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 (%) 80 83 83 85 0.51 75 75 0.37 79 79 0.71 74 73 76 77 1.02 67 72 74 74 1.40 62 62 68 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). X 30.85 Since Fo is in the rejection region, there is 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-values0

Answer 1

Notations :
A = Orifice diameter 0.37
B = Orifice diameter 0.51
C = Orifice diameter 0.71
D = Orifice diameter 1.02
E = Orifice diameter 1.40
F = Orifice diameter 1.99


1 Solve using One-way ANOVA method Observation A B C D E F 80 75 74 67 62 60 2 83 75 73 72 62 61 3 83 79 76 74 68 64 4 85 79 77 74 69 66 Solution: B C D 80 83 83 85 75 75 79 79 74 73 76 77 67 72 74 74 62 62 68 69 60 61 64 66 LA = 418 B = 426 C = 444 D = 450 B2 D2 6400 6889 6889 7225 5625 5625 6241 6241 5476 5329 5776 5929 4489 5184 5476 5476 3844 3844 4624 4761 3600 3721 4096 4356 42 = 29434 B2 = 30592 c? = 33102 D2 = 33988


Data table Group B C D Total N ή, = 6 15 = 6 13 - 6 14 = 6 η = 24 χ = 1738 Σ Σχ? = 127116 Σ Mean x; x, = 418 T, = Σχ, = 426 Τη = Σx, = 444 Τ. = Σκη = 1 = 450 = 29434 Σ = 30592 Σ = 3 = 33102 Σ= 33988 Χι = 69.6667 X) = 71 X3 = 74 X4 = 75 S, = 7.9162 = 8.3187 S, = 7.0143 S4 = 6.8993 Overall x = 72.4167 Std Dev Si S)= Let k = the number of different samples = 4 η = η, τη, +ng + na = 6 + 6 + 6 + 6 = 24 Overall x Ξ 1738 = 72.4167 24 Σx = T + T, +T,+T, = 418 + 426 + 444 + 450 = 1738 – (1) (Σx)2 17382 24 Ξ Ξ 125860.1667 – (2) n (4182 4262 4442 4502 6 6 6 6 Ε = 125972.6667 – (3) n: Σ' = Σ - Σ3 + Σ - Σ = 29434 + 30592 + 33102 + 33988 = 127116 – (4)


ANOVA: Step-1 : sum of squares between samples T SSB = (Σ = (3)-(2) ni n = 125972.6667 - 125860.1667 112.5 Or SSB = En; · (ā;-7) = 6(69.6667 - 72.4167)2 +6(71 - 72.4167)2 + 6x (74 - 72.4167)2 + 6 x (75 - 72.4167)2 112.5 Step-2: sum of squares within samples SSW = Σκ? - (Σ = (4)-(3) ni 127116 - 125972.6667 = 1143.3333 Step-3 : Total sum of squares SST = SSB + SSW = 112.5 + 1143.3333 = 1255.8333 Step-4 : variance between samples SSB MSB k-1 = 112.5 3


= 37.5 Step-5: variance within samples SSW MSW n-k 1143.3333 24 - 4 1143.3333 20 = 57.1667 Step-6: test statistic F for one way ANOVA test MSB F= MSW 37.5 = 57.1667 = 0.656 the degree of freedom between samples k - 1 = 3 Now, degree of freedom within samples n-k= 24 - 4 = 20 ANOVA table Source of Variation F p-value Between samples Within samples Total Sums of Squares Degrees of freedom Mean Squares SS DF MS SSB = 112.5 k-1 = 3 MSB = 37.5 SSW = 1143.3333 n-k = 20 MSW = 57.1667 0.656 0.5886 SST = 1255.8333 n - 1 = 23


Ho : There is no significant differentiating between samples .. H7 : There is significant differentiating between samples F(3, 20) at 0.05 level of significance = 3.0984 As calculated F = 0.656 < 3.0984 So, Ho is accepted, Hence there is no significant differentiating between samples