As I was thinking about polling while using factual data I realized that it was salient to show that the data presented in class was factual and statistically relevant. As presented there have been 90,000 troops brought home from the wars in Afghanistan and Iraq since the beginning of the conflict. Is this number statistically significant? First we would have to see what the total number of troops deployed to the aforementioned regions has been since the beginning of President Obama's term. When Obama took office in 2009, the U.S. had about 34,000 troops in
Afghanistan. Obama has initiated two major troop increases in
Afghanistan: about 20,000 additional troops were announced in February
2009, followed by the December 2009 announcement that an another 33,000
would be deployed as well; other smaller increases have brought the
total to 100,000. This number will represent our N. If we take 100,000 and divide it by 2 (we are representing total number by returning number). 50,000 is our population mean. We will then calculate the standard deviation 100,000-50,000=50,000 and 90,000-50,000=40,000. If we square both of these numbers and sum them our number is 4,100,000,000. Then we must divide the number by N (population NOT sample standard deviation) leaving us with 2,050,000,000 and then take the square root of this number which is 45,276.926. So then we will calculate the z score for the data set (we compute the z score because we are measuring the distance that a data value is from the means in terms of the number of standard deviations. The formula for this is z=x-M/Standard Deviation. So if x= 90,000-50,000=40,000/45,276.926= 0.883. Our z score is then 0.883 which means that our returned troops are 0.883 standard deviations above the mean.
Data retrieved from
http://www.fair.org/blog/2012/09/21/is-the-afghan-surge-really-over/
As I drove to work I realized that I did not provide the z score for remaining troops! 10,000-50,000=-40,000/45,276.926=
ReplyDelete-0.8834522025633984. This is showing standard deviations from the mean (on the left side of a gausian distribution.)
If this were to occur along a Gaussian distribution, these values would not be significant at a 5% significance level as they do not follow outside of a z-score of ±1.96. This being said the percentage at which they would become significant is at 18.954%, which I interpolated from my statistics book (Davis, 2002). This being said, I am not exactly sure how calculating the z score of how many troops were brought home determines whether or not the act is significant?
ReplyDeleteDavis, John. Statistics and Data Analysis in Geology. 3rd ed. New York: John Wiley & Sons, Inc, 2002. 601-602. Print.
This is true. I was simply looking at the statistical significance of the number. Not the act. As we learned in the first portion of the class unless you have agreement amongst the scientific community about an operational definition then there is little point in testing a "hypothesis" however, I don't believe that there was a hypothesis stated about the act...
ReplyDeleteI don't understand how you got 50,000 as the population mean. You divided 100,000 troops by 2. Why 2? It seems if you want to know the average number of troops deployed, it should be the average number of troops per day. You could calculate the number of troops deployed each day from say 2003 to 2011 (approximated using the numbers you give here). Then do the Z-score for the number of troops on deployment today.
ReplyDeleteI calculated the mean troops per day and came out with some interesting numbers... If we take 8x365=2920 then take the quotient of 100000/2920=34.24657534 (this being our troops per day). So if we calculate our standard deviation with this new mean then the z score formula will look like this z=90000-34/90516=.9939 for the troops returning home and 10000-34/90516=.1101 for troops remaining in country.
ReplyDelete