In yesterday’s blog, I demonstrated how to calculate the percentage strength of a dialysis mixture given an assortment of input fluids of various dextrose percentages. In today’s blog, we are going to consider the question: Given that I am using mixtures of 2.125% and 1.72% dextrose strength and that this strength is a factor in determining how much fluid (ultrafiltration) is removed during peritoneal dialysis, what is the difference, if any, in ultrafiltration between the nights when I use a 1.5% 5-liter with a 2.5% 3-liter fluid, as compared with alternative nights when I use a 2.5% 5-liter with a 1.5% 3-liter?
The hypothesis in this case is that there is no difference in the means of the two groups of observed ultrafiltration samples. This is the NULL hypothesis. A paired samples t-test based on a “matched-pairs sample” is the appropriate statistical methodology in this case (trust me!) A t-test is a type of statistical analysis used to compare the averages of two groups and determine whether the differences between them are more likely to arise from random chance.
Although I have never used my AI assistant Poe to conduct statistical analysis previously, I knew it could do so. So first, I asked Poe what format it wanted to receive data in with the following results:
To perform a t-test, you must provide the data for two groups or samples. Each group should consist of numerical values that you want to compare. Here’s an example of how you can input the data for a t-test:
Group 1: [1, 2, 3, 4, 5]
Group 2: [2, 4, 6, 8, 10]
In this example, we have two groups: Group 1 and Group 2. Group 1 contains the values [1, 2, 3, 4, 5], and Group 2 contains the values [2, 4, 6, 8, 10]. These are just sample values, and you should replace them with your own data.
I then inputted my recorded ultrafiltration data from my log sheet, using only the 18 data points since my catheter was repositioned. Group 1 is the lower percentage group, and Group 2 is the higher percentage Group. Poe responded with the following:
To perform a paired t-test on the given data, we’ll compare the paired observations between Group 1 and Group 2. Here are the two groups of data:
Group 1: [824, 1041, 542, 1337, 1163, 1180, 1174, 1184, 1131]
Group 2: [1054, 737, 935, 1017, 929, 1022, 810, 831, 1317]
I then informed Poe that I desired to have a 95% confidence level in the results, which corresponds to a p-value of 0.05, and the following resulted:
Given a significance level 0.05, we compare the calculated t-value (-1.46) with the critical t-value from the t-distribution table with 8 degrees of freedom.
Since the t-value does not fall within the critical region (beyond the critical value), we fail to reject the null hypothesis. This means that we do not have sufficient evidence to conclude that there is a significant difference between Group 1 and Group 2.
In other words, at a significance level of 0.05, the null hypothesis is supported, suggesting that there is no significant difference between the means of the paired observations in Group 1 and Group 2.
The bottom line for those of you whose eyes glazed over at the first mention of statistics is that I am obtaining the statistically same ultrafiltration whether using the 2.215% or 1.72% solution. It makes no difference in the amount of fluid we are pulling off. QED
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