Review the three ‘Non Parametric’ test videos in the Calculations section of the ‘Statistics Visual Learner’ media piece.How would you create a statistical analysis based on the scenario belo

Nonparametric Test

Chi-Square Test for Independence

The test is used to determine whether two categorical variables are independent.

Notation for the Chi-Square Test for Independence (Please note that the notation varies

depending on the text)

O represents the observed frequency of an outcome

E represents the expected frequency of an outcome

r represents the number of rows in the contingency table

c represents the number of columns in the contingency table

n represents the total number of trials

Test Statistic

 

  

2

2

1 1

O E

E

df r c

 



  



The Chi-Square test is a hypothesis test. There are seven steps for a hypothesis test.

1. State the null hypothesis

2. State the alternative hypothesis

3. State the level of significance

4. State the test statistic

5. Calculate

6. Statistical Conclusion

7. Experimental Conclusion

Example

A university is interested to know if the choice of major has a relationship to gender. A

random sample of 200 incoming freshmen students was taken (100 male and 100

female). There major and gender were recorded. The results are shown in the

contingency table below.

Major Female Male

Math 5 15

Nursing 44 10

English 10 10

Pre-Med 17 20

History 4 5

Education 15 20

Undecided 5 20

To determine if there is a relationship between the gender of a freshmen student and

thei declared major perform the hypothesis test (Use level of significance 0.05  ) .

Step 1: Null Hypothesis

0 : Gender and Major of Freshmen students are independentH

Step 2: Alternative Hypothesis

: Gender and Major of Freshmen students are not independentAH

Step 3: Level of Significance

0.05 

Step 4: Test Statistic

 

  

2

2

1 1

O E

E

df r c

 



  



Step 5: Calculations

There are several calculations for this test. We have to find the expected frequency for

each cell in the contingency table. The expected frequency is the probability under the

null hypothesis times the total frequency for the given row. Here the probability under

the null hypothesis is .5, as the probability of being male and female is equal.

rE pn

Major Female Male

Math   1 .5 20 10E     2 .5 20 10E  

Nursing   3 .5 54 27E     4 .5 54 27E  

English   5 .5 20 10E     6 .5 20 10E  

Pre-Med   7 .5 37 18.5E     8 .5 37 18.5E  

History   9 .5 9 4.5E     10 .5 9 4.5E  

Education   11 .5 35 17.5E     12 .5 35 17.5E  

Undecided   13 .5 25 12.5E     14 .5 25 12.5E  

Know calculate the test statistic.

 

       

       

       

   

2

2

2 2 2 2

2

2 2 2 2

2 2 2 2

2 2

2

5 10 15 10 44 27 10 27

10 10 27 27

10 10 10 10 17 18.5 20 18.5

10 10 18.5 18.5

4 4.5 5 4.5 15 17.5 20 17.5

4.5 4.5 17.5 17.5

5 12.5 20 12.5

12.5 12.5

2.5 2.5 10.7 10.7 0 0 .1216

obs

obs

obs

O E

E 





 

       

       

       

   

      



2

.1216 .0556 .0556 .357 .357 4.5 4.5

36.4684obs

      



The calculation for degrees of freedom is as follows:

     

  

1 1 7 1 2 1

6 1 6

df r c

df

     

 

The critical value for the Chi-Square with 6 degrees of freedom at a level of significance

0.05  is 12.592. This is found by using the Chi Square table.

Step 6: Statistical Conclusion

Since 2 2

6, 0.0536.4684 12.592obs df       then reject the null hypothesis.

Step 7: Experimental Conclusion

There is sufficient evidence to indicate that gender has an effect on choice of major for

the incoming freshmen.

Mann-Whitney

The Mann-Whitney test is a nonparametric version of the independent sample t-test.

This study is used when there are two independent samples of ordinal scores.

Test Statistic

 

 

1 1

1 1 2 1

2 2

2 1 2 2

1

2

1

2

n n U n n R

n n U n n R

   

   

Notation:

1

2

1

2

number of scores in group 1

n number of score in group 2

R sum of ranks for score in group 1

R sum of ranks for score in group 2

n 







The Mann-Whitney test is a hypothesis test. There are seven steps for a hypothesis

test.

