chapter 025

1Copyright © 2013, 2009, 2005, 2001, 1997 by Saunders, an imprint of Elsevier Inc.

Chapter 25

Using Statistics to Determine Differences


Using Statistics to Determine Differences

Differences among groups Different tests for data at the various levels of

measurement Parametric statistics (used with normally

distributed interval-level and ratio-level data): Independent samples t-test Paired t-test ANOVA


Using Statistics to Determine Differences (Cont’d)

Nonparametric statistics (used with non-normally distributed samples and with ordinal-level and nominal-level data): Mann-Whitney U Wilcoxon signed-ranks test Kruskal-Wallis H Chi-square test of independence (Nominal data

only)


Choosing Parametric Versus Nonparametric Statistics to Determine Differences

Parametric statistics are always associated with a certain set of assumptions that the data must meet

One of these is normal distribution—perform a test of normalcy on the data K² test Shapiro-Wilk test Kolmogorov-Smirnov D test


t-Tests

Independent t-test—examines difference between two independent groups

Paired (dependent) t-test—examines differences between two matched or paired groups or a comparison of pre-and post-test measurements


t-Test for Independent Samples

Samples are independent if study participants in one group are unrelated or different than those in second group

Groups are independent if study subjects are randomly assigned to either treatment or control group and are not matched on any variables

Often used to measure the difference in two similar groups After intervention


t-Test for Independent Samples (Cont’d)

Most common parametric analysis technique used in nursing

Assumptions: Normal distribution Dependent variable measured at the interval/ratio

level Equal variance, both samples All observations independent



Formula:

where: = mean of group 1 = mean of group 2 = the standard error of the difference

between the two groups



If the two groups have different sample sizes, then use this formula for the denominator:

where n1 = group 1 sample size

n2 = group 2 sample size

s1 = group 1 variance

s2 = group 2 variance



If the two groups have the same number of subjects in each group, then one can use this simplified formula for the denominator:

where n = the sample size in each group and not the

total sample of both groups


Example

GDS – Geriatric Depression Scale

No Dementia Group

Patient #

No Dementia Group

GDS Score

Severe Dementia

GroupPatient #

Severe Dementia

GroupGDS Score

1 5 9 32 5 10 63 10 11 54 11 12 45 8 13 96 8 14 77 10 15 48 10 16 7


Example (Cont’d)

The independent variable level of dementia no dementia severe dementia

Dependent variable—score on the Geriatric Depression Scale

Null hypothesis is: There are no significant differences between those with dementia and those without dementia on depression scores


Example Computations for the t-Test

Step 1: Compute means for both groups, which involves the sum of scores for each group divided by the number in the group

The mean for Group 1, No Dementia:

The mean for Group 2, Severe Dementia:


Example Computations for the t-Test (Cont’d)

Step 2: Compute the numerator of the t-test:


Example Computations for the t-Test

Step 3: Compute the standard error of the difference compute the variances for each group

• s2 for Group 1 = 5.41

• s2 for Group 2 = 3.98 Plug into the standard error of the difference formula



Step 4: Compute t value:



Step 5: Compute your degrees of freedom:

df = n1+ n2 – 2

df = 8 + 8 – 2

df = 14



Locate the critical t value in the t distribution table and compare to the obtained t value

The critical t value for 14 degrees of freedom at alpha (α) = 0.05 is 2.15

if calculated t is between –2.15 and +2.15, researcher must accept the null hypothesis

If calculated t is Less than –2.15, or More than +2.15, researcher would Reject null hypothesis


Example Interpretation of Results

t is 2.55, and it is Greater Than the critical value of 2.15. This means that the researcher will Reject the null hypothesis.


Example Interpretation of Results (Cont’d)

In the research report, the researcher’s statement would be: An independent samples t-test computed on GDS scores revealed long-term residents with no dementia had significantly higher depression scores than did those who had severe dementia, t(14) = 2.55, p < 0.05; = 8.4 versus 5.6, respectively



“t(14) = 2.55” means that a t-test was used, that the degrees of freedom were 14, and that the calculated value was 2.55

p < .005 means that the value of 2.55 obtained was so unusually high that instead of the p-value of < .05, which means that the results have a 5% chance (1 in 20) of being due to Type I error, the results have only a 0.5% chance (1 in 200) of being due to Type I error. The results are very, very convincing.


