Thursday, January 9, 2020

Statistics Test Explanation




By Dr. Sultan Muhammad Khan
              
Keywords; Using SPSS create new categorical variables for Verbal and Quant of level 3 each? Apply chi square test to new variable and interpret the result? Calculate the correlation between verbal and quant and interpret the results? Apply chi square test to Job and Experience of any cell has less than 5 expected frequencies then combine the adjacent levels and try to find chi square test, repeat until there is no cell with expected frequency less than 5 and interpret the result?



Statistics Test Explanation

Question A:  Using SPSS create new categorical variables for Verbal and Quant of level 3 each?


Answer:

To create new categorical variables use recode into different variables  from the transform menu, input the name of target variable and give new name for variable in the output variable window and press change button. Then click on old and new values and enter date by range and assign them a new value e.g. 1,2,3….

Attach are the sheets which was created by this method with name of Varbnew and Quantnew, using range 80 thru 100 for value 1 and 101 to 120 for value 2 and 131 to 140 for value 3.

In this way we get two new categorical variables Verbnew and Quantnew

 

Question B:   Apply chi square test to new variable and interpret the result?

Answer:
                                                                                                    
              After creating the new categorical variable verbnew and quantnew in question A, here we apply chi square test to these new variables using command syntax Analyze- Descriptive Statistic- Crosstab, a new window will appear, move verbnew to row window and quantnew to column window and then press on the Statistic button select chi-square in new window and press continue button and then press ok. Now Computer will calculate the chi square of Verbnew and Quantnew, the output sheet is attached.

In the result table we have the degree of freedom 4 in two cases and 1 in linear by linear association, which is good, the degree of freedom should not be 0 because then it have no freedom it will effect the whole calculation.

In the case processing summary we have N=100 mean one hundred observation for both new variables.

In the last table we have the original chi square test table which shows us that the asymp significance is .000 for all three, which mean that we can reject this hypothesis because there is a definite difference between the two. We reject the hypothesis because the difference is high if the value is less then .5 which in this case it is .000.

If the value of Asymp sig is greater than .5 then there is no difference and we accept the hypothesis which is not in this case.
                        

 Question C:  Calculate the correlation between verbal and quant and interpret the results?

Answer:
           
            To find correlation, we will use the bivariate correlation to determine if the two variables are linearly related to each other, for this purpose we will use command syntax
Analyze-correlation-bivariate to calculate the correlation between two variables.
           
            Select two variables, in this case we select verbal and quant and move it to variable pane for correlation then click on one tailed option. After that click on the option button and select mean and slandered deviation then click ok and check the output in the output viewer.

The descriptive statistics section gives the mean, slandered deviation and number of observation (n) for each variable.

            The mean for quant is 1.1232E2 and verbal 1.1163E2 and slandered deviation for variables 12.88071, 12.30926 and observation for each is 75. The correlation section gives us the values of the specified correlation test.
            It is a one 1-tailed test therefore .000 is the significance of this correlation, which tells us that the correlation is significant in verbal and quant, because the value is less than .5

The correlation coefficient is .872 and the middle number is significance of this correlation which is .000 and 75 is the number of observation in this case.  


Question D:   Apply chi square test to Job and Experience of any cell has less than 5 expected frequencies then combine the adjacent levels and try to find chi square test, repeat until there is no cell with expected frequency less than 5 and interpret the result?

Answer:
                                                                                                    
          After creating new variables job1, job2, job3, job4 and exper1, exper2, exper3, exper4 to achieve the result of no cell that have the expected frequency less than 5 and calculate chi square five times to research the end result.

In all five chi square result we have different degree of freedom and every time we combine the adjacent levels the degree of freedom decrease e.g. in first X2 test it is 30 and in the last test it is 2 with no cell that have expected frequency less than 5 which was our aim in this case to find the result till there is no cell which have expected frequency less than 5 and we achieved it on fifth chi square test and in all chi square test we have the degree of freedom 1 in linear by linear association,

In the case processing summary we have N=100 mean one hundred observation for all variables.

In all five chi square result we have find a high difference because in all chi square test the Asymp sig is .000 for all which is less than .5, Means we can reject the hypothesis and can say that there is a definite difference between them.

                                                                                              

1 comment:

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