1. State the null hypothesis

2. State the alternative hypothesis

3. State the level of significance

4. State the test statistic

5. Calculate

6. Statistical Conclusion

7. Experimental Conclusion

Example:

Suppose there was race between an Antelope and a mountain lion. The Antelope won

but you want to know if this could be extended to a general statement that Antelope will

win in a race against a mountain lion. A random sample of 10 Antelopes and 10

mountain lions are put in a race. The order in which they finish is recorded. ( A for

antelope and L for mountain lion)

AALALLLAAALAALLAALLL

Step 1: Null Hypothesis

0 : Probability of the Antelope winning is no different than the probalility of the

mountain lion wining the race.

H

Step 2: Alternative Hypothesis

: Probability of the Antelope winning is different than the probalility of the

mountain lion wining the race.

AH

Step 3: Level of Significance

0.05 

Step 4: Test Statistic

 

 

1 1

1 1 2 1

2 2

2 1 2 2

1

2

1

2

n n U n n R

n n U n n R

   

   

Step 5: Calculations

The results are ranked according to the in order in which they finished the race.

Rank

A 1

A 2

L 3

A 4

L 5

L 6

L 7

A 8

A 9

A 10

L 11

A 12

A 13

L 14

L 15

A 16

A 17

L 18

L 19

L 20

Let the Antelope be sample one and the mountain lion be sample 2.

1

2

1

2

number of scores for the Antelopes

n number of score for the mountain lions

R sum of ranks for score for the Antelopes

R sum of ranks for score for the mountain lions

n 







Find the sum of the ranks for each group.

Antelope Mountain Lion

1 3

2 5

4 6

8 7

9 11

10 14

12 15

13 18

16 19

17 20

Sum 92 118

The number of for each animal is 10 because 10 antelopes and 10 mountain lions

raced.

Now plug the information into the test statistic

 

    

 

    

1 1

1 1 2 1

1

1

1

2 2

2 1 2 2

2

2

2

1

2

10 10 1 10 10 92

2

100 55 92

63

1

2

10 10 1 10 10 118

2

100 55 118

37

n n U n n R

U

U

U

n n U n n R

U

U

U

   

   

  



   

   

  



Find the critical value for the test statistic. Use the appropriate U table found in the

appendix of the textbook. Look for alpha to be 0.05 and then the number of scores for

each category to be 10. Then you see the following critical values.

1,

2,

23

77

crit

crit

U

U





Step 6: Statistical Conclusion

The statistical conclusion tells whether to reject or fail to reject the null hypothesis. The

rejection rule is as follow.

1

2

If then reject the null hypothesis

If then reject the null hypothesis

crit

crit

U U

U U





Since 1 1,63 23 critU U   then fail to reject the null hypothesis also 2 2,37 77 critU U  

then fail to reject the null hypothesis.

Step 7: Experimental Conclusion

Since we failed to reject the null hypothesis we can say that there is not significant

evidence to support the claim that the probability of the antelope winning is any different

than the probability of the mountain lion winning the race.

Kruskal-Wallis Test

The Kruskal-Wallis Test is a nonparametric version of the one-way ANOVA. This test

however does not assume normality or homogeneity of variance. The test requires

ordinal scaling of the dependent variable and must have at least 5 data values in each

sample.

Notation

k number of sample or groups

n number of data values in each group

N number of data values in all samples combined

R sum of the ranks for each sample

Test Statistic

 

   

2

2 12 3 1

1

1

R N

N N n

df k

   

          

 



The Kruskal-Wallis test is a hypothesis test. There are seven steps for a hypothesis

test.

1. State the null hypothesis

2. State the alternative hypothesis

3. State the level of significance

4. State the test statistic

5. Calculate

6. Statistical Conclusion

7. Experimental Conclusion

Example

A gym has decided to recommend a diet to their clients. They have narrowed down the

choice to three and they want to test which diet is best. They have randomly selected 21

volunteers from the gym to participate in the diet. All the participants have similar health

and body type as well as general exercise routine. The amount of weight loss was

recorded.

Diet A Diet B Diet C

10 11 1

12 4 8

13 2 17

7 6 0

3 16 16

15 9 18

5 14 19

Step 1: Null Hypothesis

0 : There is no difference amoung the dietsH

Step 2: Alternative Hypothesis

: There is a difference between the dietsAH

Step 3: Level of Significance

0.05 

Step 4: Test Statistic

 

   

2

2 12 3 1

1

1

R N

N N n

df k

   

          

 



Step 5: Calculations

To perform the calculation, start by ranking the results.

