WK 6 DIS. DATA
Nursebeauty
NURS_8211_WK6ComparisonofCountsChi.pptxNURS 8211 Research for an evidence-based practice week 6: inferential comparison of counts (chi square)
week 6: inferential comparison of counts (chi square)
Objectives
Interpret a small sample Fisher Exact Chi Square analysis with a one-tailed test of significance for a categorical variable and compare to a t test used to test the difference in means.
Compare and contrast a one-sample Chi Square (Goodness of Fit) with a Chi Square test of independence.
Find the Chi Square in one of the three studies presented this week in optional resources and critique
Construct a bar graph in excel for sample data
Chi Square
Used for comparison of counts and percentages
Categorical data (a mean doesn’t make sense)
Goodness of fit (one sample)
Test of Independence (two samples)
Chi square goodness of fit
Allows you to test whether or not your data is the same as or different from an expected outcome.
Body composition:
Normal = 51
Overweight = 28
Obese = 21
| Test Statistics | |
| body_composition | |
| Chi-Square | 14.780a |
| df | 2 |
| Asymp. Sig. | <.001 |
| a. 0 cells (0.0%) have expected frequencies less than 5. The minimum expected cell frequency is 33.3. |
(X2 = 14.780, p < .001)
Allows you to test whether or not your data is the same as or different from an expected outcome.
That is, is the difference due to chance or not?
Use frequencies (counts)
Suppose you wanted to know about whether this distribution of body composition is statistically different from an expected outcome, that would be by chance.
Now, the predication of the null hypotheses would pose a standardized expected outcome of approximately 33.3 people would fall into these three groups. (NO difference in the counts)
You can see that there is also a note indicating that the chi sq assumption of expected frequencies of at least 5 in each cell has been met.
You actually don’t even need statistical software to display this. You can use a rather simple chi square formula and you could easily do this by hand or in excel.
The chi square, in this case 14.780 is represented by an upper case X in italics with a superscript 2 (X2 = 14.780, p < .001)
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Chi Square test of independence
Allows a comparison between two variables.
Can be 2 dichotomous variables (2x2)
Or can be more (2x3,4, or5)
All have the assumption that you must have at least 5 observations in each cell.
Small samples might use a special type of chi square called Fisher Exact.
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Practice-focused question
Was there an increase in ultrasound guided peripheral IV placements wheh compared to the number of central line placements before and after an educational intervention?
Compare line duration time in hours pre to post educational intervention
Chi Square
| Group | Total | ||||
| Preintervention | Postintervention | ||||
| LineType | CVC & Midline | Count | 9 | 11 | 20 |
| Adjusted Residual | 2.3 | -2.3 | |||
| USGPIV | Count | 0 | 8 | 8 | |
| Adjusted Residual | -2.3 | 2.3 | |||
| Total | Count | 9 | 19 | 28 |
| Chi-Square Tests | |||||
| Value | df | Asymptotic Significance (2-sided) | Exact Sig. (2-sided) | Exact Sig. (1-sided) | |
| Pearson Chi-Square | 5.305a | 1 | .021 | ||
| Continuity Correctionb | 3.443 | 1 | .064 | ||
| Likelihood Ratio | 7.639 | 1 | .006 | ||
| Fisher's Exact Test | .029 | .024 | |||
| Linear-by-Linear Association | 5.116 | 1 | .024 | ||
| N of Valid Cases | 28 | ||||
| a. 1 cells (25.0%) have expected count less than 5. The minimum expected count is 2.57. | |||||
| b. Computed only for a 2x2 table |
Note a few things about these results:
There were NO USGPIV placed in the preintervention period but 8 placed in the postintervention phase.
There were, however, 9 central lines placed pre and 11 placed post.
There were only 9 in the preintervention group and 19 in the post for a total of 28 lines placed.
There four different chi square tests performed including two sided tests, and exact tests.
