In: Math
A shoe salesman wants to see if his female customers have a preference in the color of shoe purchased. He notes the color preferences of 100 randomly selected customers. The results: Black=32, Brown=27, Red=15, Navy =13 White =13
The Chi-Square goodness of fit test is used here to test whether the sample are from same population or we can say whether the observed value of frequency are same as the expected frequency.
The Null hypothesis is defined as,
Null hypothesis: There is no significant difference between the observed frequency and the expected frequency.
The observed values are
Observed values | |
Color | Frequency |
Black | 32 |
Brown | 27 |
Red | 15 |
Navy | 13 |
White | 13 |
The expected frequencies are,
Expected values | |
Color | Frequency |
Black | 20 |
Brown | 20 |
Red | 20 |
Navy | 20 |
White | 20 |
The Chi-Square statistic is obtained using the formula,
Color |
|
Expected, | ||||
Black | 32 | 20 | 12 | 144 | 7.2 | |
Brown | 27 | 20 | 7 | 49 | 2.45 | |
Red | 15 | 20 | -5 | 25 | 1.25 | |
Navy | 13 | 20 | -7 | 49 | 2.45 | |
White | 13 | 20 | -7 | 49 | 2.45 | |
Sum | 15.8 |
The P-value for the chi square is obtained from chi square distribution table for chi square = 15.8 and degree of freedom = k - 1 = 5 - 1 = 4
Since,
at 5% significance level. it can be concluded that the null hypothesis is rejected. Hence we can conclude that female customers have a preference in the color of shoe purchased.