$ Oct 14, 2017 at 8:26
  • $\begingroup$ Yeah, it's definitely a valid answer and probably what they are looking for. I was just saying it kind of depends on the definition of the word "change". $\endgroup$ Oct 14, 2017 at 15:13

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