Question

1. The Westfall Relocation Company, located in Denver, Colorado provides services to laid-off employees of major corporations. It currently operates in three regions: the West, Southwest, and Northwest. Recently, Westfall's general manager questioned whether the company's mean billing amount differed by region. Billing data from random samples of 8 such employees in the West, 12 in the Southwest and 10 in the Northwest are shown on the Answers sheet in cells C7 to E18. Determine whether this is true applying ANOVA to this data at the 5% significance level.
	a. State the null and alternative hypotheses.
	b. Compute the sample sum, mean and standard deviation in each region.
	c Use the formulas shown in class to determine the sums of squares, mean sums of squares and degrees of freedom for your ANOVA table.
	d. Find the critical value of the F-statistic.
	e. Fill in all the cells of the ANOVA table.
	f. State the p-value and whether you should reject the null hypothesis or not.
	g. Construct 95% confidence intervals for average billing levels in each sales region.

West	Southwest	Northwest
$3,700.00	$3,300.00	$2,900.00
$2,900.00	$2,100.00	$4,300.00
$4,100.00	$2,600.00	$5,200.00
$4,900.00	$2,100.00	$3,300.00
$4,900.00	$3,600.00	$3,600.00
$5,300.00	$2,700.00	$3,300.00
$2,200.00	$4,500.00	$3,700.00
$3,700.00	$2,400.00	$2,400.00
$4,800.00		$4,400.00
$3,000.00		$3,300.00
		$4,400.00
		$3,200.00

Answer 1

a)

Null and Alternative Hypothesis:

H₀: µ_West = µ_Southwest = µ_Northwest

H₁: Not all Means are equal

b)

c)

Alpha = 0.05

N = 30

Degress of Freedom:

df_Between = a – 1 = 3-1 =2

df_Within = N-a = 30-3 = 27

df_Total = N-1 = 30-1 = 29

Decision Rule:

If F is greater than 3.21, reject the null hypothesis

Test Statistics:

SS_Betwen = ∑(∑a_i)²/n - T²/N = 5011583.33

SS_Within = ∑(Y)² - ∑(∑a_i)²/n = 21080416.67

SS_Total = SS_{Betwen ­­}+ SS_Within = 26092000

MS = SS/df

               

F = MS_effect / MS_error

Hence,

F = 2505791.67/ 780756.17 = 3.21

d)

Critical Values:

Time (df_Between, df_Within): (2,27) = 3.21

e)

f)

p-value = 0.056

Since the F is less than Critical value, we fail to reject the null hypothesis.

g)

Alpha = 0.05

Z_Critical = 1.96

West

95 % CI = Mean +/- Z_Critical * Std Dev/n^1/2 = 3950 +/- 1.96 * 1030.91/10^1/2 = {3311.03,4588.97}

SouthWest

95 % CI = Mean +/- Z_Critical * Std Dev/n^1/2 = {2334.58,3490.42}

NorthWest

95 % CI = Mean +/- Z_Critical * Std Dev/n^1/2 = {3226.85,4106.48}