This is a 2 x 2 contingency table which enables a Fisher exact test. This test (used for small samples in this 2by2 table) provides a two sided and a one-sided test. Notice that the one-sided test is just a bit more powerful, and it should be used when you have a reason to suspect that the outcome will go in a certain direction. This student launched an intensive training program on the use of the USGPIV technique with the intention of reducing the dependence on central lines. p =.024
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Difference time in hours
| Group Statistics | |||||
| Group | N | Mean | Std. Deviation | Std. Error Mean | |
| Ln Duration | Preintervention | 9 | 59.44 | 65.531 | 21.844 |
| Postintervention | 12 | 30.17 | 46.861 | 13.528 |
| Ranks | ||||
| Group | N | Mean Rank | Sum of Ranks | |
| Ln Duration | Preintervention | 9 | 13.06 | 117.50 |
| Postintervention | 12 | 9.46 | 113.50 | |
| Total | 21 |
| Test Statisticsa | |
| Ln Duration | |
| Mann-Whitney U | 35.500 |
| Wilcoxon W | 113.500 |
| Z | -1.318 |
| Asymp. Sig. (2-tailed) | .188 |
| Exact Sig. [2*(1-tailed Sig.)] | .193b |
| a. Grouping Variable: Group | |
| b. Not corrected for ties. |
Line duration pre M=59.44 SD 65.531
Line duration post M=30.17 SD
Nonparametric test chosen to compare means because of the very small sample. The Mann Whitney U test was used because it is the nonparametric equivalent of the independent samples t test.
You can see by the signifance levels, both the asymptotic 2 tailed test and the exact 1 tailed test are both much greater than .05. Thus, we must accept the null hypothesis that there is no difference in the means.
The nonparametric tests use ranks but they are still tests used to evaluate for differences in the means. The reason is that in samples that are not normally distributed, there is a fair amount of violation of normality. The distributions are badly skewed and often have outliers. Outliers cause the mean to be distorted. The median (which relys on ranking) is a better way to evaluate a dataset.
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Creating a bar chart
There is a pretty good chapter in Salkind & Frey on making various graphs and charts.
Here are the steps to create a 2D column chart for showcasing the number of central lines pre and postintervention and the number of USGPIV lines pre and post.
First, create a row for the labels. Then add the numbers right underneath.
Next, highlight both rows: the row of labels and the row of counts.
Go to the insert menu at the top of excel, right next to home. Choose the icon for a 2D column and click.
The chart appears.
You can add the data labels, too, by going to the “add chart element” icon (you still have to be in the insert tab) and click on data labels…..then the numbers will appear on the bars.
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Association between medication adherence and health-related quality of life in patients with chronic obstructive pulmonary disease
Moradkhani et al. (2021)
The purpose of the study was to investigate the relationship between medication adherence and Health-Related Quality of Life (HRQoL) in COPD patients referred to the pulmonologist’s office
N=100 patients
Morisky Medication Adherence Scale 8-item (MMAS-8 item)
St George’s Respiratory Questionaire for COPD (SGRQ-C)
MMAS 8 item scale was used to measure medication adherence and resulted in three categories: high adherence, medium adherence and low adherence.
Authors used a variety of statistical methods including measurement with both categorical and continuous variables.
Pearson’s Chi Sq, or Fisher’s exact for categorical comparisons.
ANOVA or Kruskal-Wallis.
Predictors evaluated using logistic regression.
For this discussion we will focus only on one chi sq comparisons.
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Categorical vs. continuous data
Two demographics were measured comparing low, medium and high adherence patients.
The number of administered drugs is a categorical variable and so is the MMAS-8item variable. These are both categorical variables and so a 2x3 Chi Sq test of independence was used. It shows a statistically significant outcome with a p value of <0.001. High adherence patients had fewer drugs (3) to take as compared to low and medium adherence patients.
On the other hand, disease duration was measured in years. Although years were measured on a continuous basis, the researchers also collapsed these data into another category. There were 9 patients with longer duration (4 to 17 years) and 4 patients with high adherence (1 to 8 years) suggesting that less time with the diagnosis is linked with higher adherence (p=.04).
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Key Points
Fisher exact a type of Chi Sq used for small samples; must be a 2x2 design
Chi Sq for comparisons test of independence, can be 2x2 or 2x3
Bar charts are fairly easy to do in excel
Moradkhani et al. (2021) used Chi Sq along with several other statistical tests.
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