Question

Use Minitab to answer the questions. Make sure to copy all output from the Minitab:

1.   Followings Tables shows previous 11 months stock market returns.

Date	Monthly SP500 Return	Monthly DJIA Return
12/7/2007	-0.8628	-0.7994
1/8/2008	-6.1163	-4.6323
2/8/2008	-3.4761	-3.0352
3/8/2008	-0.5960	-0.0285
4/8/2008	4.7547	4.5441
5/8/2008	1.0674	-1.4182
6/8/2008	-8.5962	-10.1937
7/8/2008	-0.9859	0.2468
8/8/2008	1.2191	1.4548
9/8/2008	-9.2054	-6.0024
10/8/2008	-16.8269	-4.8410

Let consider we know the variance of monthly return for all stock were 25 percent in 2008, perform the following hypothesis for each index:

Ho :  µ = -5

Ha :  µ ≠ -5

Considering the population variance is unknown, perform a hypothesis test if the average stock return is 0 or not for each index:

Ho :  µ = 0

Ha :   µ ≠  0

Perform following hypothesis based on all assumption we had in 2)

Ho :  µ ≥ -3

Ha :   µ <  -3

Let’s consider the population mean of SP500 as µ1 and that of DJIA as µ2 while none of population variance is known. Test following hypothesis:

  Ho : µ1  =  µ2

Ha :   µ1 ≠  µ2

Let’s consider the population variance of SP500 as σ²₁and that of DJIA as σ²₂, and none of them are known. Test following hypothesis:

  Ho :  σ²₁  =  σ²₂

  Ha :  σ²₁≠  σ²₂

6) Perform the following hypothesis test

  Ho :  σ²₁  ≤  σ²₂

  Ha :  σ²₁>  σ²₂

2.  The U.S. Bureau of Labor Statistics publishes a variety of unemployment statistics, including the number of individuals who are unemployed and the mean length of time the individuals have been unemployed. For November 1998, the Bureau of Labor Statistics reported that the national mean length of time of unemployment was 14.5 weeks.

The mayor of Chicago has requested the study on the status of unemployment in City of Chicago. A sample of 60 unemployed residents shows the sample mean is 15.7 and the sample standard deviation is 9.0. Test whether the length of time in Chicago is long than national average.

Let's think about the house price. According to the Case-Shiller Home Price Indices in August 2009, Chicago and San Francisco have following sample mean and population standard deviations (the sample mean was calculated by daily base, so the sample size was 30):

	CHICAGO	San Francisco
Sample Mean	130.55	132.47
Population Standard Deviation	9	12

                  

Using hypothesis test, prove if these house price indices are same.  (Setup a hypothesis, show your works to perform the test, and state your verdict)

2)             Some people argue that San Francisco has higher house price than that of Chicago. Prove/disprove the argument using a hypothesis test.

3)             Let’s assume the population standard deviations are unknown, and the sample standard deviation of for Chicago is 9.2 and that of San Francisco is 11.5. Some people argue that San Francisco has higher variability (higher variance) in house prices than that of Chicago. Setup a hypothesis, perform the test and prove/disprove the argument.

4.     Let’s consider a company’s growth rate of sales.

Year	Annual Growth Rate (%)
1993	6.80
1994	6.10
1995	5.60
1996	5.40
1997	4.90
1998	4.50
1999	4.20
2000	4.00
2001	4.80
2002	5.80
2003	6.20
2004	5.50
2005	5.00
2006	6.10

Find the sample mean and sample standard deviation using Minitab using descriptive statistics.

The store manager found that the average growth rate in 80s was 6.00%. Using the data, prove if the average growth rate for the sample period was same as that of 80s.

Prove if the average growth rate was less than 6%.

Answer 1

ANSWER:

The U.S. Bureau of Labor Statistics publishes a variety of unemployment statistics, including the number of individuals who are unemployed and the mean length of time the individuals have been unemployed. For November 1998, the Bureau of Labor Statistics reported that the national mean length of time of unemployment was 14.5 weeks.