Non-Parametric Alternative

Mann-Whitney U test Difference between rankings of members of

both groups Often used to measure difference in two

similar groups After intervention


Non-Parametric Alternative (Cont’d)

Very commonly used in nursing Assumptions:

Distribution may be normal or non-normal Dependent variable measured at interval/ratio or

ordinal level


t-Tests for Paired Samples

Assumptions Differences between the paired scores are

independent Normally or approximately normal distribution Dependent variable measured at the interval/ratio

level


t-Tests for Paired Samples (Cont’d)

Paired samples types Repeated measures design Case-control designs with matched controls Crossover design


t-Tests for Paired Samples calculation

where = the mean difference of the paired data = the standard error of the difference


t-Tests for Paired Samples calculation (Cont’d)

where sD = the standard deviation of the differences

between the paired data N = the number of subjects in the sample


Example

Level of functional impairment among 10 adults receiving rehabilitation for a painful injury

Independent variable: treatment over time (three weeks of rehab)

Dependent variable: functional impairment (measured twice) (lower score is healthier)

Null hypothesis is: There is no significant reduction in functional impairment from baseline to post-treatment for patients in a rehabilitation program


Example (Cont’d)

Functional Impairment Levels at Baseline and Post-Treatment

Subject #

Baseline FunctionalImpairment

Scores

Post-Treatment Functional

Impairment Scores Difference

1 2.9 1.7 1.22 5.7 2.9 2.83 2.3 2.9 –0.64 3.9 3 0.95 3.8 3.1 0.76 3.3 3.2 0.17 2.9 3.2 –0.38 4.7 3.2 1.59 3.2 2.1 1.1

10 4.9 3.4 1.5


Example t-Test Computations

Step 1: Compute difference between each subject's data pair (see last column of Table)

Step 2: Compute mean of the difference scores, which becomes the numerator of the t-test:


Example t-Test Computations (Cont’d)

Step 3: Compute the standard error of the difference Compute the standard deviation of the difference scores

S = 0.99 Plug into the standard error of the difference formula



Step 4: Compute t value:



Step 5: Compute degrees of freedom:

df = n –1

df = 10 –1

df = 9



Step 6: Locate critical t value on the t distribution table and compare to the obtained t value

The critical t value for 9 degrees of freedom at alpha (α) = 0.05 is 2.26. If calculated t is between –2.26 and +2.26,

researcher must Accept the null hypothesis If calculated t is Less than –2.26, or More than

+2.26 researcher would Reject the null hypothesis



The obtained t is 2.84, which is Greater Than the critical value of 2.26. This means that the researcher will Reject the null hypothesis

Thus during the three-week rehabilitation program, patients successfully reduced their functional impairment levels



In the research report, the researcher’s statement would be: A paired samples t-test computed on MPI functional impairment scores revealed that the patients undergoing rehabilitation had significantly lower functional impairment levels from baseline to post-treatment, t(9) = 2.84, p < 0.05; = 3.8 versus 2.9, respectively.


Non-parametric Alternative

Wilcoxon signed-rank test The difference between the rankings of the

members of both groups Often used to measure the difference in two

similar groups After intervention A very common nonparametric analysis

technique used in nursing


Non-parametric Alternative (Cont’d)

Assumptions: Distribution may be normal or non-normal Dependent variable measured at the interval/ratio

or ordinal level


One-way Analysis of Variance (ANOVA)

Statistical procedure that compares data between two or more groups or conditions

(Rarely used for two groups—t-test is more common with two)

Purpose: to investigate the presence of differences between those groups

Dependent variable continuous (interval or ratio)

Computes two estimates of variance—between and among


One-way Analysis of Variance (Cont’d)

Produces the F statistic

Mean Square = Variance This value is compared with values in an F

distribution table, to determine whether the groups’ dependent variables differ significantly from one another


Example

Medical costs among 15 females receiving treatment for a chronic pain condition, monthly medical costs for one year incurred after treatment ended were examined

Independent variable—type of treatment or intervention

Dependent variable—average medical costs incurred per month

Null hypothesis: There is no significant difference between the treatment groups on post-treatment monthly medical costs


Example (Cont’d)

Costs were U.S. Dollars, averaged monthly

Multidisciplinary Group Costs

Standard Care

Group CostsPharmacotherapy

Group Costs

74 168 603

748 328 707

433 186 868

422 199 286

297 154 919

Total 1974 1035 3383

Grand Total (G) = 6392


Example (Cont’d)

