In the case “Creatinine Levels in Hospitalized Patients” we visualized each variable in the data set individually using the JMP Distribution platform as shown in Figure 6.5. The average length of stay for these patients is 14.1 days with a minimum of 0 and a maximum of 29. Of the 372 patients, 25% have acute kidney injury. The problem statement asks us to consider inpatient length of stay. Patients having a length of stay of 0 would have been treated at the hospital on an outpatient basis; their records should not be included in this analysis. Select those rows with LOS = 0 using the JMP data filter. Select Rows > Data Filter, highlight LOS and click Add. Drag the right hand slider to 0. The completed Data Filter dialog is shown in Figure 7.2 Completed Data Filter Dialog to Select Rows Where LOS=0.

The rows where LOS=0 will be highlighted in the data table. Right-click over one of the highlighted rows and select the Exclude and Hide option. This will exclude these observations from subsequent analyses and hide them in subsequent graphs.

Since this is a comparative analysis, it is beneficial to describe the data separately for each group, those with AKI and those without AKI. JMP Graph Builder can easily create two length of stay histograms, one for those patients with AKI and one for those patients without AKI. Drag LOS into the X drop zone and Outcome into the Y Group drop zone. Select the histogram icon from the Control Panel. The resulting data visualization is shown in Figure 7.3 Histograms of Length of Stay by Outcome.

Notice that the resulting chart includes the count of excluded rows.

Graph Builder uses the same axis scales for both histograms allowing an accurate visual comparison. This is an example of the use of small multiples, which are series of graphs plotted on the same scale. This is a best practice in data visualization.

An alternate data visualization that facilities outlier identification is an outlier box plot which can be selected from the Graph Builder Control Panel. The resulting plot is shown in Figure 7.4 Outlier Box Plots for Length of Stay by Outcome.

The box indicates the middle 50% of the data (the first quartile to the third quartile). The first quartile corresponds the 25th percentile, where 25% of the length of stays are below that value. For the AKI group, the 25th percentile is 23 days. The vertical line inside the box is the median. The “whiskers” are the first quartile minus 1.5 times the interquartile range (third quartile – first quartile) and the third quartile plus 1.5 times the interquartile range. Outliers are indicated as dots that lie beyond the end of the whiskers. Box plots are a compact way to visualize a data distribution including its center, spread, skewness, and outliers.

In the No AKI box plot we observe an outlier with a length of stay of 29 days. Click on the dot to highlight the corresponding patient record in the JMP data table. This is an example of JMP’s dynamic data linking feature, where an observation or group of observations highlighted in either a data table or graph will be highlighted in all other data tables and graphs. The highlighted record is for Patient_ID = 7581, a 92 year old African-American man with no co-morbidities. Further investigation should be conducted with the help of a subject matter expert to determine if this outlier should be removed from the data set. Outliers are removed, not based on their influence on the statistical results, but on an understanding of the data in the domain context, and should be dispositioned accordingly. For example, if investigation revealed a data collection or recording error, then either a corrected value should be entered or the observation removed. Exclusion of observations should be documented in accordance with the practices of reproducible research.

Finally, Figure 7.5 Descriptive Statistics for Length of Stay by Outcome shows a table of descriptive statistics for length of stay by outcome. This can be accomplished using Tabulate where LOS and the desired statistics are placed in the drop zone for Rows and the nominal variable Outcome is placed into the drop zone for Columns.

This research question can be answered by conducting a two-independent samples t-test that will determine if the mean length of stay for the AKI group differs from that of the group without AKI. The two-independent samples t-test is an appropriate statistical method to apply when the dependent variable is continuous and the independent variable is nominal with two levels. This test assumes that the lengths of stay for both groups are normally distributed. The Normal quantile plots for these two groups do not show serious departures from normality. In the problems at the end of this case, you will be asked to create and assess these plots.

To begin, select Analyze > Fit Y by X and enter LOS into the Y field and Outcome into the X field. There are two different two-independent samples t-test available in JMP. The test to apply depends on whether the length of stay variances of the two groups (AKI and No AKI) are equal or unequal. Several equality of variance tests are available from the Fit Y by X drop-down menu option Unequal Variances. The JMP output is shown in Figure 7.6 Equality of Variance Tests.

The Levene test is a good general equality of variance test. The null hypothesis is that the variances of the two groups are equal versus the alternative that they are not equal. Small p-values (less than 0.05) cause the null hypothesis to be rejected. For this data, the Levene test p-value is <0.0001 which is significant at the 5% level, so the variance of the length of stay for the AKI group is significantly different from the group without AKI.

