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What is your conclusion

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

E

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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. 

Erc = nr*nc/n, where nr is total observed frequency in column c, nc 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.

 

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