The California Transit system would like to see if the type of transportation used depended on the time of day: Please use the following table to answer questions:
Transportation Total |
Bus |
Carpool |
Own Car |
Total |
||||
Time of Day |
Observed |
Expected |
Observed |
Expected |
Observed |
Expected |
||
Morning |
60 |
30 |
30 |
120 |
||||
Night |
20 |
20 |
40 |
80 |
||||
Total |
80 |
50 |
70 |
200 |
Solution:
Here, we have to use chi square test for independence of two categorical variables.
Null hypothesis: H0: The type of transportation used not depended on the time of day.
Alternative hypothesis: Ha: The type of transportation used depended on the time of day.
We assume level of significance = α = 0.05
Test statistic formula is given as below:
Chi square = ∑[(O – E)^2/E]
Where, O is observed frequencies and E is expected frequencies.
E = row total * column total / Grand total
What is the critical value of χ2?
We are given
Number of rows = r = 2
Number of columns = c = 3
Degrees of freedom = df = (r – 1)*(c – 1) = 1*2 = 2
α = 0.05
Critical value = 5.991465
(by using Chi square table or excel)
Calculation tables for test statistic are given as below:
Observed Frequencies |
||||
Type of transportation |
||||
Time of Day |
Bus |
Carpool |
Own Car |
Total |
Morning |
60 |
30 |
30 |
120 |
Night |
20 |
20 |
40 |
80 |
Total |
80 |
50 |
70 |
200 |
Expected Frequencies |
||||
Type of transportation |
||||
Time of Day |
Bus |
Carpool |
Own Car |
Total |
Morning |
48 |
30 |
42 |
120 |
Night |
32 |
20 |
28 |
80 |
Total |
80 |
50 |
70 |
200 |
Transportation Total |
Bus |
Carpool |
Own Car |
Total |
||||
Time of Day |
Observed |
Expected |
Observed |
Expected |
Observed |
Expected |
||
Morning |
60 |
80*120/200 = 48 |
30 |
50*120/200 = 30 |
30 |
70*120/200 = 42 |
120 |
|
Night |
20 |
80*80/200 = 32 |
20 |
50*80/200 = 20 |
40 |
70*80/200 = 28 |
80 |
|
Total |
80 |
80 |
50 |
50 |
70 |
70 |
200 |
The variable time of the day represents rows.
The variable type of transportation represents columns.
Calculations |
||
(O - E) |
||
12 |
0 |
-12 |
-12 |
0 |
12 |
(O - E)^2/E |
||
3 |
0 |
3.428571 |
4.5 |
0 |
5.142857 |
What is the value of χ2, for independent samples?
Chi square = ∑[(O – E)^2/E] = 16.07143
P-value = 0.000324
(By using Chi square table or excel)
P-value < α = 0.05
So, we reject the null hypothesis
Conclusion:
There is sufficient evidence to conclude that the type of transportation used depended on the time of day.
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