In: Statistics and Probability
The contingency table to the right shows counts of the types of gasoline bought during the week and during weekends. (a) Find the value of chi-squared and Cramer's V for this table. (b) Interpret these values. What do these tell you about the association in the table? |
|
(a)
chiχsquared2equals=nothing
(Round to two decimal places as needed.)
Vequals=nothing
(Round to two decimal places as needed.)
(b) Interpret the values from part (a). What do they indicate about the association?
Since
▼
Upper VV
chi squaredχ2
is
▼
somewhat large
rather small
large
, there is
▼
strong
weak
moderate
association between the two variables.
chart with given information
Premium - Weekday 127 Weekend 124 Total 251
Plus- Weekday 100 Weekend 203 Total 303
Regular-Weekday 487 Weekend 128 Total 615
Total-Weekday 714 Weekend 455 Total 1169
Since V or X to the second is somewhat large rather small or large there is strong weak or moderate association between the two variables.
a)
Applying chi square test of independence: |
Observed | weekday | weekend | Total | |
Premium | 127 | 124 | 251 | |
Plus | 100 | 203 | 303 | |
regular | 487 | 128 | 615 | |
total | 714 | 455 | 1169 | |
Expected | Ei=row total*column total/grand total | weekday | weekend | Total |
Premium | 153.305 | 97.695 | 251.00 | |
Plus | 185.066 | 117.934 | 303.00 | |
regular | 375.629 | 239.371 | 615.00 | |
total | 714.00 | 455.00 | 1169.00 | |
chi square χ2 | =(Oi-Ei)2/Ei | weekday | weekend | Total |
Premium | 4.514 | 7.083 | 11.5967 | |
Plus | 39.101 | 61.358 | 100.4587 | |
regular | 33.021 | 51.817 | 84.8380 | |
total | 76.6352 | 120.2583 | 196.893 | |
test statistic X2 = | 196.89 |
Cramer V=φc=√(X2/(n*(k-1)))=sqrt(196.89/(1169*(2-1)) = | 0.41 (here k =minimum of number of rows and columns) |
b)
Since V is somewhat large there is strong association between the two variables.