In: Statistics and Probability
A restaurant chain that has 3 locations in Portland is trying to determine which of their 3 locations they should keep open on New Year’s Eve. They survey a random sample of customers at each location and ask each whether or not they plan on going out to eat on New Year’s Eve. The results are below. Run a test for independence to decide if the proportion of customers that will go out to eat on New Year’s Eve is dependent on location. Use α=0.05.
NW Location |
NE Location |
SE Location |
|
Will Go Out |
66 |
40 |
45 |
Won’t Go Out |
20 |
25 |
20 |
Can it be concluded that the choice to go out on New Year's Eve is dependent on restaurant location?
No, it cannot be concluded that the choice to go out on New Year's Eve is dependent on restaurant location because the p-value = 0.8706
Yes, it can be concluded that the choice to go out on New Year's Eve is dependent on restaurant location because the p-value = 0.8706
Yes, it can be concluded that the choice to go out on New Year's Eve is dependent on restaurant location because the p-value = 0.1294.
No, it cannot be concluded that the choice to go out on New Year's Eve is dependent on restaurant location because the p-value = 0.1294.
This is a test for independent between two categorical variables. So we will use chi-square dist.
Observed | NW Location | NE Location | SE Location | Total |
Will Go Out | 66 | 40 | 45 | 151 |
Won’t Go Out | 20 | 25 | 20 | 65 |
Total | 86 | 65 | 65 | 216 |
Expected | NW Location | NE Location | SE Location | Total |
Will Go Out | 60.12037 | 45.43981 | 45.43981 | 151 |
Won’t Go Out | 25.87963 | 19.56019 | 19.56019 | 65 |
Total | 86 | 65 | 65 | 216 |
Expected Value of a cell =R total * c total / overall total ...................... r = no. of rows and c = no. of columns
NW Location | NE Location | SE Location | |
Will Go Out | 0.57501 | 0.65123 | 0.00426 |
Won’t Go Out | 1.33580 | 1.51285 | 0.00989 |
Chi -square test STat =
Test Stat = 4.08904
c. df = (r -1) * (c -1) ...................... r = no. of rows and c = no. of columns
= 1* 2
df = 2
p - value = P( > T.S.)
= P( > 4.09) ....using chi -square dist tables
p-value = 0.129
Null hypo: The choice to go out on New Year's Eve is not dependent on restaurant location
Alternative hypo: The choice to go out on New Year's Eve is dependent on restaurant location
Since p-value > 0.05
We do not reject the null hypothesis
The choice and location are independent.
D.
No, it cannot be concluded that the choice to go out on New Year's Eve is dependent on restaurant location because the p-value = 0.1294.