Customer Distribution by Weekday: A drop-in auto repair shop staffs the same number of mechanics on...
Customer Distribution by Weekday: A drop-in auto repair shop staffs the same number of mechanics on every weekday (weekends are not counted here). One of the mechanics thinks this is a bad idea because he suspects the number of customers is not evenly distributed across these days. For a sample of 289 customers, the counts by weekday are given in the table.
Number of Customers by Day (n = 289)
| Monday | Tuesday | Wednesday | Thursday | Friday | |
| Count | 43 | 48 | 57 | 75 | 66 |
The Test: Test the claim that the number of
customers is not evenly distributed across the five weekdays. Test
this claim at the 0.01 significance level.
(a) The table below is used to calculate the test statistic.
Complete the missing cells.
Round your answers to the same number of decimal places as
other entries for that column.
| Week | Observed | Assumed | Expected | ||||
| i | Day | Frequency (Oi) | Probability (pi) | Frequency Ei |
|
||
| 1 | Monday | ? | 0.2 | 57.8 | 3.790 | ||
| 2 | Tuesday | 48 | ? | 57.8 | 1.662 | ||
| 3 | Wednesday | 57 | 0.2 | ? | 0.011 | ||
| 4 | Thursday | 75 | 0.2 | 57.8 | ? | ||
| 5 | Friday | 66 | 0.2 | 57.8 | 1.163 | ||
| Σ | n = 289 | χ2 = ? | |||||
(b) What is the value for the degrees of freedom?
(c) What is the critical value of χ2
(Use the answer found in the
χ2-table or round to 3 decimal
places.)
tα =
(d) What is the conclusion regarding the null hypothesis?
reject H0
fail to reject H0
(e) Choose the appropriate concluding statement.
We have proven that the number of customers is evenly distributed across the five weekdays.
The data supports the claim that the number of customers is not evenly distributed across the five weekdays.
There is not enough data to support the claim that the number of customers is not evenly distributed across the five weekdays.
Homework Answers
a)
| Week | Observed | Assumed | Expected | ||
| i | Day | Frequency (Oi) | Probability (pi) | Frequency Ei | (Oi − Ei)2 |
| Ei | |||||
| 1 | Monday | 43 | 0.2 | 57.8 | 3.790 |
| 2 | Tuesday | 48 | 0.2 | 57.8 | 1.662 |
| 3 | Wednesday | 57 | 0.2 | 57.8 | 0.011 |
| 4 | Thursday | 75 | 0.2 | 57.8 | 5.118 |
| 5 | Friday | 66 | 0.2 | 57.8 | 1.163 |
| Σ | n = 289 | χ2 =11.744 | |||
b)
degrees of freedom =5-1=4
c)
critical value of χ2 =13.277
d)
fail to reject H0
e) There is not enough data to support the claim that the number of customers is not evenly distributed across the five weekdays.
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