Hello everyone!
This week was a bit slower than other weeks, but that was because I hit a complication in the project. When I say complication, I mean standard deviation...
I was in contact with my ASU advisor and I had a difficult time understanding what he was trying to get me to do. Really, that was the complication because finding the standard deviation itself is not hard. Thank you chemistry!! Just in case you are not familiar with the term standard deviation, it refers to how far data is from the mean (average). It's on a scale from 0 to 1. If your data gives you a standard deviation of 1, it means your data is pretty scattered.
After a reading a great deal about standard deviation, I realized what he was asking me to do. He wanted me to find the standard deviation of the data and the standard deviation of the data with a fitted line (trendline). I had to do all of this in Excel, for it's the easiest way to graph, use trendlines, and input equations. In case you didn't read my last post, I am using two scenarios for my data. I am graphing the peak magnitudes (maximum brightness) of the supernovae (explosions) against the time it takes for the supernovae to reach 95% of their peak magnitudes. I am also graphing the peak magnitudes of the supernovae against the time it takes for the supernovae to gain 1 magnitude from its peak magnitude. Since we are dealing with negatives, the magnitude will be heading towards zero.
For the Peak Magnitude vs 95% Magnitude data, I had to find two standard deviations. I found a standard deviation of 0.817 for the fitted trendline and a standard deviation of 0.941 for the data without the fitted line. As you can see, the data is pretty scattered.
For the Peak Magnitude vs 1 + Peak Magnitude data, I had to find two more standard deviations. I found a standard deviation of 0.922 for the trendline and a standard deviation of .941 for the data without the fitted line. Similarly, the data is scattered.
That's pretty much what I have been up to this week. In summary, I had to find the standard deviation of my data. This would show me how scattered my data is. However, with supernovae type II, there is usually a great deal of scatter, as the explosions are usually varied.
Now, you may be asking yourself, "what does this all mean?" Great Question!!
See you next week!
This week was a bit slower than other weeks, but that was because I hit a complication in the project. When I say complication, I mean standard deviation...
I was in contact with my ASU advisor and I had a difficult time understanding what he was trying to get me to do. Really, that was the complication because finding the standard deviation itself is not hard. Thank you chemistry!! Just in case you are not familiar with the term standard deviation, it refers to how far data is from the mean (average). It's on a scale from 0 to 1. If your data gives you a standard deviation of 1, it means your data is pretty scattered.
After a reading a great deal about standard deviation, I realized what he was asking me to do. He wanted me to find the standard deviation of the data and the standard deviation of the data with a fitted line (trendline). I had to do all of this in Excel, for it's the easiest way to graph, use trendlines, and input equations. In case you didn't read my last post, I am using two scenarios for my data. I am graphing the peak magnitudes (maximum brightness) of the supernovae (explosions) against the time it takes for the supernovae to reach 95% of their peak magnitudes. I am also graphing the peak magnitudes of the supernovae against the time it takes for the supernovae to gain 1 magnitude from its peak magnitude. Since we are dealing with negatives, the magnitude will be heading towards zero.
For the Peak Magnitude vs 95% Magnitude data, I had to find two standard deviations. I found a standard deviation of 0.817 for the fitted trendline and a standard deviation of 0.941 for the data without the fitted line. As you can see, the data is pretty scattered.
For the Peak Magnitude vs 1 + Peak Magnitude data, I had to find two more standard deviations. I found a standard deviation of 0.922 for the trendline and a standard deviation of .941 for the data without the fitted line. Similarly, the data is scattered.
That's pretty much what I have been up to this week. In summary, I had to find the standard deviation of my data. This would show me how scattered my data is. However, with supernovae type II, there is usually a great deal of scatter, as the explosions are usually varied.
Now, you may be asking yourself, "what does this all mean?" Great Question!!
See you next week!
No comments:
Post a Comment