Organize the data into a two-by-two contingency table, with buses above and below the median of each variable. Determine whether the age of the bus is related to the amount of the maintenance cost. Use 0.05 significance level. (chi-saqure)
Median | |||
Maint. Cost | 4178.5 | ||
Age | 7 | ||
Contingency Table (Observed) | |||
Old | New | Total | |
High Maint. | 31 | 9 | 40 |
Low Maint. | 7 | 33 | 40 |
Total | 38 | 42 | 80 |
We first compute the expected value of each cell here as:
Old | New | |
High Maintenance | 40*38/80 = 19 | 40*42/80 = 21 |
Low Maintenance | 40*38/80 = 19 | 40*42/80 = 21 |
The chi square test statistic now is computed from the given observed frequencies and expected frequencies computed above as:
Df = (num of col - 1)(num of row - 1) = 1
Therefore the p-value here is computed from the chi square distribution tables as:
As the p-value here is approx. 0, therefore the test is significant and we can conclude here that we have sufficient evidence that age of the bus is related to the amount of the maintenance cost
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