A study was conducted to determine the effects of two factors on the latherabili
ID: 3395155 • Letter: A
Question
A study was conducted to determine the effects of two factors on the latherability of soap. The two factors were type of water (tap or distilled) and glycerol (present or absent). The outcome measured was the amount of foam produced (in milliliters). The experiment was repeated three times for each combination of the factors. The results are as follows.
No Glycerol
Glycerol
Distilled Water
165 181 168
170 197 190
Tap Water
155 142 139
139 160 160
Analyze the data using a=.05.
Mintab may be used, include output
No Glycerol
Glycerol
Distilled Water
165 181 168
170 197 190
Tap Water
155 142 139
139 160 160
Explanation / Answer
Ho:there is no effect of two factors i.e. water (tap or distilled) and glycerol (present or absent) on the latherability of soap. H1:there is a effect of two factors i.e. water (tap or distilled) and glycerol (present or absent) on the latherability of soap. SST: total sum of squares, SSA:sum of squares of factor water (tap or distilled) , SSB:sum of squares of factor glycerol (present or absent), SSAB:sum of squares of two factors water (tap or distilled) and glycerol (present or absent) i.e., interaction effect & SSE:sum of squares of error, then SST=326050-[(1966)2/(2*2*12)]=326050-80524.0833=245525.9167, SSA=[((1071)2+(895)2)/24]-80524.0833=81169.3334-80524.0833=645.3334. SSB=[((950)2+(1016)2)/24]-80524.0833=80614.8333-80524.0833=90.75, SSAB=[((514)2+(436)2+(557)2+(459)2)/12]-80524.0833-645.3334-90.75=8.3333, & SSE=SST-SSAB-SSA-SSB=245525.9167-8.3333-645.3334-90.75=244781.5003. The table value of F(1,40) is 4.08 & F(1,60) is 4.
The ANOVA table of two factors i.e. water (tap or distilled) and glycerol (present or absent) on the latherability of soap.
Here Fcal<Ftab at 0.05, so we accept null hypothesis.
Therefore, there is no effect of two factors i.e. water (tap or distilled) and glycerol (present or absent) on the latherability of soap.
Source of variation sum of squares degrees of freedom mean sum of squares F test Water(A) 645.3334 2-1=1 645.3334/1=645.3334 645.3334/5563.2159=0.1160 Glycerol(B) 90.75 2-1=1 90.75/1=90.75 90.75/5563.2159=0.0163 AB 8.3333 (2-1)(2-1)=1 8.3333/1=8.3333 8.3333/5563.2159=0.0015 Error 244781.5003 2*2*(12-1)=44 244781.5003/44=5563.2159 Total 245525.9167 (2*2*12)-1=47Related Questions
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