Step 1: Compute correction term, C Square the grand sum (G), and divide by total N

Step 2: Compute Total Sum of Squares Square every value in dataset, sum, and subtract

C

742 + 7482 + + 9192 – 2723844.3 = 1071877.7


Example (Cont’d)

Step 3: Compute Between Groups Sum of Squares Square the sum of each column and divide by N. Add each, and then subtract C

(779335.2 + 214245 + 2288937.8) – 2723844.3

= 558673.7


Example (Cont’d)

Step 4: Compute within Groups Sum of Squares Subtract the Between Groups Sum of Squares

(Step 3) from Total Sum of Squares (Step 2) 1071877.7 – 558673.7 = 513204

Calculate the degrees of freedom (df) Mean square between groups df = number of groups – 1 Mean square within groups df =(number of groups –1) (n–1)


Example (Cont’d)

Step 5: Calculate F by dividing the MS between by the MS within


Example (Cont’d)

Step 6: Create ANOVA Summary Table

Source of Variation

Sum of Squares

Degrees of Freedom

Mean Square

F

Between Groups

558673.73 2 279336.9 6.53

Within Groups

513204 12 42767

Total 1071877.7 14


Example (Cont’d)

Step 7: Locate the critical F value on the F distribution table and compare to our obtained F value with it. The critical F value for 2 and 12 degrees of

freedom at alpha (α) = 0.05 is 3.88. Our obtained F is 6.53, which exceeds the critical value



The obtained F= 6.53 exceeds the critical value in the table, which means that the F is statistically significant and that the population means are not equal. Therefore, the researcher rejects the null hypothesis that the three groups have the same monthly post-treatment medical costs.



However, the F does not reveal which treatment groups differ from one another. Further testing, termed multiple comparison tests or post-hoc tests, are required to complete the ANOVA process and determine all the significant differences among the study groups


Post Hoc Tests

Developed specifically to determine the location of group differences after ANOVA is performed on data from more than two groups

Frequently used post hoc tests Newman-Keuls test Tukey Honestly Significant Difference (HSD) test Scheffé test Dunnett test


Post Hoc Tests (Cont’d)

All divide the alpha (usually p < .05) over all groups

After Tukey HSD testing was computed for the example: Pharmacotherapy group and Standard care group

costs were significantly different (Pharm group costs higher by a wide margin)


Non-parametric Alternative

Kruskal-Wallis test Dependent variable may be ordinal Data converted to ranks, which are then

compared Used for answering the question, “Are these

all the same?”


Other ANOVA Procedures

Randomized complete block Repeated measures Factorial


Chi-square Test of Independence

Compares differences in proportions (percentages) of nominal level variables

One-way chi-square: compares different levels of one variable only

Two-way chi-square: tests whether proportions in levels of one variable are significantly different from proportions of the second variable


Assumptions of the Chi-Square Test

One data point for each subject (no repeated measures)

Mutually exclusive categories Mutually exhaustive categories Distribution-free, or nonparametric—no

requirement of normal distribution


Formula (Two-way Chi-Square)

where the Contingency table is labeled as such:

A B

C D


Example (Two-way Chi-Square)

Presence of candiduria among 97 adults with a spinal cord injury

Candiduria

Antibiotic Use

Yes No

Yes 15 43

No 0 39


Calculation


Degrees of Freedom (df)

Must be calculated to determine the significance of the value of the statistic

df = (R –1)(C – 1)

where R = Number of rows C = Number of columns

df = (2 –1)(2 –1) = 1


Interpretation of Results

The chi-square statistic is compared with the chi-square values in the table in Appendix X. The table includes the critical values of chi-square for specific degrees of freedom at selected levels of significance. If the value of the statistic is equal to or greater than the value identified in the chi-square table, the difference between the two variables is statistically significant.


Interpretation of Results (Cont’d)

The critical 2 for df = 1 is 3.84, and the researcher’s obtained 2 is 11.93, thereby exceeding the critical value and indicating a significant difference between antibiotic users and non-users on the presence of candiduria.


Interpretation of Results (Cont’d)

A research report would read: Antibiotic users had significantly higher rates of candiduria than those who did not use antibiotics (26% versus 0%, respectively). This finding suggests that antibiotic use may be a risk factor for developing candiduria, and further research is needed to investigate candiduria as a direct effect of antibiotics.

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