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	6.2	4.3	4.1
6.6	7.4	5.3	5.4
7.6	7.5	4.4	6.5
6.9	7.8	5.8
6.9		4.8

(a) What is the level of significance?


State the null and alternate hypotheses.

H_o: μ₁ = μ₂ = μ₃ = μ₄; H₁: All four means are different.H_o: μ₁ = μ₂ = μ₃ = μ₄; H₁: Exactly two means are equal.    H_o: μ₁ = μ₂ = μ₃ = μ₄; H₁: Not all the means are equal.H_o: μ₁ = μ₂ = μ₃ = μ₄; H₁: Exactly three means are equal.

(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) Find the P-value of the sample test statistic.

P-value > 0.1000.050 < P-value < 0.100    0.025 < P-value < 0.0500.010 < P-value < 0.0250.001 < P-value < 0.010P-value < 0.001

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

Since the P-value is greater than the level of significance at α = 0.05, we do not reject H₀.Since the P-value is less than or equal to the level of significance at α = 0.05, we reject H₀.    Since the P-value is greater than the level of significance at α = 0.05, we reject H₀.Since the P-value is less than or equal to the level of significance at α = 0.05, we do not reject H₀.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance there is insufficient evidence to conclude that the means are not all equal.At the 5% level of significance there is sufficient evidence to conclude that the means are not all equal.    At the 5% level of significance there is insufficient evidence to conclude that the means are all equal.At the 5% level of significance there is sufficient evidence to conclude that the means are all equal.

(f) 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					---Select--- p-value > 0.100 0.050 < p-value < 0.100 0.025 < p-value < 0.050 0.010 < p-value < 0.025 0.001 < p-value < 0.010 p-value < 0.001	---Select--- Do not reject H0. Reject H0.
Within groups
Total

Answer 1

Applying 1 way ANOVA( from excel:data-data analysis: 1 way ANOVA)

a)

level of significance =0.05

H_o: μ₁ = μ₂ = μ₃ = μ₄; H₁: Not all the means are equal.

b)

SSTOT	=	24.681
SSBET	=	18.147
SSW	=	6.534

dfBET	=	3
dfW	=	13
MSBET	=	6.049
MSW	=	0.503

value of the sample F statistic =12.035

df(numerator) =		3
df(denominator)=		13

P-value < 0.001

d)

Since the P-value is less than or equal to the level of significance at α = 0.05, we reject H₀.

e) At the 5% level of significance there is sufficient evidence to conclude that the means are not all equal.

f)

Source of Variation	SS	df	MS	F	P-value
Between Groups	18.147	3	6.049	12.035	0.0005
Within Groups	6.534	13	0.503
Total	24.681	16