Can someone tell me if this is done correctly or if I did it wrong and if it is can you please show me the right way to solve this.
I ended up with
x2>x20.05 = 7.815
df = 4-1 = 3
test statistics = 4.011
Since the calculated value of chi square is 4.011 is less than the crital value which is 7.815 we do not reject ho at 5% level of significance. So we conclude that men and women not select different breakfast.
Do men and women select different breakfasts? The breakfasts ordered by randomly selected men and women at a popular breakfast place is shown in Table 11.55. Conduct a test for homogeneity at a 5% level of significance.
|
|
French Toast
|
Pancakes
|
Waffles
|
Omelettes
|
Total
|
|
Men
|
47
|
35
|
28
|
53
|
163
|
|
Women
|
65
|
59
|
55
|
60
|
239
|
|
Total
|
112
|
94
|
83
|
113
|
402
|