A marketing organization wishes to study the effects of four sales methods on we

Question

A marketing organization wishes to study the effects of four sales methods on weekly sales of a product. The organization employs a three salesman use each sales method. The results obtained are given in the following table, along with the Excel The organization employs a randomized block design in which output of a randomized block ANOVA of these data 40 40 31 23 33 22 15 35 19 15 26 2 Method 1 Method 2 Method 3 Method4 Salesman A Salesman B Salesman C 94 31.3333 57.3333 3 94 31.3333 72.3333 3 70 23 3333 82 3333 3 6923.0000 1120000 37.00 12.8007 30.0000 19.75 31.5833 4 148 100 79 4 2500 MS 68.7500 F crit 00224757 0000 5.1433 Rows 200.2500 625.5000 22 5000 848.2500 17.80 312.7500 83.40 83.75000 Total (a Test the null hypothesis H that no differences ex st between the effects of the sales methods (treatments) on mean weekly sales Set?,05 Can we conclude that the different sales methods have different effects on mean weekly sales? 17.80·p-value-0022. C e onieco Ho-there is (Clek tett egl in etects ofthe sales methods (treatments) on mean weekly sales. (b) Test the nulhypothesis Ho that no differences exist between th effects of the salesmen (blocks) on mean weekly sales Sea 05. Can we conelude that the different salesmen have different effects on mean weekly saies? F- 83 40. p-value 0000.Click to select Ho salesman (Click to select have an eflecr on sales (c) Use Tukey simultaneous 95 percent confidence intervals to make pairwise comparisons of the sales meshod effects an mean weeky sales Which sales methodie maximize mean weskly sales? (Round your answers to 2 decimal places Negative amounts shouid be ind cated by a minus sign.) Method 1- Method 2 Method 1-Method 3 Method 1-Method 4 Method 2-Method 3 Method 2 -Method 4 Method 3-Method

Explanation / Answer

1

(a) Reject H0 , there is difference between the effects of sales methods on mean weekly sales because p-value is less than 0.05.

(b) Reject H0 , Salesman difference have an effect on sales

(c) We can use R to solve this. The command is

xx=data.frame(x=c(40,28,26,40,31,23,33,22,15,35,19,15),
Salesman=factor(rep(c("A", "B", "C"), 4)),
Sales=factor(rep(1:4, rep(3, 4))))
an=aov(xx$x~xx$Salesman+xx$Sales)
tt=TukeyHSD(x=an, 'xx$Sales', conf.level=0.95)

Tukey multiple comparisons of means
    95% family-wise confidence level

Fit: aov(formula = xx$x ~ xx$Salesman + xx$Sales)

$`xx$Sales`
             diff        lwr       upr     p adj
2-1 -1.065814e-14 -5.473446 5.473446 1.0000000
3-1 -8.000000e+00 -13.473446 -2.526554 0.0091981
4-1 -8.333333e+00 -13.806780 -2.859887 0.0075281
3-2 -8.000000e+00 -13.473446 -2.526554 0.0091981
4-2 -8.333333e+00 -13.806780 -2.859887 0.0075281
4-3 -3.333333e-01 -5.806780 5.140113 0.9963257

2

(a,b,c)

xx=data.frame(x=c(16,33,23,14,30, 21, 1,19,18,6,23,12),
Bottle=factor(rep(c("A", "B", "C"), 4)),
Super=factor(rep(1:4, rep(3, 4))))
an=aov(xx$x~xx$Bottle+xx$Super)
summary(an)

          Df Sum Sq Mean Sq F value   Pr(>F)  
xx$Bottle    2 579.5 289.75   35.84 0.000461 ***
xx$Super     3 290.0   96.67   11.96 0.006078 **
Residuals    6   48.5    8.08          
tt=TukeyHSD(x=an, 'xx$Bottle', conf.level=0.95)
tt

Tukey multiple comparisons of means
    95% family-wise confidence level

Fit: aov(formula = xx$x ~ xx$Bottle + xx$Super)

$`xx$Bottle`
     diff        lwr       upr     p adj
B-A 17.00 10.831573 23.168427 0.0003667
C-A 9.25   3.081573 15.418427 0.0087821
C-B -7.75 -13.918427 -1.581573 0.0196989

Same conclusion are obtained because p-value is less than 0.05.

Bottle design BC is the maximize the sales.

Similarly we can get the result for three table.

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