BUS 308 Week 2 Problem Set. Get an A++.

T-Testing and Anova

BUS 308: Statistics for Managers

Suppose that an automotive parts and accessories chain is experimenting with a new sales promotion. Two similar stores are selected for the experiment. For Store 1, nothing changes. This store constitutes the control group. For Store 2, the treatment group, the promotion is implemented. Sales in hundreds of dollars over a five-day period are as follows:

a. Control: 6, 6, 7, 10, 12, 9, 6, 5, 5, 7

b. Treatment: 2, 5, 2, 4, 7, 1, 2, 3, 4, 5

 

Hypothesis test:

Null Hypothesis :(1((2 against H1 : (1<(2 (One tailed)

t-Test: Two-Sample Assuming Equal Variances
	control	treatment
Mean	7.3	3.5
Variance	5.3444	3.3889
Observations	10	10
Pooled Variance	4.3667
Hypothesized Mean Difference	0
Df	18
t Stat	4.06625
P(T<=t) one-tail	0.00036
t Critical one-tail	1.73406
P(T<=t) two-tail	0.00072
t Critical two-tail	2.10092

The null hypothesis is declined based on :the P(T<=t) one tail = p-value = 0.00036 which is less than 0.05 level of significance.

Suppose that a home builder is approached by a customer who wants to move in as soon as possible. The customer chooses three home designs that she likes and asks the home builder which one could be completed the fastest.  To compare the three designs on speed of completion, the builder randomly selects 10 homes that he built in the past based on each of the three designs. The data for the number of days to build each home are as follows:

c. Design A: 15, 17, 19, 21, 23, 25, 27, 29, 31, 33

d. Design B: 29, 34, 39, 44, 49, 54, 59, 64, 69, 74

e. Design C: 22, 24, 25, 27, 28, 28, 29, 31, 33, 34

· (1 :Symbolizes the mean number of days to build home with usage of design A.

· (2 :Symbolizes e the mean number of days to build home with usage design B.

· (3 :Symbolizes the mean number of days to build home with usage design C.

Hypothesis test:

Null Hypothesis :(1 = (2 = (3 against H1 : Two will be different at the least.

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Design A	10	240	24	36.66667
Design B	10	515	51.5	229.1667
Design C	10	281	28.1	14.76667

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	4402.067	2	2201.033	23.53207	1.21E-06	3.354131
Within Groups	2525.4	27	93.53333
Total	6927.467	29

ANOVA table the p-value = 1.21E-06 (also equals 0.00000121) this would be less than 0.05.

This is not a null hypothesis and at least two means are different.

An insurance company is reviewing its current policy rates. When originally setting the rates they believed that the average claim amount was $1,800. Now there are concerns that if the true mean is actually higher than this they could potentially lose a lot of money. They randomly select 40 claims, and calculate a sample mean of $1,950. Assuming that the standard deviation of claims is $500, and set significance level = :05, test to see if the insurance company should be concerned.

Hypothesis test:

Null Hypothesis :

( = $1,800 against H1: (> $1,800.

n = 40, ( = $ 500. The level of significance ( = 0.05.

If we test hypothesis above↑

One tailed critical value of Z at 0.05 level of significance is 1.645.

Z –test statistics falls in the critical region, the null hypothesis is declined.

True mean higher than $1,800.