BME100 f2014:Group1 L2

From OpenWetWare
Jump to: navigation, search
Owwnotebook icon.png BME 100 Fall 2014 Home
People
Lab Write-Up 1 | Lab Write-Up 2 | Lab Write-Up 3
Lab Write-Up 4 | Lab Write-Up 5 | Lab Write-Up 6
Course Logistics For Instructors
Photos
Wiki Editing Help
BME494 Asu logo.png


OUR TEAM

Name: Hawley Helmbrecht
Name: Sarah Fakhoury
Role(s)
Name: Prerna Gupta
Role(s)
Name: Jonathan Riecker
Name: Michael Pineda
Role(s)
Name: Timothy Black

LAB 2 WRITE-UP

Descriptive Statistics

Experiment 1
Human Study

Inflammotin Mean:

0mg:3.834 pg/mL   
5mg:8.932 pg/mL    
10mg:61.622 pg/mL   
15mg:657.941 pg/mL

Standard Deviation:

0mg:1.523 pg/mL
5mg:1.594 pg/mL
10mg:30.111 pg/mL
15mg:212.943 pg/mL

Standard Error:

0mg:0.482 pg/mL
5mg:0.504 pg/mL
10mg:9.522 pg/mL
15mg:67.338 pg/mL


Experiment 2
Rat Study

Inflammotin Mean:

0mg:10.516 pg/mL
10mg:11.112 pg/mL

Standard Deviation:

0mg:2.226 pg/mL
10mg:7.403 pg/mL

Standard Error:

0mg:0.995 pg/mL
10mg:3.311 pg/mL




Results

Experiment 1

BME100 WG1 Human Study Graph.png


Experiment 2

Description of image 




Analysis

Experiment 1
Human Study

Use the ANOVA test to analyze the data. The ANOVA test is the best way to analyze the human study data because it is a one-way analysis of variance and it is use to determine significant differences between the means of three or more independent groups that are unrelated (0mg, 5mg, 10mg, 15mg groups).

After using the ANOVA test to analyze the data, the p-value obtained was 1.4E-16

When analyzing the p-value, since it is less than .05, there is a significant difference.


Experiment 2
Rat Study

Use the t-test in order to analyze the data. The t-test is the best way to analyze the rat study data because there are only two different groups of rats receiving separate treatment (0mg and 10mg doses).

After using the T-test with Two Sample assuming equal variance, the p-value obtained was .867403

When analyzing the p-value, since the value is more than .05, there is no significant difference.




Summary/Discussion

When analyzing the data, the p-value found in the human study by performing the ANOVA test showed that there was a significant difference. However, the p-value found in the rat study by performing the t-test showed that there is not a significant difference. Since there is a significant difference in the human study, it can be concluded that the implementation of the lipopolysaccharide in order to increase the inflammatory protein (Inflammotin) in the elderly is greatly supported by the data. However, since the p-value proved that there is no significant difference in the Rat Study the lipopolysaccharide does not increase the inflammation within the rats and does not support the original hypothesis. This could be the case due to the small size of the rats and the efficiency of the quick metabolism did not allow enough time for the drug to take affect. Therefore, even though the two studies provide different results, the human study still supports the use of the lipopolysaccharide in the elderly in order to increase Inflammotin.