Do NOT use the finite population corrections for standard deviation calculation in this problem. Just assume we're
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Question:
Do NOT use the "finite population" corrections for standard deviation calculation in this problem. Just assume we're dealing with an infinite population.
- Provide the formal null and alternative hypotheses for a hypothesis test of the question of whether or not the mean of tub sales is really $93,500.
Assume that the underlying population of tub sales is normally distributed.
- Conduct this hypothesis test based on the 95% significance level. What is the p-value? Do you reject or fail to reject your null hypothesis? Why?
- How do I create a 90% confidence interval for the mean tub sales. Explain in a sentence, in layman's terms, what this confidence interval implies.
- Compare your hypothesis test in part "b" to your confidence interval in part "c". Do these results conflict with each other? (i.e., is it OK to find these two answers simultaneously?...if so, why?...OR, is there a conflict where it should be impossible for these results to occur simultaneously?)
- If I told you that the underlying population of tub sales was NOTnormally distributed, how would this change your response to parts "b" and "c"?
- If I told you that the underlying population of tub sales was NOTnormally distributed, BUT, your final calculations (p-value for "b", confidence interval for "c") were based on double the sample size (a sample of 40 salesmen, rather than 20), would this change your response to part "e"? If so, why? If not, why not?
Tub Sales ($) | College Degree (1 = Yes) |
107000 | 1 |
87300 | 0 |
103000 | 1 |
97700 | 1 |
111600 | 1 |
95700 | 0 |
91200 | 0 |
119600 | 1 |
94200 | 0 |
91300 | 1 |
88700 | 0 |
109700 | 1 |
110400 | 0 |
90800 | 0 |
118300 | 1 |
91500 | 0 |
88800 | 0 |
97400 | 1 |
71800 | 0 |
106900 | 1 |
87000 | 0 |
84500 | 0 |
91100 | 1 |
95800 | 0 |
104100 | 1 |
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