1) a)

Two-sample T for Monthly vs Monthly DJIA Return

N Mean StDev SE Mean

Monthly 11 -3.60 6.13 1.8

Monthly DJIA Return 11 -2.25 4.05 1.2

Difference = μ (Monthly) - μ (Monthly DJIA Return)

Estimate for difference: -1.36

95% CI for difference: (-5.98, 3.26)

T-Test of difference = -5 (vs ≠): T-Value = 1.64 P-Value = 0.116 DF = 20

Both use Pooled StDev = 5.1956

b)

Two-sample T for Monthly vs Monthly DJIA Return

N Mean StDev SE Mean

Monthly 11 -3.60 6.13 1.8

Monthly DJIA Return 11 -2.25 4.05 1.2

Difference = μ (Monthly) - μ (Monthly DJIA Return)

Estimate for difference: -1.36

95% CI for difference: (-5.98, 3.26)

T-Test of difference = 0 (vs ≠): T-Value = -0.61 P-Value = 0.547 DF = 20

Both use Pooled StDev = 5.1956

c)

Two-sample T for Monthly vs Monthly DJIA Return

N Mean StDev SE Mean

Monthly 11 -3.60 6.13 1.8

Monthly DJIA Return 11 -2.25 4.05 1.2

Difference = μ (Monthly) - μ (Monthly DJIA Return)

Estimate for difference: -1.36

95% lower bound for difference: -5.18

T-Test of difference = -3 (vs >): T-Value = 0.74 P-Value = 0.233 DF = 20

Both use Pooled StDev = 5.1956

d)

Two-Sample T-Test and CI: Monthly, Monthly DJIA Return

Two-sample T for Monthly vs Monthly DJIA Return

N Mean StDev SE Mean

Monthly 11 -3.60 6.13 1.8

Monthly DJIA Return 11 -2.25 4.05 1.2

Difference = μ (Monthly) - μ (Monthly DJIA Return)

Estimate for difference: -1.36

95% lower bound for difference: -5.21

T-Test of difference = -3 (vs >): T-Value = 0.74 P-Value = 0.234 DF = 17

e)

Test and CI for Two Variances: Monthly, Monthly DJIA Return

Method

Null hypothesis σ(Monthly) / σ(Monthly DJIA Return) = 1

Alternative hypothesis σ(Monthly) / σ(Monthly DJIA Return) ≠ 1

Significance level α = 0.05

F method was used. This method is accurate for normal data only.

Statistics

95% CI for

Variable N StDev Variance StDevs

Monthly 11 6.131 37.584 (4.284, 10.759)

Monthly DJIA Return 11 4.050 16.404 (2.830, 7.108)

Ratio of standard deviations = 1.514

Ratio of variances = 2.291

95% Confidence Intervals

CI for

CI for StDev Variance

Method Ratio Ratio

F (0.785, 2.918) (0.616, 8.516)

Tests

Test

Method DF1 DF2 Statistic P-Value

F 10 10 2.29 0.207

f)

Test and CI for Two Variances: Monthly, Monthly DJIA Return

Method

Null hypothesis σ(Monthly) / σ(Monthly DJIA Return) = 1

Alternative hypothesis σ(Monthly) / σ(Monthly DJIA Return) > 1

Significance level α = 0.05

Statistics

95% Lower

Bound for

Variable N StDev Variance StDevs

Monthly 11 6.131 37.584 3.973

Monthly DJIA Return 11 4.050 16.404 2.832

Ratio of standard deviations = 1.514

Ratio of variances = 2.291

95% One-Sided Confidence Intervals

Lower Bound Lower Bound

for StDev for Variance

Method Ratio Ratio

Bonett 0.710 0.504

Levene 0.620 0.384

Tests

Test

Method DF1 DF2 Statistic P-Value

Bonett 1 — 1.06 0.151

Levene 1 20 0.77 0.196