What is your conclusion

Question

The following data were collected on the number of emergency ambulance calls for an urban county and a rural county in Florida. Is County type independent of the day of the week in receiving the emergency ambulance calls? Use α = 0.005. What is your conclusion?

Day of the Week

Sun Mon Tue Wed Thu Fri Sat

County

Urban 62 47 48 51 61 74 41

Rural 6 10 18 17 11 13 12

Answer 1

Chi square test for independence is applied to test the independence of county from day of the weak.

 

Hypothesis Formation

H₀: County is independent from day of the week.

H₁: County is Not independent from day of the week.

Test Statistic

Χ² = ∑(O – E)²/E

Expected Frequency table

Using the below formula, Expected frequencies for relevant columns and rows are calculated. 

E_rc = n_r*n_c/n, where n_r is total observed frequency in column c, n_c is the total observed frequency in row r where n is total sample size.

 

       Sun        Mon        Tues       Wed       Thurs     Friday   Saturday

Urban  55.4       46.5       53.8       55.4       58.7       70.9       43.2   

Rural  12.6       10.5       12.2       12.6       13.3       16.1       9.79 

 

Critical value of χ²

d.f = (2-1)*(7-1)

      = 6

Upper Critical value χ² with d.f=6 at significance level of 0.005 = 18.6

Lower Critical value χ² with d.f=6 at significance level of 0.005 = 0.676

Critical Region

Reject null hypothesis if χ² is greater than upper critical value of 18.6 or if less than lower critical value of 0.676.

Computation

χ² = (62-55.4)²/55.4 + (47-46.5)²/46.5 + (48-53.8)²/53.8 + (51-55.4)²/55.4 + (61-58.7)²/58.7 + (74-70.9)²/70.9 + (41-43.2)²/43.2 + (6-12.6)²/12.6 + (10-10.5)²/10.5 + (18-12.2)²/12.2 + (17-12.6)²/12.6 + (11-13.3)²/13.3 + (13-16.1)²/16.1 + (12-9.79)²/9.79

= 11.4

Decision

As χ² is neither less than 0.676 nor greater than 18.6, so we can’t reject null hypothesis. Therefore county is independent from day of the week.