Reliability Statistics

Reliability Statistics

  1. A RELIABILITY procedure was performed on the 10 items of the AUDIT scale resulting in the attached output from SPSS. Please comment fully on the results of this procedure (shown below).

                          Case Processing Summary 

N %
Cases Valid 98 92.5
Excluded(a) 8 7.5
Total 106 100.0

a  Listwise deletion based on all variables in the procedure.

              Reliability Statistics 

Cronbach’s Alpha N of Items
.855 10

                                         Item Statistics 

Mean Std. Deviation N
audit1 1.74 1.115 98
audit2 1.10 1.272 98
audit3 1.16 1.155 98
audit4 .21 .662 98
audit5 .18 .544 98
audit6 .17 .626 98
audit7 .44 .826 98
audit8 .71 1.015 98
audit9 .88 1.548 98
audit10 .39 1.136 98

                                                         Item-Total Statistics 

Scale Mean if Item Deleted Scale Variance if Item Deleted Corrected Item-Total Correlation Cronbach’s Alpha if Item Deleted
audit1 5.26 36.110 .678 .830
audit2 5.90 33.804 .744 .822
audit3 5.84 34.550 .777 .819
audit4 6.79 42.954 .351 .856
audit5 6.82 43.286 .398 .854
audit6 6.83 41.629 .546 .846
audit7 6.56 40.620 .488 .847
audit8 6.29 36.722 .705 .828
audit9 6.12 35.016 .492 .857
audit10 6.61 37.498 .549 .842

                                           Scale Statistics 

Mean Variance Std. Deviation N of Items
7.00 46.433 6.814 10
  1. Self ratings of problem behaviour of 2 groups of adolescents were classified into the ranges ‘normal’ ‘borderline’ and ‘clinical’ (labelled 1,2 & 3 respectively) based on a standard clinical measure. An SPSS crosstabs procedure with a Pearson Chi-square statistical test was used to test whether there was an association between which school an adolescent belonged to and the assessment category allocated to them.  Please fully interpret the output shown below.

                                                                            Case Processing Summary 

Cases
Valid Missing Total
N Percent N Percent N Percent
school * problembeh 95 99.0% 1 1.0% 96 100.0%

                                                           school * problembehCrosstabulation 

problembeh Total
1.00 2.00 3.00
School 1.00 Count 19 12 16 47
Expected Count 31.2 7.4 8.4 47.0
Adjusted Residual -5.3 2.6 4.1
2.00 Count 44 3 1 48
Expected Count 31.8 7.6 8.6 48.0
Adjusted Residual 5.3 -2.6 -4.1
Total Count 63 15 17 95
Expected Count 63.0 15.0 17.0 95.0

                                                  Chi-Square Tests

Value df Asymp. Sig. (2-sided)
Pearson Chi-Square 28.549(a) 2 .000
Likelihood Ratio 31.931 2 .000
Linear-by-Linear Association 26.768 1 .000
N of Valid Cases 95

a  0 cells (.0%) have expected count less than 5. The minimum expected count is 7.42 

  1. In the attached SPSS output a t-test has been performed to test the hypothesis that there are no differences in reported alcohol problems between two sets of adolescents. Please interpret the output information fully.

                                                            Group Statistics 

school N Mean Std. Deviation Std. Error Mean
Audit 1.00 48 10.7708 7.97400 1.15095
2.00 48 3.2500 3.27141 .47219

Independent Samples Test 

Levene’s Test for Equality of Variances t-test for Equality of Means
F Sig. t df Sig. (2-tailed) Mean Difference Std. Error Difference 95% Confidence Interval of the Difference
Lower Upper
audit Equal variances assumed 31.162 .000 6.045 94 .000 7.52083 1.24404 5.05076 9.99091
Equal variances not assumed 6.045 62.386 .000 7.52083 1.24404 5.03433 10.00733

Solution 

  1. A RELIABILITY procedure was performed on the 10 items of the AUDIT scale resulting in the attached output from SPSS. Please comment fully on the results of this procedure (shown below).

                          Case Processing Summary

N %
Cases Valid 98 92.5
Excluded(a) 8 7.5
Total 106 100.0

a  Listwise deletion based on all variables in the procedure.

Case processing summary suggests that out of total 106 samples, 8 elements are excluded and only 98 elements are included. This means only 98 elements completed/responded sufficient items to be included in the analysis.

              Reliability Statistics

Cronbach’s Alpha N of Items
.855 10

The reliability statistics table determineCronbach’s Alpha. The table indicates that Cronbach’s Alpha is 0.855. This indicates that there is high level of internal consistency for the considered scale. A Cronbach’s alpha of 0.7 or high consider as a metric for relatively higher degree of internal consistency. Further, number of items is only 10 which meansnumber of items in scale is 10.

                                         Item Statistics

Mean Std. Deviation N
audit1 1.74 1.115 98
audit2 1.10 1.272 98
audit3 1.16 1.155 98
audit4 .21 .662 98
audit5 .18 .544 98
audit6 .17 .626 98
audit7 .44 .826 98
audit8 .71 1.015 98
audit9 .88 1.548 98
audit10 .39 1.136 98

Above table gives means and standard deviations for each of your question items. Highest mean for audit 1 is 1.74 while highest standard deviation for audit 9 is 1.548.

