BME100 s2017:Group5 W8AM L3

From OpenWetWare
Jump to navigationJump to search
BME 100 Spring 2017 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

OUR TEAM

Name: Luis Paez
Name: Anisa Ahamed
Name: Matthew Grudza
Name: Fernanda Nunez
Name: Jenna Forrey
Name: Smita Gopalakrishnan

LAB 3 WRITE-UP

Descriptive Stats and Graph

Heart Rate:

Graph I: Average and Standard Deviation for Heart Rate Measures of Gold Standard and Spree
Description of image

Table I: Average, Standard Deviation, and p-value Calculations for Heart Rate Data

Gold Standard Spree
Average 98.090 98.954
SD 23.031 24.878
Pearson's Coefficient 0.691

Results Analysis
- From the Standard Deviation results, the values of about 23.0 and 24.9 indicate that there was a high variability among the data points.
- With the data for heart rate, we conducted a paired t-test. From the results of the t-test, we concluded that the data was not significant since the p-value was very high. Since less than a 0.05 value on the t-test indicates statistical significance, there is a greater divergence between Spree and Gold Standard temperature devices, and therefore not much confidence in the results.
- From the Pearson’s coefficient, we saw that the value was around 0.69, which infers that there is not much correlation between both arrays of data (Spree and Gold Standard temperature records). This is because a value of 1 or -1 shows a positive or negative correlation, and 0 indicates no linear correlation.


Temperature:

Graph II: Average and Standard Deviation for Temperature Measures of Gold Standard and Sprees
Description of image

Table II: Average, Standard Deviation, and p-value Calculations for Temperature Calculations

Gold Standard Spree
Average 96.647 95.531
SD 1.923 0.870
Pearson's Coefficient 0.193

Results Analysis
- From the Standard Deviation results, the values of 1.9 and 0.8 indicate that most of the data points are very close to the mean value, which increases confidence in the precision of the temperature measuring device.
- From the results of the t-test, we concluded that the p-value was very high and the results were not statistically significant.
- From Pearson’s coefficient, we saw that the value was around 0.19, displaying that there is not much correlation between both arrays of data.





Inferential Stats

Heart Rate:
Paired T-Test: 0.427 > 0.05
Results Analysis
This result is higher than our critical value for the student test, which is 0.05, and this indicates the rejection of the null hypothesis which states that there is no relationship between the two measured sets of data. This means that there is a relationship between the data obtain from both companies, and since the t-test is an inferential statistic test, shows the tendency of future measurements to have the same relationship.

Temperature:
Paired T-Test: 1.097E-21 < 0.05
Results Analysis
This result is much lower than our critical value for the t-test, which indicates no relationship between the two measured sets of data. This is not a favorable result for the reliability of the temperature measuring device, as it indicates that in the future this measures will not have a relationship even though they are intended two measured the same dependent variable (temperature of the user).




Design Flaws and Recommendations

One of the flaws of this design was that the products did not have consistent accuracy. The data collected had a couple of blank spots for both the Gold Standard and Spree Headband, displaying that the product failed to produce accurate results many times. This was especially prevalent with the Spree device and its the temperature readings. To improve this, these devices could develop more advanced technology for measuring heart rate and temperature.



Experimental Design of Own Device

Experimental Design
To test the accuracy of the HIV detector, two groups will be tested using the device and the results will be compared. The first group will be composed of 100 people who have been tested in a hospital and diagnosed with HIV. The other group of people will be composed of 100 people that have been tested in a hospital and are known to not have HIV. This is the negative control group. Once both of these test groups have used the device, the accuracy of the device will be calculated along with the chances of a false positive and a false negative. A false positive reading indicates that the device did not accurately obtain an HIV-negative person’s accurate diagnosis and a false negative indicates that the device did not accurately obtain an HIV-positive person’s accurate diagnosis. This experiment design allows us to obtain accurate statistics about our device and will allow potential users to see if it is reliable or not.

Controls
- The number of patients being tested in both groups.
- Number of HIV-positive vs HIV-negative patients.
- Swabbing technique

Variables
Independent Variable: HIV positive VS HIV-negative patients
Depedent Variable: : Results of nanotube testing device

Procedure
1. Obtain patient information to correctly identify the individual and prevent any mislabeling.
2. Take swab from inside mouth going along top gum from left to right, them bottom gums from right to left with nanotube testing device.
3. Place back in the cap to prevent any contamination.
4. Wait for device readings.
5. Compare the reading of the device to the patient’s actual HIV status.
6. Determine accuracy and reliability of the device.

Expected Results

The HIV-positive patients tested should have positive results. The HIV-negative patients should have negative results. In order for the experiment to be deemed as successful, the accuracy of the device should be precise and reliable with the already known diagnosis of the patient. Ideally, the standard deviation of the patient data in both groups should show a lower value, indicating the close spreading of the values compared to the mean value. This will show more consistency in the measurements taken by the device. The student test should yield a result lower than 0.05, indicating that the statistical difference between the control and the testing group is 95% different. This will indicate the device is diagnosing properly and therefore will elevate the confidence in the accuracy of the diagnose. Lastly, the Pearson’s coefficient should yield a value close to -1, indicating a negative correlation between both groups, as it is expected since they should yield different binary results.