##### SCA-13
##### 
##### One-Sample Variance Procedures
##### 

### This gives a few examples of the analysis process for testing
### for a population variance



### Preamble

# Import extra functionality
source("http://rfs.kvasaheim.com/stat200.R")






### Example I: Risk of IBM
#
#   I would like to estimate the risk (variance) of an investment
#   in IBM stock. To do this, I measure closing price of IBM for
#   10 random days in the past month.
#

IBM = c(152.64, 153.45, 149.00, 153.94, 152.14, 153.84, 152.69, 
	158.07, 160.91, 148.79)

shapiroTest(IBM)

onevar.test(IBM)

#   Conclusion: There is no evidence that the stock price of IBM
#   does not follow a Normal distribution (p-value = 0.2521). 
#   Because of this, I use the Chi-square provedure for variance. 
#   According to that procedure, a 95% confidence interval for the 
#   risk (variance) of an investment in IBM is between 6.4 and 45.1, 
#   with a point estimate of 13.5.





### Example II: Risk of Microsoft
#
#   I would like to estimate the risk (variance) of an investment
#   in Microsoft stock. To do this, I measure closing price of 
#   Microsoft for 10 random days in the past month.
#

MSFT = c(89.39, 88.52, 92.33, 90.77, 91.86, 93.08, 96.07, 96.11, 95.35, 92.31)

shapiroTest(MSFT)

onevar.test(MSFT)

#   Conclusion: There is no evidence that the stock price of MSFT
#   does not follow a Normal distribution (p-value = 0.4989). 
#   Because of this, I use the Chi-square provedure for variance. 
#   According to that procedure, a 95% confidence interval for the 
#   risk (variance) of an investment in IBM is between 3.3 and 23.4, 
#   with a point estimate of 7.0.






### Example III: Relative Risk of Microsoft
#
#   I would like to estimate the risk (variance) of an investment
#   in Microsoft stock and compare it to the expected risk for a
#   stock in the technology sector, which is 25. To do this, I 
#   measure closing price of Microsoft for 10 random days in the 
#   past 1 month.
#

MSFT = c(89.39, 88.52, 92.33, 90.77, 91.86, 93.08, 96.07, 96.11, 95.35, 92.31)

shapiroTest(MSFT)

onevar.test(MSFT, s2=25)

#   Conclusion: There is no evidence that the stock price of MSFT
#   does not follow a Normal distribution (p-value = 0.4989). 
#   Because of this, I use the Chi-square provedure for variance. 
#   According to that procedure, there is significant evidence that 
#   Microsoft has a significantly lower risk than what is expected 
#   for technology stocks (p-value = 0.0397). 
#
#   A 95% confidence interval for the risk (variance) of an investment 
#   in IBM is between 3.3 and 23.4, with a point estimate of 7.0.






### Example IV: Relative Risk of Gold
#
#   I would like to estimate the risk (variance) of an investment
#   in gold metal and compare it to the expected risk for a
#   stock in the NYSE, which is 15. To do this, I measure closing 
#   price of gold for 10 random days in the past 1 month.
#

Au = c(1310.3, 1319.6, 1366.8, 1315.1, 1340.4, 1366.4, 1348.6, 
	1369.1, 1367.0, 1316.0)

shapiroTest(Au)

st = numeric()
for(i in 1:1e4) {
  x = sample(Au, replace=TRUE)
  st[i] = var(x)
}
mean(st<=15)
quantile(st, c(0.025,0.5,0.975))

#   Conclusion: There is evidence that the gold prices do not
#   follow a Normal distribution (p-value = 0.0351). Because of 
#   this, I use the non-parametric bootstrapping procedure for 
#   the variance. According to that procedure, there is significant 
#   evidence that gold has a significantly higher risk than what is 
#   expected for the (p-value = 0.0003). 
#
#   A 95% confidence interval for the risk (variance) of a gold 
#   investment is between 282.1 and 762.9, with a point estimate 
#   of 560.3.