Question

The quantity of dissolved oxygen is a measure of water pollution in lakes, rivers, and streams. Water samples were taken at four different locations in a river in an effort to determine if water pollution varied from location to location. Location I was 500 meters above an industrial plant water discharge point and near the shore. Location II was 200 meters above the discharge point and in midstream. Location III was 50 meters downstream from the discharge point and near the shore. Location IV was 200 meters downstream from the discharge point and in midstream. The following table shows the results. Lower dissolved oxygen readings mean more pollution. Because of the difficulty in getting midstream samples, ecology students collecting the data had fewer of these samples. Use a 5% level of significance. Do we reject or not reject the claim that the quantity of dissolved oxygen does not vary from one location to another?

Location I	Location II	Location III	Location IV
7.2	6.7	4.5	4.3
6.1	7.6	5.5	5.9
7.5	7.8	4.1	6.3
6.8	8.1	5.3
6.5		4.9

(a) What is the level of significance?

(b) Find SS_TOT, SS_BET, and SS_W and check that SS_TOT = SS_BET + SS_W. (Use 3 decimal places.)

SSTOT	=
SSBET	=
SSW	=

Find d.f._BET, d.f._W, MS_BET, and MS_W. (Use 3 decimal places for MS_BET, and MS_W.)

dfBET	=
dfW	=
MSBET	=
MSW	=

Find the value of the sample F statistic. (Use 3 decimal places.)

What are the degrees of freedom?
(numerator)
(denominator)

(c) Make a summary table for your ANOVA test.

Source of Variation	Sum of Squares	Degrees of Freedom	MS	F Ratio	P Value	Test Decision
Between groups
Within groups
Total

Answer 1

a)

level of significance =0.05

b)

	location I	location II	location III	location IV
count, ni =	5	4	5	3
mean , x̅ i =	6.820	7.550	4.860	5.500
std. dev., si =	0.554	0.603	0.573	1.058
sample variances, si^2 =	0.307	0.363	0.328	1.120
total sum	34.1	30.2	24.3	16.5	105.1	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =				6.182353
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	0.406593772	1.87045848	1.748617301	0.46560554
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	2.032968858	7.48183391	8.743086505	1.39681661	19.655
SS(within ) = SSW = Σ(n-1)s² =	1.228	1.09	1.312	2.24	5.870

SST=Σ( x - x̅̅ )² = 25.525

SSTOT	=	19.655+5.870=25.525
SSBET	=	19.655
SSW	=	5.870

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no. of treatment , k =4

df between = k-1 = 3

N = Σn =17

df within = N-k =13

mean square between groups , MSB = SSB/k-1 = 6.552

mean square within groups , MSW = SSW/N-k = 0.452

dfBET	=	3
dfW	=	13
MSBET	=	6.552
MSW	=0.452

-----------------------------

F statistic = MSB/MSW = 14.509

------------

What are the degrees of freedom?
(numerator) = 3
(denominator)=13

-----------------------

anova table
	SS	df	MS	F	p-value	F-critical
Between:	19.655	3	6.552	14.509	0.0002	3.4105
Within:	5.870	13	0.452
Total:	25.525	16
					α =	0.05
			conclusion :	p-value<α , reject null hypothesis