Question

1. Lin Xiang, a young banker at Bank of America, where she has recently been promoted and made the operations manager. After a few weeks, she has discovered that there appears to be a difference in customer wait lines in three branches in: Boston, New York and Philadelphia. She wonders if the differences are statically significant. Below are the wait times in seconds for 10 randomly selected customer at each branch. Please conduct a hypothesis test and advise Lin.
   

Boston		New York	Philadelphia
182		52	180
181		192	187
175		181	187
188		180	174
167		172	188
168		71	179
176		168	182
186		176	185
182		168	172
176		186	165

a.         State the null and research hypothesis

b.        Set the level of risk

c.         Select the appropriate test

d.        Compute the obtained value.

e.         Find the critical value

f.         Compare the obtained value and the critical value.

g.        State your decision.

h.        Write the result

Answer 1

Solution

This is a direct application of One-way ANOVA (Analysis of Variance) with equal number of observations per group.

Answers as stipulated in the question are presented below. Back-up Theory and

Detailed Calculations follow at the end.

1. Null and research hypotheses:

Null hypothesis: H₀: Mean customer wait lines are the same in three branches Vs

Alternative: (Research hypothesis) H₁: The three branches differ with respect to mean customer wait lines.

b.   Level of risk (Significance Level) = 5% i.e., 0.05

c.    Appropriate test statistic: F

d.    Obtained value of F = 2.307

e.    Critical value = 3.354

f.    Obtained value < the critical value.

g.   Decision: Accept H₀

h.    The result: There is not enough evidence to support Research hypothesis that the three branches differ with respect to mean customer wait lines. Answer

Back-up Theory

One-WAY CLASSIFICATION EQUAL # OBSNS PER CELL

Suppose we have data of a 1-way classification ANOVA, with r rows, and n observations per cell.

Let xij represent the jth observation in the ith row, j = 1,2,…,n; i = 1,2,……,r

Then the ANOVA model is: xij = µ + αi + εij, where µ = common effect, αi = effect of ith row, and εij is the error component which is assumed to be Normally Distributed with mean 0 and variance σ².

Hypotheses:

Null hypothesis: H₀₁: α₁ = α₂ = ….. = α_r = 0 Vs Alternative: H₁₁: at least one αi is different from other αi’s.

Now, to work out the solution,

Terminology:

Row total = xi..= sum over j of xij

Grand total = G = sum over i of xi.

Correction Factor = C = G²/N, where N = total number of observations = r x n

Total Sum of Squares: SST = (sum over i,j of xij²) – C

Row Sum of Squares: SSR = {(sum over i of xi.²)/n} – C

Error Sum of Squares: SSE = SST – SSR

Mean Sum of Squares = Sum of squares/Degrees of Freedom

Degrees of Freedom:

Total: N (i.e., rn) – 1;

Rows: (r - 1);

Error: Total - Row

F_obs: MSR/MSE;

F_crit: upper α% point of F-Distribution with degrees of freedom n1 and n2, where n1 is the DF for the numerator MS and n2 is the DF for the denominator MS of F_obs

Significance: F_obs is significant if F_obs > F_crit

Detailed Calculations

Row represents the branch – 1 for Boston, 2 for New York and 3 for Philadelphia

	Row		j = 1	2			3			4		5		6			10		xi.		xi.^2		Sumxij^2
	1		182	181			175			188		167		168			176		1781		3171961		317639
	2		52	192			181			180		172		71			186		1546		2390116		261374
	3		180	187			187			174		188		179			165		1799		3236401		324157
																	Total		5126		8798478		903170
r		3				n			10
N		30
Grand Total G									5126
Correction Factor C									875863
SST									27307
SSR									3985.3
SSE									23322
ANOVA TABLE								α			0.05
Source					DF			SS			MS				F			Fcrit		p-value
Row					2			3985.3			1992.633				2.306862			3.354		0.11887
Error					27			23322			863.7852
Total					29			27307			941.6368

DONE