A local franchise of a national chain of day-old pastry stores recorded the numb
ID: 3159418 • Letter: A
Question
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 – 15%, Monday – 9%, Tuesday – 12%, Wednesday – 11%, Thursday – 12%, Friday – 16%, and Saturday – 25%.The manager doesn’t trust the franchisor and wants to show that they are wrong.Use a ² Goodness-of-Fit test to determine if the manager has evidence that the national franchisor is wrong.
Customers per day First five weeks after September Sun 32 Mon 23 Tue 30 Wed 30 Thu 38 Fri 40 Sat 84 Sun 35 Mon 15 Tue 32 Wed 32 Thu 37 Fri 44 Sat 68 Sun 48 Mon 30 Tue 36 Wed 29 Thu 34 Fri 48 Sat 72 Sun 53 Mon 30 Tue 38 Wed 37 Thu 39 Fri 54 Sat 83 Sun 50 Mon 32 Tue 40 Wed 32 Thu 36 Fri 58 Sat 81Explanation / Answer
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 – 15%, Monday – 9%, Tuesday – 12%, Wednesday – 11%, Thursday – 12%, Friday – 16%, and Saturday – 25%.The manager doesn’t trust the franchisor and wants to show that they are wrong. Use a ² Goodness-of-Fit test to determine if the manager has evidence that the national franchisor is wrong.
Week1
week2
week3
week4
week5
total
%
Expected
32
35
48
53
50
218
15
225
23
15
30
30
32
130
9
135
30
32
36
38
40
176
12
180
30
32
29
37
32
160
11
165
38
37
34
39
36
184
12
180
40
44
48
54
58
244
16
240
84
68
72
83
81
388
25
375
1500
Goodness of Fit Test
observed
expected
O - E
(O - E)² / E
218
225.000
-7.000
0.218
130
135.000
-5.000
0.185
176
180.000
-4.000
0.089
160
165.000
-5.000
0.152
184
180.000
4.000
0.089
244
240.000
4.000
0.067
388
375.000
13.000
0.451
1500
1500.000
0.000
1.250
1.25
chi-square
6
df
.9744
p-value
DF=7-1=6
Chi square value at 0.05 level =12.592
Calculated chi square =1.25 < 12.592 the table value.
The null hypothesis is not rejected.
The manager has no evidence that the national franchisor is wrong.
Week1
week2
week3
week4
week5
total
%
Expected
32
35
48
53
50
218
15
225
23
15
30
30
32
130
9
135
30
32
36
38
40
176
12
180
30
32
29
37
32
160
11
165
38
37
34
39
36
184
12
180
40
44
48
54
58
244
16
240
84
68
72
83
81
388
25
375
1500
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