You are the owner of three local sandwich shops. Recently you\'ve received compl

Question

You are the owner of three local sandwich shops. Recently you've received complaints from customers that their "footlong" sub sandwiches were less than 12 inches long. As a measure of quality assurance, data was collected on all footlong sub sandwiches made over the period of an hour at each store. Determine which, if any, of your stores are making sub sandwiches that are significantly shorter than 12 inches. You will conduct three t-tests, one for each store. You want to be 95% confident in your results. 1. What is the null hypothesis for these tests? 2. What is the alternative hypothesis for these tests? 3. Given the desired confidence level, what alpha should be used? 4. Can you say with confidence that Store 1 is producing sub sandwiches with an average length that is less than 12 inches? 5. Can you say with confidence that Store 2 is producing sub sandwiches with an average length that is less than 12 inches? 6. Can you say with 95% confidence that Store 3 is producing sub sandwiches with an average length that is less than 12 inches?

Explanation / Answer

1. Null Hypothesis - The length of footlong subsandwich is equal to 12 inches.

2. Alternative Hypothesis - The length of footlong subsandwich is less than 12 inches.

3. Confidence level = 95%, so alpha = 100 - 95 = 5 % = 0.05

4. Difference in mean = -1.6

Standard error = standard deviation / sqrt(n) = 1.52/sqrt(18) = 0.3582

degree of freedom, df = n-1= 18-1 = 17

t-statistic = Difference in mean/Standard error = -1.6 / 0.3582 = -4.466

Using t-table, p-value for t=-4.466 and df = 17, p = 0.0001.

As p is less than the alpha (0.05), we reject the null hypothesis and accept the alternative hypothesis that the length of footlong subsandwich is less than 12 inches.

5. Difference in mean = -1

Standard error = standard deviation / sqrt(n) = 1.5/sqrt(8) = 0.5303

degree of freedom, df = n-1= 8-1 = 7

t-statistic = Difference in mean/Standard error = -1 / 0.5303 = -1.885

Using t-table, p-value for t=-1.885 and df = 7, p = 0.0507.

As p is greater than the alpha (0.05), we fail to reject the null hypothesis and infer that the length of footlong subsandwich is equal to12 inches.

6. Difference in mean = 0.1

Standard error = standard deviation / sqrt(n) = 2.18/sqrt(20) = 0.4874

degree of freedom, df = n-1= 20-1 = 19

t-statistic = Difference in mean/Standard error = 0.1 / 0.4874 = 0.2051

Using t-table, p-value for t=0.2051 and df = 19, p = 0.5801.

As p is greater than the alpha (0.05), we fail to reject the null hypothesis and infer that the length of footlong subsandwich is equal to12 inches.

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