Question

Price	Adver	staff
28.20871609	21.97500534	65.07690664
32.59666738	41.51570177	33.1777398
26.73039949	6.920529801	44.29288614
59.29822687	59.55671865	37.07861568
52.74620808	52.85088656	13.92544328
31.61610767	66.05960265	13.39564196
32.18436232	54.1134373	14.47874386
28.08694723	9.449903867	54.78209784
31.64174322	16.83294168	39.88357189
5.534073916	37.02307199	16.0339671
24.57274087	60.28519547	33.38068789
42.16727195	45.67506943	21.88741722
51.27002777	6.993163854	67.19183935
64.4103824	11.40888699	38.51634266
72.56248665	31.62038026	5.892971587
37.48664815	47.56782739	48.82183294
7.279427473	58.56547746	53.58790857
59.76821192	51.82973724	61.44520402
23.4725486	8.339030122	14.78423414
10.83635975	61.94936979	23.55586413
33.35718863	52.69280068	69.06109806
13.34864345	60.25101474	57.36060671
72.03482162	38.46079897	49.44776757
28.5953856	66.86925871	40.38132878
19.94979705	40.19974364	24.58769494
55.85879086	31.74214911	69.4840846
69.11664174	11.54133732	14.63682974
25.48280282	15.35676138	58.64238411
43.04956206	5.320444349	30.89831233
15.71779534	28.23435163	10.47105319
20.50096133	52.43430891	19.2362743
23.30378124	19.17645802	51.49433882
22.91070284	49.46272164	57.2153386
57.81136509	12.25699637	21.16321299
25.61738945	5.034180731	35
69.71480453	42.56889554	14.64537492

Question No 2: For the given data set, used in question 1; test the hypothesis that the means of variables sales, price and staff are same by using any 0.02 level of significance.    Report the Excel output as well. (Marks 1.5)

Answer 1

(The problem is manually solved and excel output is given thereafter.)

Let and be the mean of treatment 1, treament 2 and treatment 3 i.e mean of variables sales, price and staff

The null hypothesis for an anova always assumes the population means are equal.

Hence, the null hypothesis as:

i.e all treatment means are equal.

and the alternative hypothesis is given by

i.e all treatment means are not equal.

The given observation are

	Price	Adver	staff
	28.209	21.975	65.077
	32.597	41.516	33.178
	26.73	6.9205	44.293
	59.298	59.557	37.079
	52.746	52.851	13.925
	31.616	66.06	13.396
	32.184	54.113	14.479
	28.087	9.4499	54.782
	31.642	16.833	39.884
	5.5341	37.023	16.034
	24.573	60.285	33.381
	42.167	45.675	21.887
	51.27	6.9932	67.192
	64.41	11.409	38.516
	72.562	31.62	5.893
	37.487	47.568	48.822
	7.2794	58.565	53.588
	59.768	51.83	61.445
	23.473	8.339	14.784
	10.836	61.949	23.556
	33.357	52.693	69.061
	13.349	60.251	57.361
	72.035	38.461	49.448
	28.595	66.869	40.381
	19.95	40.2	24.588
	55.859	31.742	69.484
	69.117	11.541	14.637
	25.483	15.357	58.642
	43.05	5.3204	30.898
	15.718	28.234	10.471
	20.501	52.434	19.236
	23.304	19.176	51.494
	22.911	49.463	57.215
	57.811	12.257	21.163
	25.617	5.0342	35
	69.715	42.569	14.645
Total	1318.84	1282.134	1324.916
Mean	36.63444	35.61484	36.80321

Let Xij be the observations.

i = 1,2 ...r. r is total number of rows

j= 1,2,...t t is total number of treatment.

Here

r= 36

t=3

N= Total observation

= r*t

=108

The mean of these treatment can be calculate as

= 36.6344

= 35.61484

= 36.80321

Now obtaining the grand mean ( )

= 36.35083

Now calculating Sum of Squares.

Total Sum of Squares (TSS)

= ( 28.209- 36.35083)2 + ( 32.597- 36.35083)2 + ( 26.73- 36.35083)2 +.... ( 35- 36.35083)2 + ( 14.645- 36.35083)2  

= 41376.37

Treatment Sum of Squares ( SST)

= 36*(36.6344- 36.35083)2 +36 ( 35.61484- 36.35083)2 +36 (36.35083- 36.35083)2  

= 36*0.826771

= 29.76374

Error Sum of Square ( SSE)

SSE = TSS - SST

=  41376.37 - 29.76374

= 41.346.61

Obtaining Degree of Freedom

df( Treatment )= t-1

= 2

df(error)= N-t

= 105

df(Total)=N-1

= 107

Calculating Mean Square of Error

Treatment Mean Square ( MST)

= 29.76374/2

= 14.88187

= 41376.37/107

= 386.6951

= 41346/405

=393.7772

Now obtaining the test statistic( F- statistic)

= 14.88187/393.7772

= 0.037793

The anova table is

Source of Variation	SS	df	MS	F	P-value
Treatment	29.76374	2	14.88187	0.037793	0.962926
Error	41346.61	105	393.7772
Total	41376.37	107

Obtaining the p-value for F statistic =   0.037793

and df =( t-1,N-1)

= (2,105)

Therefore

p-value = 0.962926  

Decision rule:

Reject H0 p-value < level of significance

Here

p-value> 0.02

We failed to reject the null hypothesis.

There is no sufficient evidence to conclude the means of variables sales, price and staff are different 0.02 level of significance.

The excel out put is given as

From the excel output, it is evident that p-value> 0.02 hence we failed to reject the null hypothesis. Hence means are not significantly different.

Excel output is obtained as follows