an ambulance service responds to emergency calls for two cou

an ambulance service responds to emergency calls for two countries in virginia. one country is an urban country the other is a rural country. a sample of 471 ambulances calls over the past two years showed the country and the day of the week for each emergency call. data are as follows

SUN

MON

TUE

WED

THU

FRI

SAT

URBAN

61

48

50

55

63

73

43

RURAL

7

9

16

13

9

14

10

Test for independence of the county and the day of the week. Using a .05 level of significance, what is teh p-value and what is your conclusion?

	SUN	MON	TUE	WED	THU	FRI	SAT
URBAN	61	48	50	55	63	73	43
RURAL	7	9	16	13	9	14	10

Solution

First find the row totals and column totals, and the grand total for the table:

SUN

MON

TUE

WED

THU

FRI

SAT

URBAN

61

48

50

55

63

73

43

393

RURAL

7

9

16

13

9

14

10

78

total

68

57

66

68

72

87

53

471

Then find the expected frequencies. They are found for each cell by taking its corresponding row total times column total and dividing by the table total. For example, for Urban and Sunday: 393*68/471 = 56.738854, and another example, for rural and Thursday: 78*72/471 = 11.923567.

Expected frequencies:

SUN

MON

TUE

WED

THU

FRI

SAT

URBAN

56.738854

47.560510

55.070064

56.738854

60.076433

72.592357

44.222930

RURAL

11.261146

9.439490

10.929936

11.261146

11.923567

14.407643

8.777070

Next, we compute the chi square test statistic by taking each observed frequency from the first table subtracting the corresponding expected frequency, squaring that and dividing by the expected frequency. We do that for each of the 14 cells and add them up.

Chi square = (61

	SUN	MON	TUE	WED	THU	FRI	SAT
URBAN	61	48	50	55	63	73	43	393
RURAL	7	9	16	13	9	14	10	78
total	68	57	66	68	72	87	53	471