Systemic complications in acute pancreatitis are largely responsible for mortali

Question

Systemic complications in acute pancreatitis are largely responsible for mortality associated with the disease. Proinflammatory cytokines, particularly TNFa, may play a role in acute pancreatitis by mediating the systemic sequelae. One research team used a bile-infusion model of acute pancreatitis to show amelioration of disease severity as well as an improvement in overall survival by TNFa inhibition. Experimental material consisted of adult male Sprague- Dawley rats weighing between 250 and 300 grams divided into three groups: untreated (bile solution infused without treatment); treated (bile solution infused preceded by treatment with polyclonal anti-TNF antibody); and sham (saline infused). Among the data collected were the following hematocrit (%) values for animals surviving more than 48 hours: Sham Untreated Treated 56 60 50 50 50 38 40 32 36 40 40 38 40 38 40 40 42 38 46 36 35 40 40 35 36 40 40 35 45 Was there any effect of treatment? Give p-value. If there was a significant effect overall, determine which pairs were significant.

Explanation / Answer

In order to check wiether there is any effect of treatment, we will one way ANOVA to this data. Excel gives following output:

In order to identify exactly which pairs differ significantly, we run two sample t test for each pair.

Anova: Single Factor Anova: Single Factor SUMMARY Groups Count Sum Average Variance Sham 10 382 38.2 6.622222 Untreated 5 266 53.2 21.2 Treated 15 603 40.2 28.6 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 817.5 2 408.75 20.25743 4.23E-06 3.354131 Within Groups 544.8 27 20.17778 Total 1362.3 29

Observe that the p-value corresponding to overall model significance is 4.23E-06, i.e. 0.00000423 < 0.05, hence we reject the null hypothesis and conclude that at least two groups means differ significantly.

In order to identify exactly which pairs differ significantly, we run two sample t test for each pair.

t-Test: Two-Sample Assuming Equal Variances Sham Untreated Mean 38.2 53.2 Variance 6.622222 21.2 Observations 10 5 Pooled Variance 11.10769 Hypothesized Mean Difference 0 df 13 t Stat -8.2171 P(T<=t) one-tail 8.33E-07 t Critical one-tail 1.770933 P(T<=t) two-tail 1.67E-06 t Critical two-tail 2.160369 Observe that the p value = 0.00000167 < 0.05, hence there is significant difference between Sham and Untreated t-Test: Two-Sample Assuming Equal Variances Sham Treated Mean 38.2 40.4 Variance 6.622222 14.8 Observations 10 5 Pooled Variance 9.138462 Hypothesized Mean Difference 0 df 13 t Stat -1.3287 P(T<=t) one-tail 0.1034 t Critical one-tail 1.770933 P(T<=t) two-tail 0.2068 t Critical two-tail 2.160369 Observe that the p value = 0.2068 > 0.050.05, hence there is no significant difference between Sham and treated t-Test: Two-Sample Assuming Equal Variances Untreated Treated Mean 53.2 40.4 Variance 21.2 14.8 Observations 5 5 Pooled Variance 18 Hypothesized Mean Difference 0 df 8 t Stat 4.770278 P(T<=t) one-tail 0.000704 t Critical one-tail 1.859548 P(T<=t) two-tail 0.001408 t Critical two-tail 2.306004 Observe that the p value = 0.001408 < 0.050.05, hence there is significant difference between Untreated and treated

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