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## Answer:

No answer provided yet.NOTE: THIS QUESTION WAS REMOVED DUE TO REQUEST FROM A UNIVERSITY LEGAL DEPARTMENT

Original Question
A potato chip company packages its potato chips into 12.0 ounce bags. You find it hard to believe that the bag contains enough potato chips to weigh 12.0 ounces and would like to make an official complaint. Before doing so, you decide to run an experiment so that you can have some confidence that the company’s claim is incorrect. Over the next several months you buy 30 bags of potato chips and weigh the contents of each one. You discover that the mean weight is 11.9 ounces with a standard deviation of 0.4 ounces. You decide that you will only complain if you can be 95% sure that the bags do not contain at least 12.0 ounces of potato chips. You decide to construct a hypothesis test.

Task:
A.  Determine if this is a one-tailed or two-tailed test. Justify your decision.
B.  State the null hypothesis and alternative hypothesis, including the parameters on both statements. Your null hypothesis should assume the company’s claim is correct.
C.  In a discussion of type I errors, do the following:
1.  Define the term type I error.
2.  Explain what a type I error is in terms of this problem.

D.  In a discussion of level of significance, do the following:
1.  Define the term level of significance.
2.  Identify the level of significance for this problem.
E.  Calculate the test statistic as a z-score by using the z-test formula. Show all relevant work.
F.  Using a standard table, you determine that the critical value is –1.645.
1.  Determine if the test statistic allows you to reject the null hypothesis.
2.  Explain how you reached this conclusion, including a comment relating the results to the original problem.

Answer

A. This is a 1-tailed test since we are testing whether the average is less than the advertised amount.

B. The Null hypothesis is that the mean weight of all potato chip bags is 12 ounces. Ho: μ = 12.

The alternative hypothesis is that the mean weight of all potato chip bags is less than 12 ounces: Ha: μ < 12

C1: A type I error is the probability of stating there is a difference, when the difference doesn't really exists. Type I errors refer to the probability of incorrectly rejecting the NULL hypothesis. It is expressed as α and by convention is set to .05.

C2: The Type I error would be incorrectly saying the mean weight of potato chips is less than 12 ounces, if in actuality they are equal to or greater than 12 ounces.  Since we need to be 95% sure of our decision, we're setting the Type I error threshold (α) to .05.

D1: The level of significance refers to the degree of confidence we can have that our observed results are not due to chance alone. The level of significance is 1- α, and by convention, since α usually = .05, the level of significance = .95.

D2: The level of significance has been defined as 95%.

E: The test statistic z* is calculated using the following steps.

1. Subtract the sample mean from the test mean = 11.9-12 = -1

2. Calculate the standard error of the mean (SE), which is the standard deviation divided by the square root of the sample size = .04/SQRT(30) = .07303

3. The test statistic z* is Step 1 divided by Step 2's results = -1/.07303 = -1.36931

F1 : The test-statistic z* of -1.36931 is less than the critical value of -1.645 so we would NOT reject the NULL hypothesis.

F2: We would fail to reject the NULL hypothesis since our test statistics fails to exceed the critical value for a 95% level of confidence (α of .05). This means that there is not enough evidence, given the sample size and variability in the sample, for us to conclude the mean difference of 1 ounce is not due to chance alone.  Since the test statistic is reasonably close to the critical value (the p-value is .085), perhaps a larger sample size might provide more evidence for us to reject the NULL and conclude the claim made about the mean bag weight is incorrect. For now, with the evidence collected, we cannot reject the company's claim.

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