Answer:
(a) The sample variance for the daily price change is 0.2501.
(b) The sample standard deviation for the daily price change is 0.5001.
(c) The 95% confidence interval estimates of the population variance is (0.1255, 0.7210).
Step-by-step explanation:
Let the random variable X denote the stock price changes for a sample of 12 companies on a day.
The data provided is:
X = {0.82 , 1.44 , -0.07 , 0.41 , 0.21 , 1.33 , 0.97 , 0.30 , 0.14 , 0.12 , 0.42 , 0.15}
(a)
The formula to compute the sample variance for the daily price change is:
[tex]s^{2}=\frac{1}{n-1}\sum\limits^{12}_{i=1}{(X_{i}-\bar X)^{2}}[/tex]
The sample mean is computed using the formula:
[tex]\bar X=\frac{1}{n}\sum\limits^{12}_{i=1}{X_{i}}[/tex]
Consider the Excel output attached below.
In Excel the formula to compute the sample mean and sample variance are:
[tex]\bar X[/tex] =AVERAGE(A2:A13)
[tex]s^{2}[/tex] =VAR.S(A2:A13)
Thus, the sample variance for the daily price change is 0.2501.
(b)
The formula to compute the sample standard deviation for the daily price change is:
[tex]s=\sqrt{\frac{1}{n-1}\sum\limits^{12}_{i=1}{(X_{i}-\bar X)^{2}}}[/tex]
Consider the Excel output attached below.
In Excel the formula to compute the sample standard deviation is:
[tex]s[/tex] =STDEV.S(A2:A13)
Thus, the sample standard deviation for the daily price change is 0.5001.
(c)
The (1 - α)% confidence interval for population variance is:
[tex]CI=[\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2} } \leq \sigma^{2}\leq \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2} } ][/tex]
Compute the critical value of Chi-square for α = 0.05 and (n - 1) = (12 - 1) = 11 degrees of freedom as follows:
[tex]\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2,11}=21.920[/tex]
[tex]\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{(1-0.05/2),11}=\chi^{2}_{0.975,11}=3.816[/tex]
*Use a Chi-square table.
Compute the 95% confidence interval estimates of the population variance as follows:
[tex]CI=[\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2} } \leq \sigma^{2}\leq \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2} } ][/tex]
[tex]=[\frac{(12-1)\times 0.2501}{21.920 } \leq \sigma^{2}\leq \frac{(12-1)\times 0.2501}{3.816} ][/tex]
[tex]=[0.125506\leq \sigma^{2}\leq 0.720938]\\\approx [0.1255, 0.7210][/tex]
Thus, the 95% confidence interval estimates of the population variance is (0.1255, 0.7210).
