Question 1064093: Can some please show me how to use the chi-squared distribution to test the hypothesis of this problem?
A restaurant owner wants to see if the business is good enough for him to purchase a restaurant. He asks the present owner for a breakdown of how many customers that come in for lunch each day and the results are as follows: Monday - 20, Tuesday - 30, Wednesday - 25, Thursday - 40 and Friday - 55. The prospective owner observes the restaurant and finds the following number of customers coming for lunch each day: Monday- 30, Tuesday - 15, Wednesday- 7, Thursday 40, and Friday - 33. At a 95% confidence level determine whether the present owner reported the correct number of customers for lunch each day.
Thank you in advance
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! restaurant owner's sample is 20+30+25+40+55 = 170
p1 = 20/170 = 0.12 for Monday
p2 = 30/170 = 0.18 for Tuesday
p3 = 25/170 = 0.15 for Wednesday
p4 = 40/170 = 0.23 for Thursday
p5 = 55/170 = 0.32 for Friday
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prospective owner's sample is 30+15+7+40+33 = 125
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H0: The distribution of the prospective owners sample is 0.12, 0.18, 0.15, 0.23, 0.32
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H1: H0 is false
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strictly speaking in order to use chi-squared distribution we must check for
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min(0.12(125) +0.18(125), 0.15(125), 0.23(125), 0.32(125)) > 5
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min(15, 22, 19, 29, 40) > 5, yes since 15 > 5
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degrees of freedom = 5 - 1 = 4 and a 95% confidence level
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the X^2 table for 4 degrees of freedom and a = 0.05 ( 1 - 95/100)
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the critical value is 9.49
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we reject H0 if X^2 > or = 9.49
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X^2 = summation from 1 to 5 ( observed value - expected value)^2 / expected value
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Note the expected values are (15, 22, 19, 29, 40) and observed values are (30, 15, 7, 40, 33)
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X^2 = 30.02
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we reject H0 - the owner did not report the correct number of customers
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