A local franchise of a national chain of day-old pastry stores recorded the number of customers that came in to the store for five weeks this fall. The national franchisor states that the percentage of customers is distributed as follows:
Sunday – 20%,
Monday – 8%,
Tuesday – 6%,
Wednesday – 7%,
Thursday – 12%,
Friday – 22%,
Saturday – 25%.
The manager doesn’t trust the franchisor and wants to show his store has different percentages. Use a χ² Goodness-of-Fit test to determine if the manager has evidence that his store is varies from the national percentages.
First five weeks after September Customers per day | |
Sun | 53 |
Mon | 22 |
Tue | 18 |
Wed | 24 |
Thu | 36 |
Fri | 51 |
Sat | 83 |
Sun | 69 |
Mon | 14 |
Tue | 19 |
Wed | 24 |
Thu | 37 |
Fri | 57 |
Sat | 68 |
Sun | 60 |
Mon | 23 |
Tue | 16 |
Wed | 22 |
Thu | 34 |
Fri | 60 |
Sat | 72 |
Sun | 68 |
Mon | 22 |
Tue | 17 |
Wed | 25 |
Thu | 39 |
Fri | 64 |
Sat | 83 |
Sun | 84 |
Mon | 25 |
Tue | 15 |
Wed | 19 |
Thu | 36 |
Fri | 60 |
Sat | 81 |
1500 |
The Hypothesis:
H0: The sample has a distribution that agrees with the national percentage.
Ha: The sample has a distribution that is different from the national percentage.
The Test Statistic:
Each Expected value = (% / 100) * 1500
Observed | Expected % | Expected | (O-E)2 | (O-E)2/E | |
Sunday | 334 | 20 | 300 | 1156 | 3.85333 |
Monday | 106 | 8 | 120 | 196 | 1.63333 |
Tuesday | 85 | 6 | 90 | 25 | 0.27778 |
Wednesday | 114 | 7 | 105 | 81 | 0.77143 |
Thursday | 182 | 12 | 180 | 4 | 0.02222 |
Friday | 292 | 22 | 330 | 1444 | 4.37576 |
Saturday | 387 | 25 | 375 | 144 | 0.38400 |
Total | 1500.00 | 100.00 | 1500.00 | 11.318 |
test = 11.318
The Critical Value: The critical value at = 0.05, df= n - 1 = 6
critical = 11.07
The p value: The p value: The p value at test = 11.318, df = 6; P value = 0.0454
The Decision Rule:
The Critical Value Method: If test is > critical, then Reject H0.
The p - value Method: If p value is < , Then Reject H0.
The Decision:
The Critical Value Method: Since test (11.318) is > critical (11.07), We Reject H0.
The p - value Method: Since p value (0.0454) is < (0.05), We Reject H0.
The Conclusion: There is sufficient evidence at the 95% significance level to conclude that sample has a distribution that is different from the national percentage.
A local franchise of a national chain of day-old pastry stores recorded the number of customers that came in to the store for five weeks this fall. The national franchisor states that the percentage o...
A local franchise of a national chain of day-old pastry stores recorded the number of customers that came in to the store for five weeks this fall. These data are recorded in the attached file. The national franchisor states that the percentage of customers is distributed as follows: Sunday – 20%, Monday – 8%, Tuesday – 6%, Wednesday – 7%, Thursday – 12%, Friday – 22%, and Saturday – 25%. The manager doesn’t trust the franchisor and wants to show...
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