In: Statistics and Probability
Is the transportation mode used to ship goods independent of
type of industry? Suppose the following contingency table
represents frequency counts of types of transportation used by the
publishing and the computer hardware industries. Analyze the data
by using the chi-square test of independence to determine whether
type of industry is independent of transportation mode. Let α =
0.05.
Transportation Mode | |||||||||||||
Industry |
|
(Round the intermediate values to 2 decimal places.
Round your answer to 2 decimal places.)
Observed χ2 =
Transportation mode is
independentnot independent of
industry.
Hypotheses are:
H0: Transportation mode is independent of industry.
Ha: Transportation mode is not independent of industry.
Followng table shows the row and column total:
Air | Train | Truck | Total | |
Publishing | 34 | 12 | 40 | 86 |
Computer Hardwares | 5 | 6 | 24 | 35 |
Total | 39 | 18 | 64 | 121 |
Expected frequencies will be calculated as follows:
Following table shows the expected frequencies:
Air | Train | Truck | Total | |
Publishing | 27.72 | 12.79 | 45.49 | 86 |
Computer Hardwares | 11.28 | 5.21 | 18.51 | 35 |
Total | 39 | 18 | 64 | 121 |
Following table shows the calculations for chi square test statistics:
O | E | (O-E)^2/E |
34 | 27.719 | 1.42 |
5 | 11.281 | 3.5 |
12 | 12.793 | 0.05 |
6 | 5.207 | 0.12 |
40 | 45.488 | 0.66 |
24 | 18.512 | 1.63 |
Total | 7.38 |
Following is the test statistics:
Degree of freedom: df =( number of rows -1)*(number of columns-1) = (2-1)*(3-1)=2
The p-value is: 0.0250
Conclusion: Since p-value is less than 0.05 so we reject the null hypothesis. That is we can conclude that transportation mode is not independent of industry..
Excel function used for p-value: "=CHIDIST(7.38, 2)"