Question

1. To analyze the risk, or volatility, associated with investing in Chevron Corporation common stock, a sample of the monthly total percentage return for 12 months was taken. The returns for the 12 months of 2005 are shown here (CompuStat, Feb 24, 2006). Total return is price appreciation plus any dividend paid. togel sgp seni takiem. The returns to Month Jan Feb Fi Apr May Return (%) 3.60 14.86 -6.07 -10.82 4.29 3.98 3.74 6.62 5.42 -11.83 1.21 -0.94 | Jun Jul Aug Sep Oct Nov Dec a. Compute the sample variance and sample standard deviation as a measure of volatility of monthly total return for Chevron. b. Construct a 95% confidence interval for the population variance. c. Construct a 95% confidence interval for the population standard deviation.

Answer 1

a.

x: Monthly return for a month

Sample mean

$\overline{x}=\frac{\Sigma x}{n}$

n: Sample size = 12

sample Variance

T-u (2-x)3 2

sample standard deviation

S Σ(x – )2 TV η -1 = 1

Month	x:Return %	(x-$\overline{x}$)	(x-$\overline{x}$)2
Jan	3.6	2.4283	5.896803
Feb	14.86	13.6883	187.3705
Mar	-6.07	-7.2417	52.44174
Apr	-10.82	-11.9917	143.8001
May	4.29	3.1183	9.724003
Jun	3.98	2.8083	7.886736
Jul	3.74	2.5683	6.596336
Aug	6.62	5.4483	29.68434
Sep	5.42	4.2483	18.04834
Oct	-11.83	-13.0017	169.0433
Nov	1.21	0.0383	0.001469
Dec	-0.94	-2.1117	4.459136
Total: $\Sigma x$	14.06		Σr – ) =634.9528
Sample mean: $\overline{x}$	14.06/12 =1.1717

sample Variance

152 - %(v – 7)? _ 6349528 – 87.7230 -1 11

Sample standard deviation

s = 11 = V57.7230 = 7.5976

b.

Confidence interval for population variance

(n-1)s Xia/2) (n-1)s X{1-a/2)

$\alpha$ for 95% confidence level = (100-95)/100 =0.05

$\alpha$/2 =0.025

Degrees of freedom = n-1 =12-11

For 11 degrees of freedom

xi-a/2 = xỉ-0,025 = x3 975 = 3.816

AB2 = 13.025 = 21.92

95% Confidence interval for population variance

((n − 1) X..025 (n - 1) x..975 (12-1) 57.723 (12 - 1) x 57.723 21.92 3.816 = (28.9668, 166.3923)

95% Confidence interval for population variance = ( 28.9668 , 166.3923)

c.95% Confidence interval for population standard deviation

(n-1) s2 27.025 )(n-1) s2 V X6.975 (12-1) x 57.723 21.921 /(12 - 1) x 57.723 3.816 = (v28.9668, V166.3923) = (5.3821, 12.8993)

95% Confidence interval for population standard deviation =  (5.3821,12.8993)