In: Statistics and Probability
Researchers are investigating the number of car crashes occurring on a certain stretch of a highway. They classify the crashes as early in the week (Monday/Tuesday), mid-week (Wednesday/Thursday) or on the weekend (Friday/Saturday/Sunday). The number of crashes for each time period is given in the table below:
Early | Mid-Week | Weekend |
28 | 34 | 46 |
Use the appropriate hypothesis test at the 5% level to determine
if the proportions of crashes occurring on this stretch of highway
are different for the 3 time periods.
Null hypothesis: Ho: proportions of crashes occurring on this stretch of highway are same for the 3 time periods.
Alternate hypothesis: Ha: the proportions of crashes occurring on this stretch of highway are different for the 3 time periods.
degree of freedom =categories-1= | 2 | |||
for 0.05 level and 2 df :crtiical value X2 = | 5.991 | from excel: chiinv(0.05,2) | ||
Decision rule: reject Ho if value of test statistic X2>5.991 |
applying chi square goodness of fit test: |
relative | observed | Expected | Chi square | ||
Category | frequency(p) | Oi | Ei=total*p | R2i=(Oi-Ei)2/Ei | |
Early | 0.3333 | 28 | 36.00 | 1.78 | |
Mid-Week | 0.3333 | 34 | 36.00 | 0.11 | |
Weekend | 0.3333 | 46 | 36.00 | 2.78 | |
total | 1.00 | 108 | 108 | 4.67 | |
test statistic X2= | 4.667 |
since test statistic does not falls in rejection region we fail to reject null hypothesis |
we do not have have sufficient evidence to conclude that the proportions of crashes occurring on this stretch of highway are different for the 3 time periods. |