In an experiment to assess the effect of the angle of pull on the force required

Question

In an experiment to assess the effect of the angle of pull on the force required to cause separation in electrical connectors, four different angles (factor A) were used, and each of a sample of five connectors (factor B) was pulled once at each angle. The data appears in the accompanying table 45.3 44.4 42.5 43.7 2 42.3 44.3 42.8 45.5 3 39.7 38.4 42.9 47.9 4 36.9 38.2 42.2 37.9 5 45.7 47.4 48.9 56.4 Does the data suggest that true average separation force is affected by the angle of pull? State the appropriate hypotheses 0' "1 "2 "3 "4 0'"1 "2 "3 "4 H,: at least one ,t 0 Ha: at least one ß,-0 Ha: no p, 0 Test the hypotheses at level 0.01 by first constructing an ANOVA table (SST 391.61, SSA 54.91, and SSB 244.26). (Round your answers to two decimal places.) Source df MS 0.01 Error Total State the conclusion in the problem context. Fail to reject Ho. The data suggests that there is an angle of pull effect. O Reject H The data suggests that there is an angle of pull effect Fail to reject Ho The data does not suggest that there is an angle of pull effect O Reject Ho. The data does not suggest that there is an angle of pull efect.

Explanation / Answer

ANOVA METHOD

step 1

null hypothesis ho : µ1 =µ2 =µ3 =µ4

alternative hypothesis : µ1 µ2 µ3 µ4 [ atleast one is not equals ]

step 2

degrees of freedom between = k - 1 = 4 - 1 = 3

degrees of freedom within = n - k = 20 - 4 = 16

degrees of freedom total f( k-1,n - k,) at 0.01 is = f crit = 5.292

step 3

grand mean = g / n = 41.98+42.54+43.86+46.28 / 4 = 43.665

sst = ( xi - grandmean)^2 = (45.3-43.665)^2 + (42.3-43.665)^2 + (39.7-43.665)^2 + ……..& so on = 391.606

ss within = (xi - mean of xi ) ^2 =,(45.3-41.98)^2 + (42.3-41.98)^2 + (39.7-41.98)^2 + ……..& so on = 336.7

ss between = sst - ss within = 391.606 - 336.7 = 54.906

step 4

mean square between = ss between / df between = 54.906/3 = 18.302

mean square within = ss within / df within = 336.7/16 = 21.044

step 5

f cal = ms between / ms within = 18.302/21.044 = 0.87

we got |f cal| = 0.87 & |f crit| =5.292

make decision

hence value of |f cal| < |f crit|and here we failed to reject ho, the data does not suggest that there is an angle of pull effect

One factor ANOVA

Treatments ONE WAY ANOVA Mean = X /n 0 45.3 42.3 39.7 36.9 45.7 41.98 2 44.4 44.3 38.4 38.2 47.4 42.54 4 42.5 42.8 42.9 42.2 48.9 43.86 6 43.7 45.5 47.9 37.9 56.4 46.28

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