The standard deviation of daily returns of a stocks price is

The standard deviation of daily returns of a stock’s price is used as a measure of the risk of that stock. Suppose that in a sample of 101 days, the standard deviation of a particular stock is 1.15%.

a) Find the 90% confidence interval of the population variance for this stock.

b) In the past, the standard deviation of the daily returns of this stock has been

1.56%. Test the hypothesis at the 1% level of significance that the standard

deviation has decreased from its previous level. Please show work.

Solution

A)

As              
              
df = n - 1 =    100          
alpha = (1 - confidence level)/2 =    0.05          
              
Then the critical values for chi^2 are              
              
chi^2(alpha/2) =    124.3421134          
chi^2(alpha/2) =    77.92946517          
              
Thus, as              
              
lower bound = (n - 1) s^2 / chi^2(alpha/2) =    1.063597814          
upper bound = (n - 1) s^2 / chi^2(1 - alpha/2) =    1.697047448          
              
Thus, the confidence interval for the variance is              
              
(   1.063597814   ,   1.697047448   ) [ANSWER]

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B)

Formulating the null and alternative hypotheses,          
          
Ho:   sigma   >=   1.56
Ha:    sigma   <   1.56
          
As we can see, this is a    left   tailed test.  
          
Thus, getting the critical chi^2, as alpha =    0.01   ,  
alpha =    0.01      
df = N - 1 =    100      
chi^2 (crit) =    70.06489493        
          
Getting the test statistic, as          
s = sample standard deviation =    1.15      
sigmao = hypothesized standard deviation =    1.56      
n = sample size =    101      
          
          
Thus, chi^2 = (N - 1)(s/sigmao)^2 =    54.34335963      
          
Comparing chi^2 < chi^2(crit), we REJECT THE NULL HYPOTHESIS.          

Thus, there is significant evidence that the standard deviation decreased from its previous level. [conclusion]