# BME100 f2014:Group5 L2

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# OUR TEAM

 Name: Keaton Sorenson Name: Megan Danforth Name: Jad Jazzar Name: Zahra Khuraidah Name: Christopher Rojas Name: Sahar Mohamed

# LAB 2 WRITE-UP

## Descriptive Statistics

Experiment 1 - Humans

Experiment 2 - Rats

## Results

Experiment 1
For Experiment 1: The comparison in T.test p-value for each two groups in humans are:
- (0 mg VS. 5 mg) p-value = 8.59 E-07
- (0 mg VS. 10mg) p-value = 9.94 E-06
- (0mg VS. 15mg) p-value = 1.39 E-08
- (5mg VS. 10mg) p-value = 3.02 E-05
- (5mg VS.15mg) p-value = 1.57 E-08
- (10mg VS.15mg)p-value = 6.48 E-08
The T.test p-values for the six groups are less than 0.0167 which means that the results are statistically significant.

Experiment 2
For Experiment 2: The p-value in Rats= 0.867
Since our p-value is P > 0.05 that means, there is no significant difference between groups.

## Analysis

Experiment 1

The p-value in the human study shows a significant difference in inflammotin protein produced.

Experiment 2

The p-value shows no significant difference in inflammotin protein produced.

## Summary/Discussion

After analyzing the data, it was found that the p-value of the ANOVA test in the human study showed a significant difference. Unlike the human study, the t-test from the rats showed a p-value with no significant difference. Because of the significant difference in the human study, it can be concluded that implementation of lipopolysaccharide in elderly people can increase the inflammatory protein inflammotin. Since the p-value from the rat study shows no significant difference, it can be concluded that lipopolysaccharide does not increase inflammation within rats. This does not support the original hypothesis of lipopolysaccharide increasing inflammation in rats. Although results from the two tests are different, the human study proves that lipopolysaccharide does increase inflammotin.