Note Though two of the 15 cells will not show expected frequ

Note: Though two of the 15 cells will not show expected frequency at least 5, the approximation of the test statistic by a chi-square distribution could be considered good enough. So you can use chi-square test. This question could be answered either by manually or preparing a SAS data set and using SAS.

Solution

Null Hypothesis: There is no association between physician specialty and recommended treatment.

Alternate Hypothesis: There is a association between physician specialty and recommended treatment

Chi-Square Test

Observed Frequencies

Column variable

Calculations

R

CR

C

Total

fo-fe

Int

6

22

42

70

0.75

4.5

-5.25

sur

23

61

127

211

7.175

8.25

-15.425

Rad

2

3

54

59

-2.425

-11.75

14.175

onc

1

12

43

56

-3.2

-2

5.2

Gyn

1

12

31

44

-2.3

1

1.3

Total

33

110

297

440

Expected Frequencies

Column variable

R

CR

C

Total

(fo-fe)^2/fe

Int

5.25

17.5

47.25

70

0.1071

1.1571

0.5833

sur

15.825

52.75

142.425

211

3.2531

1.2903

1.6706

Rad

4.425

14.75

39.825

59

1.3290

9.3602

5.0453

onc

4.2

14

37.8

56

2.4381

0.2857

0.7153

Gyn

3.3

11

29.7

44

1.6030

0.0909

0.0569

Total

33

110

297

440

Data

Level of Significance

0.05

Number of Rows

5

Number of Columns

3

Degrees of Freedom

8

Results

Critical Value

15.507

Chi-Square Test Statistic

28.9860

p-Value

0.0003

Reject the null hypothesis

Calculated chi square =28.986

Table value of chi square =15.507

Calculated chi square 28.986 > 15.507, table value

The null hypothesis is rejected.

We conclude that there is association between physician specialty and recommended treatment.

Chi-Square Test

Observed Frequencies

Column variable

Calculations

R

CR

C

Total

fo-fe

Int

6

22

42

70

0.75

4.5

-5.25

sur

23

61

127

211

7.175

8.25

-15.425

Rad

2

3

54

59

-2.425

-11.75

14.175

onc

1

12

43

56

-3.2

-2

5.2

Gyn

1

12

31

44

-2.3

1

1.3

Total

33

110

297

440

Expected Frequencies

Column variable

R

CR

C

Total

(fo-fe)^2/fe

Int

5.25

17.5

47.25

70

0.1071

1.1571

0.5833

sur

15.825

52.75

142.425

211

3.2531

1.2903

1.6706

Rad

4.425

14.75

39.825

59

1.3290

9.3602

5.0453

onc

4.2

14

37.8

56

2.4381

0.2857

0.7153

Gyn

3.3

11

29.7

44

1.6030

0.0909

0.0569

Total

33

110

297

440

Data

Level of Significance

0.05

Number of Rows

5

Number of Columns

3

Degrees of Freedom

8

Results

Critical Value

15.507

Chi-Square Test Statistic

28.9860

p-Value

0.0003

Reject the null hypothesis

Note: Though two of the 15 cells will not show expected frequency at least 5, the approximation of the test statistic by a chi-square distribution could be cons
Note: Though two of the 15 cells will not show expected frequency at least 5, the approximation of the test statistic by a chi-square distribution could be cons
Note: Though two of the 15 cells will not show expected frequency at least 5, the approximation of the test statistic by a chi-square distribution could be cons
Note: Though two of the 15 cells will not show expected frequency at least 5, the approximation of the test statistic by a chi-square distribution could be cons
Note: Though two of the 15 cells will not show expected frequency at least 5, the approximation of the test statistic by a chi-square distribution could be cons

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