Now we are ready to perform the t-test. When the variances are not equal, the option to select from the Fit Y by X drop-down menu is t Test. The null hypothesis for the two sample t-test is that the means of the two groups are equal with an alternative that the means are not equal. The results for the length of stay data are shown in Figure 7.7 Two-independent sample t-test for Unequal Variances.

In the plot showing LOS by Outcome, the points have been “jittered” which spreads out the markers to avoid overplotting and gives you a better sense of the data density (From the Oneway Analysis of LOS by Outcome drop-down, select Display Options > Points Jittered).

The length of stay for patients without AKI is on average 14.1 days lower than for patients with AKI. Is this difference statistically significant? The null hypothesis for the t-test is that the length of stay means of the two patient groups (with and without AKI) are equal. The p-value associated with the t-test determines if the observed difference (-14.1) is statistically significant. P-values less than the chosen significance level indicate that the two patient groups are on average significantly different. The JMP output for the t-test gives three possible p-values, two associated with the one-sided alternatives (greater than and less than) and one associated with the two-sided alternative (not equal). The correct p-value depends on how the alternative hypothesis was specified and is derived from the research question. For this research question we want to see if there is a difference between the two patient groups which corresponds to a two-sided (≠) alternative. The correct p-value is Prob > |t| <0.0001 as shown in Figure 7.7 Two-independent sample t-test for Unequal Variances, which indicates a statistically significant difference at the 5% level. This means that on average patients without AKI have length of stay 14.1 days shorter than patients with AKI.

A statistical model quantifying the relationship between length of stay and AKI diagnosis will address this research question. When the dependent variable (Outcome) is nominal and there is a continuous independent variable (LOS), a logistic regression will yield such a statistical model. A logistic regression expresses the natural log odds of the dependent variable as a linear function of the independent variable. Odds expresses the likelihood of an event occurring and is calculated as the ratio of the number of occurrences of the event to the number of times the event did not occur. For this data, the odds of having AKI are 93/263 = 0.354; the odds of not having AKI for hospitalized patients is 263/93 = 2.828. This means that a hospitalized patient is almost three times more likely to not have AKI than to have AKI. It is easier to understand the likelihood when expressed in the form that is greater than one. Odds can be expressed in terms of probability as p/(1-p). The probability of having AKI, p, is 93/356 = 0.261.

A logistic regression equation can be estimated with the Fit Y by X platform by entering Outcome in the Y field and LOS in the X field. The JMP output is shown in Figure 7.8 JMP Logistic Regression Output.

The Whole Model Test uses a Chi-square test to determine whether the logistic regression model is significant in explaining the relationship between the likelihood of AKI and length of stay. In this case, the Chi-square test yields a p-value of <0.0001 which is significant at the 5% level.

The fitted logistic regression coefficients are given in the Parameter Estimates table. The logistic regression equation relating LOS and log odds of having AKI to not having AKI is:

ln odds(AKI/No AKI) = −19.052 + 0.883*LOS

This is the estimated log odds ratio of having AKI to not having AKI for a given length of stay. A chi-square test is used to determine if the regression coefficient for LOS is significantly different from zero. For this data, LOS is a significant predictor of AKI (p<0.0001) at the 5% significance level. The plot in Figure 7.8 JMP Logistic Regression Output shows the logistic regression model as a blue curve. Notice that the y-axis scale is in terms of the probability, not odds.

It is not intuitive to think in terms of a log odds ratio. Exponentiating the regression coefficient for LOS gives what is referred to as the unit odds ratio which is the odds ratio associated with a one-unit increase in the LOS.

The odds ratio can be obtained from the drop-down menu associated Logistic Fit Outcome by LOS as shown in Figure 7.9 Obtaining Odds Ratio Output for Logistic Regression.

The odds ratio output is added to the Parameter Estimates table as shown in Figure 7.10 Odds Ratio Output for Logistic Regression.

The odds ratio of AKI to No AKI associated with an increase in one day of length of stay is 2.42.

RSquare (U) gives a measure of goodness-of-fit on a scale of 0 to 1. An RSquare (U) of 0 indicates that the model does not predict the outcome (AKI) and an RSquare (U) of 1 indicates that the logistic regression model is a perfect predictor of the outcome (AKI). For this data the Rsquare (U) is 0.7924 indicating relatively good predictability.