Question

The table showing the stock price changes for a sample of 12 companies on a day is contained in the Excel file below.

Price Change ($)
0.82
1.44
-0.07
0.41
0.21
1.33
0.97
0.30
0.14
0.12
0.42
0.15

Construct a spreadsheet to answer the following questions.
a. Compute the sample variance for the daily price change (to 4 decimals).
b. Compute the sample standard deviation for the price change (to 4 decimals).
c. Provide 95% confidence interval estimates of the population variance (to 4 decimals).

Answer 1

(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).

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:

`s^(2)=(1)/(n-1)\sum\limits^(12)_(i=1){(X_(i)-\bar X)^(2)}`

The sample mean is computed using the formula:

`\bar X=(1)/(n)\sum\limits^(12)_(i=1){X_(i)}`

Consider the Excel output attached below.

In Excel the formula to compute the sample mean and sample variance are:

`\bar X` =AVERAGE(A2:A13)

`s^(2)` =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:

`s=\sqrt{(1)/(n-1)\sum\limits^(12)_(i=1){(X_(i)-\bar X)^(2)}}`

Consider the Excel output attached below.

In Excel the formula to compute the sample standard deviation is:

`s` =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:

`CI=[((n-1)s^(2))/(\chi^(2)_(\alpha/2) ) \leq \sigma^(2)\leq ((n-1)s^(2))/(\chi^(2)_(1-\alpha/2) ) ]`

Compute the critical value of Chi-square for α = 0.05 and (n - 1) = (12 - 1) = 11 degrees of freedom as follows:

`\chi^(2)_(\alpha/2, (n-1))=\chi^(2)_(0.05/2,11)=21.920`

`\chi^(2)_(1-\alpha/2, (n-1))=\chi^(2)_((1-0.05/2),11)=\chi^(2)_(0.975,11)=3.816`

*Use a Chi-square table.

Compute the 95% confidence interval estimates of the population variance as follows:

`CI=[((n-1)s^(2))/(\chi^(2)_(\alpha/2) ) \leq \sigma^(2)\leq ((n-1)s^(2))/(\chi^(2)_(1-\alpha/2) ) ]`

`=[((12-1)* 0.2501)/(21.920 ) \leq \sigma^(2)\leq ((12-1)* 0.2501)/(3.816) ]`

`=[0.125506\leq \sigma^(2)\leq 0.720938]\\\approx [0.1255, 0.7210]`

Thus, the 95% confidence interval estimates of the population variance is (0.1255, 0.7210).