                                                         Item-Total Statistics

Scale Mean if Item Deleted Scale Variance if Item Deleted Corrected Item-Total Correlation Cronbach’s Alpha if Item Deleted
audit1 5.26 36.110 .678 .830
audit2 5.90 33.804 .744 .822
audit3 5.84 34.550 .777 .819
audit4 6.79 42.954 .351 .856
audit5 6.82 43.286 .398 .854
audit6 6.83 41.629 .546 .846
audit7 6.56 40.620 .488 .847
audit8 6.29 36.722 .705 .828
audit9 6.12 35.016 .492 .857
audit10 6.61 37.498 .549 .842

This table can really help you to decide whether any items need to be removed. The analysis is based on the fact that if any items gets removed, Cronbach’s alpha gets improved. Analysis suggests if ‘audit 4’ and ‘audit 9’ be removed, Cronbach’s Alpha gets improved. Hence ‘audit 4’ and ‘audit 9’ may be removed for improving reliability.

                                           Scale Statistics 

Mean Variance Std. Deviation N of Items
7.00 46.433 6.814 10

This final table in the output gives you the descriptive statistics for the questionnaire as a whole.

  1. Self ratings of problem behaviour of 2 groups of adolescents were classified into the ranges ‘normal’ ‘borderline’ and ‘clinical’ (labelled 1,2 & 3 respectively) based on a standard clinical measure. An SPSS crosstabs procedure with a Pearson Chi-square statistical test was used to test whether there was an association between which school an adolescent belonged to and the assessment category allocated to them.  Please fully interpret the output shown below.

                                                                            Case Processing Summary

Cases
Valid Missing Total
N Percent N Percent N Percent
school * problembeh 95 99.0% 1 1.0% 96 100.0%

Case processing summary suggests that out of total 96 samples, 1 element is excluded and only 95 elements are included. This means only 95 elements completed/responded sufficient items to be included in the analysis.

                                                           school * problembehCrosstabulation 

problembeh Total
1.00 2.00 3.00
School 1.00 Count 19 12 16 47
Expected Count 31.2 7.4 8.4 47.0
Adjusted Residual -5.3 2.6 4.1
2.00 Count 44 3 1 48
Expected Count 31.8 7.6 8.6 48.0
Adjusted Residual 5.3 -2.6 -4.1
Total Count 63 15 17 95
Expected Count 63.0 15.0 17.0 95.0

This table helps in establishing the fact that maximum problem behaviour of both group classified into normal

                                                  Chi-Square Tests 

Value df Asymp. Sig. (2-sided)
Pearson Chi-Square 28.549(a) 2 .000
Likelihood Ratio 31.931 2 .000
Linear-by-Linear Association 26.768 1 .000
N of Valid Cases 95

a  0 cells (.0%) have expected count less than 5. The minimum expected count is 7.42.

The table above is quite useful in inferring the results of Chi-Square Tests.  “Pearson Chi-Square” row     indicates that χ(2) = 28.549, p = .000. The degree of freedom may be consider as 2 [ (r-1) x (c-1)].  This indicates    there are enough statistical evidence to reject the null hypothesis. Hence there is statistically significant association between   groups and problem behaviourthat is, both groupsdo not  associate similar to different category of  problem behaviour namely  ‘normal’ ‘borderline’ and ‘clinical’.

  1. In the attached SPSS output a t-test has been performed to test the hypothesis that there are no differences in reported alcohol problems between two sets of adolescents.  Please interpret the output information fully.

The null and alternate hypothesis is described below:

H0: There are no differences in reported alcohol problems

Ha: There are significant statistical difference in reported alcohol problems 

                                                            Group Statistics 

school N Mean Std. Deviation Std. Error Mean
audit 1.00 48 10.7708 7.97400 1.15095
2.00 48 3.2500 3.27141 .47219

As a first step, descriptive statistics are estimated. The number of samples in both the groups is 48. The mean of first group is 10.7708 while mean of second group is 3.2500. The standard deviation of first group is 7.97400 while standard deviation of second group is 3.27141

                                                                                                                          Independent Samples Test 

Levene’s Test for Equality of Variances t-test for Equality of Means
F Sig. t df Sig. (2-tailed) Mean Difference Std. Error Difference 95% Confidence Interval of the Difference
Lower Upper
audit Equal variances assumed 31.162 .000 6.045 94 .000 7.52083 1.24404 5.05076 9.99091
Equal variances not assumed 6.045 62.386 .000 7.52083 1.24404 5.03433 10.00733

The above table comprises of two important test. First, Levene’s Test for Equality of Variances and second, t-test for Equality of Means.Levene’s test is used to test if k samples have equal variances. Equal variances across samples is called homogeneity of variance. Some statistical tests, for example the analysis of variance, assume that variances are equal across groups or samples. The Levene test can be used to verify that assumption.

The null and alternate hypothesis for Levene’s Test for Equality of Variances is shown below:

H0: σ12

Ha: σ1≠σ2

Test Statistic:  Given a variable Y with sample of size N divided into k subgroups, where Ni is the sample size of the ith subgroup, the Levene test statistic is defined as:

The p-value of test is less than 0.05 which is the level of significance of test. Hence there are sufficient statistical evidence to reject the null hypothesis. Hence, it can be inferred that variances of two groups are unequal.

Post that t-test is conducted for independent samples having un-equal variances. The null and alternate hypothesis of the test is already defined above. Degree of freedom for the tests is n1 + n2 -2 = 48 + 48 -2 = 96.The t- static is defined as follows:

Where symbols have standard meanings

The p-value of the tests is 0.00 which means there is sufficient statistical evidence to reject null hypothesis which means there are significant statistical difference in reported alcohol problems of the two samples.