Question

An article in Environment International (Vol. 18, No. 4, 1992) describes an experiment in which the amount of radon released in showers was investigated. Radon-enriched water was used in the experiment, and six different orifice diameters were tested in shower heads. The data from the experiment are shown in the following table. Orifice Diameter Radon Released (%) 0.37 80 83 83 85 dy. 0.51 75 75 79 79 0.71 74 73 76 77 1.02 67 72 74 74 1.40 62 62 66 69 1.99 60 61 64 66 (a) Does the size of the orifice affect the mean percentage of radon released? Use a = 0.05. Round all intermediate calculations to four decimal places (e.g. 12.3456). 32.92 Since Fo in the rejection region, there sufficient evidence to conclude that the mean percentage varies according to orifice diameter. (b) Find bounds on the P-value for the F-statistic in part (a). Use only Table VI of Appendix A. P-values

Answer 1

Solve using One-way ANOVA method Observation A B C D 80 84 83 85 1 2 75 75 79 79 3 4 74 73 76 77 67 72 74 74 62 62 65 69 60 61 64 66 ΟΙ 6 Solution: A B C D E F 80 75 74 67 62 60 84 75 73 72 62 61 83 79 76 74 65 64 85 79 77 74 69 66 ΣΑ= 332 ΣΒ = 308 Σc= 300 ΣD = 287 = 287 ΣΕ = 258 ΣF = 251 B2 D? Ε? F2 6400 5625 5476 4489 3844 3600 7056 5625 5329 5184 3844 3721 6889 6241 5776 5476 4225 4096 7225 6241 5929 5476 4761 4356 ΣΑ? = = 27570 ΣΒ' = 23732 Σc = 22510 ΣD? = 20625 ΣΕ’ = 16674 ΣF = 15773


Data table Group N A B C D E F Total 11 = 4 n, = 4 n, = 4 14 = 4 25 4 ης = 4 η = 24 Σ», Τ, = Σχ, = 332 Τ, = Σχ = 308 Τη = Σχ, = = 300 To = Σχ. = X4 = 287 T, = Σας = 258 To = Σχε = 251 Σ Σκη = 27570 Σ3 = 23732 Σ = 22510 Σ = 20625 Σ = 16674 Σε = 15773 Mean x X, = 83 X3 = 75 X4 = 71.75 Xς = 64.5 Xς = 62.75 Std Dev S, S, = 2.1602 S, = 2.3094 S, = 1.8257 S. = 3.304 S, = 3.3166 S= 2.7538 Σχ= 1 = 1736 Σ? = 126884 Overall x = 72.3333 X) = 77 Let k = the number of different samples = 6 η = 11 +12 +15 + n4+ ης + n = 4 + 4 + 4 + 4 + 4 + 4 = 24 1736 Overall x = = 72.3333 24 Σx = T, +T,+T,+T,+ T + To = 332 + 308 + 300 + 287 + 258 + 251 = 1736 – (1) (Σx)17362 24 = 125570.6667 – (2) (3322 3082 3002 2s72 2582 2512 Ε τ = 126755.5 – (3) n: 4 4 4 4 4 4 Σ? Σ - Σ+ Σ+Σ+Σ+ Σ = 27570 + 23732 + 22510 + 20625 + 16674 + 15773 = 126984 – (4)


ANOVA: Step-1 : sum of squares between samples SSB = = (3)-(2) n n = 126755.5 - 125570.6667 = 1184.8333 Or SSB = En; · (; -7) = 4 x (83 - 72.3333)2 + 4x (77 - 72.3333)2 + 4x (75 - 72.3333)2 + 4x(71.75 - 72.3333)2 + 4x(64.5 - 72.3333)2 + 4 x (62.75 - 72.3333)2 = 1184.8333 Step-2: sum of squares within samples ssw=E==(23) = (4)-(3) = 126884 - 126755.5 = 128.5 Step-3: Total sum of squares SST = SSB + SSW = 1184.8333 + 128.5 = 1313.3333 Step-4 : variance between samples SSB MSB = k-1 1184.8333 5


= 236.9667 Step-5: variance within samples SSW MSW = n-k 128.5 24 - 6 128.5 18 = 7.1389 Step-6 : test statistic F for one way ANOVA test MSB F= MSW 236.9667 7.1389 = 33.1938 the degree of freedom between samples k- 1 = 5 Now, degree of freedom within samples n-k= 24 - 6 = 18 p-value ANOVA table Source of Variation Sums of Squares Degrees of freedom Mean Squares F SS DF MS Between samples SSB = 1184.8333 k- 1 = 5 MSB = 236.9667 33.1938 Within samples SSW = 128.5 n-k = 18 MSW = 7.1389 Total SST = 1313.3333 n-1 = 23 0


Ho : There is no significant differentiating between samples H1 : There is significant differentiating between samples F(5, 18) at 0.05 level of significance = 2.7729 As calculated F = 33.1938 > 2.7729 So, H, is rejected, Hence there is significant differentiating between samples

therefore F=33.1938 and it is in the rejection region, there is enough evidence to conclude that the mean percentage varies according to orifice diameter

P-value ~ 0 as F is so high.