Diet A Rank Diet B Rank Diet C Rank

10 11 11 12 1 2

12 13 4 5 8 9

13 14 2 3 17 19

7 8 6 7 0 1

3 4 16 17 16 17

15 16 9 10 18 20

5 6 14 15 19 21

Notice that some data values have the same ranks. This is how you account for data

values that are the same.

k = 3

n = 7

N = 21

Plug these values into the test statistic

 

   

 

       

 

  

2

2

2 2 2

2

2

2

2

2

12 3 1

1

72 69 8912 3 21 1

21 21 1 7 7 7

12 740.571 680.143 1131.571 66

462

.025974 2552.285 66

66.293 66

0.2931

1 3 1 2

obs

obs

obs

obs

obs

obs

R N

N N n

df k













       

     

         

     

        

 

 



    



The critical value can be found using the Chi-Square table for 0.05  and 2 degrees of

freedom.

2

0.05, 2 5.991df   

Step 6: Statistical Conclusion

Since 2 2

0.05, 20.2931 5.991obs df      then we will fail to reject the null hypothesis.

Step 7: Experimental Conclusion

The evidence does not support the claim that the there is a significant difference

between the three diets at a level of significance of 0.05.

Nonparametric Test

Chi-Square Test for Independence

The test is used to determine whether two categorical variables are independent.

Notation for the Chi-Square Test for Independence (Please note that the notation varies

depending on the text)

O represents the observed frequency of an outcome

E represents the expected frequency of an outcome

r represents the number of rows in the contingency table

c represents the number of columns in the contingency table

n represents the total number of trials

Test Statistic

 

  

2

2

1 1

O E

E

df r c

 



  



The Chi-Square test is a hypothesis test. There are seven steps for a hypothesis test.

1. State the null hypothesis

2. State the alternative hypothesis

3. State the level of significance

4. State the test statistic

5. Calculate

6. Statistical Conclusion

7. Experimental Conclusion

Example

A university is interested to know if the choice of major has a relationship to gender. A

random sample of 200 incoming freshmen students was taken (100 male and 100

female). There major and gender were recorded. The results are shown in the

contingency table below.

Major Female Male

Math 5 15

Nursing 44 10

English 10 10

Pre-Med 17 20

History 4 5

Education 15 20

Undecided 5 20

To determine if there is a relationship between the gender of a freshmen student and

thei declared major perform the hypothesis test (Use level of significance 0.05  ) .

Step 1: Null Hypothesis

0 : Gender and Major of Freshmen students are independentH

Step 2: Alternative Hypothesis

: Gender and Major of Freshmen students are not independentAH

Step 3: Level of Significance

0.05 

Step 4: Test Statistic

 

  

2

2

1 1

O E

E

df r c

 



  



Step 5: Calculations

There are several calculations for this test. We have to find the expected frequency for

each cell in the contingency table. The expected frequency is the probability under the

null hypothesis times the total frequency for the given row. Here the probability under

the null hypothesis is .5, as the probability of being male and female is equal.

rE pn

Major Female Male

Math   1 .5 20 10E     2 .5 20 10E  

Nursing   3 .5 54 27E     4 .5 54 27E  

English   5 .5 20 10E     6 .5 20 10E  

Pre-Med   7 .5 37 18.5E     8 .5 37 18.5E  

History   9 .5 9 4.5E     10 .5 9 4.5E  

Education   11 .5 35 17.5E     12 .5 35 17.5E  

Undecided   13 .5 25 12.5E     14 .5 25 12.5E  

Know calculate the test statistic.

 

       

       

       

   

2

2

2 2 2 2

2

2 2 2 2

2 2 2 2

2 2

2

5 10 15 10 44 27 10 27

10 10 27 27

10 10 10 10 17 18.5 20 18.5

10 10 18.5 18.5

4 4.5 5 4.5 15 17.5 20 17.5

4.5 4.5 17.5 17.5

5 12.5 20 12.5

12.5 12.5

2.5 2.5 10.7 10.7 0 0 .1216

obs

obs

obs

O E

E 





 

       

       

       

   

      



2

.1216 .0556 .0556 .357 .357 4.5 4.5

36.4684obs

      



The calculation for degrees of freedom is as follows:

     

  

1 1 7 1 2 1

6 1 6

df r c

df

     

 

The critical value for the Chi-Square with 6 degrees of freedom at a level of significance

0.05  is 12.592. This is found by using the Chi Square table.

Step 6: Statistical Conclusion

Since 2 2

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