Question

a store manager wants to see if the proportion of sales for an item is different in 5 locations around their store. the manager observes the number of items sold in each location for a week. test to see if there is a difference in the proportions using $\alpha = 0.05$
a. complete the table. round answers to at least 4 decimal places.

location	observed	expected	$\frac{(o - e)^2}{e}$
b	19
c	24
d	21
e	11
total	86

b. calculate the chi - square test - statistic.

$\chi^2 = $

c. calculate the p - value.

p - value =

d. this p - value leads to a decision to...

Explanation:

Step1: Calculate expected values

Expected value for each location: $E = \frac{\text{Total}}{5} = \frac{86}{5} = 17.2$

Step2: Compute $\frac{(O-E)^2}{E}$ for each location

• Location A: $\frac{(11-17.2)^2}{17.2} = \frac{(-6.2)^2}{17.2} = \frac{38.44}{17.2} ≈ 2.23488372$
• Location B: $\frac{(19-17.2)^2}{17.2} = \frac{(1.8)^2}{17.2} = \frac{3.24}{17.2} ≈ 0.18837209$
• Location C: $\frac{(24-17.2)^2}{17.2} = \frac{(6.8)^2}{17.2} = \frac{46.24}{17.2} ≈ 2.68837209$
• Location D: $\frac{(21-17.2)^2}{17.2} = \frac{(3.8)^2}{17.2} = \frac{14.44}{17.2} ≈ 0.83953488$
• Location E: $\frac{(11-17.2)^2}{17.2} = \frac{(-6.2)^2}{17.2} = \frac{38.44}{17.2} ≈ 2.23488372$

Step3: Sum to get chi-square statistic

$\chi^2 = 2.23488372 + 0.18837209 + 2.68837209 + 0.83953488 + 2.23488372 ≈ 8.1860465$

Step4: Find degrees of freedom

$df = 5 - 1 = 4$

Step5: Calculate p-value

Using chi-square distribution with $df=4$, p-value for $\chi^2≈8.186$ is between 0.05 and 0.10 (exact value ≈ 0.085)

Answer:

a. Completed table:

Location	Observed	Expected	$\frac{(O-E)^2}{E}$
B	19	17.2	0.1884
C	24	17.2	2.6884
D	21	17.2	0.8395
E	11	17.2	2.2349
Total	86	86	8.1861

b. $\chi^2 ≈ 8.1860$

c. p-value ≈ 0.085

d. Fail to reject the null hypothesis (since p-value > 0.05)