Exercise 11.7 METHODS AND APPLICATIONS A consumer preference study compares the
ID: 3227641 • Letter: E
Question
Exercise 11.7 METHODS AND APPLICATIONS A consumer preference study compares the effects of three different bottle designs (A, B, and C) on sales of a popular fabric softener. A completely randomized design is employed. Specifically, 15 supermarkets of equal sales potential are selected, and 5 of these supermarkets are randomly assigned to each bottle design. The number of bottles sold in 24 hours at each supermarket is recorded. The data obtained are displayed in the following table. Bottle Design Study Data A B C 13 34 26 17 31 23 13 34 28 14 32 27 16 30 26 The Excel output of a one-way ANOVA of the Bottle Design Study Data is shown below. SUMMARY Groups Count Sum Average Variance Design A 5 73 14.6 3.3 Design B 5 161 32.2 3.2 Design C 5 130 26.0 3.5 ANOVA Source of Variation SS df MS F P-Value F crit Between Groups 796.9333 2 398.4667 119.54 3.23E-06 3.88529 Within Groups 40.0 12.0 3.3333 Total 836.9333 14 (a) Test the null hypothesis that A, B, and C are equal by setting = .05. Based on this test, can we conclude that bottle designs A, B, and C have different effects on mean daily sales? (Round your answers to 2 decimal places. Leave no cells blank - be certain to enter "0" wherever required.) F 119.54 p-value 3.23 H0: bottle design have an impact on sales. (b) Consider the pairwise differences B – A, C – A , and C – B. Find a point estimate of and a Tukey simultaneous 95 percent confidence interval for each pairwise difference. Interpret the results in practical terms. Which bottle design maximizes mean daily sales? (Round your answers to 2 decimal places. Negative amounts should be indicated by a minus sign.) Point estimate Confidence interval B –A: , [ , ] C –A: , [ , ] C –B: , [ , ] Bottle design maximizes sales. (c) Find a 95 percent confidence interval for each of the treatment means A, B, and C. Interpret these intervals. (Round your answers to 2 decimal places. Negative amounts should be indicated by a minus sign.) Confidence interval A: [ , ] B: [ , ] C: [ , ]
Explanation / Answer
(a) The F value (calculated) is 119.54 while the critical value of F distribution at 0.05 level of significance with 2 and 12 df is 3.89. Since the calculated F-value is higher than the critical value, we reject the null hypothesis that the means of te three design are equal.
(b) We carry out the Tukey HSD method in R software:
> tt <- read.csv("clipboard",sep=" ")
> class(tt$Design) <- 'factor'
> sales_aov <- aov(Sales~Design,tt)
> summary(sales_aov)
Df Sum Sq Mean Sq F value Pr(>F)
Design 2 797 398 120 1.2e-08 ***
Residuals 12 40 3
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> TukeyHSD(sales_aov)
Tukey multiple comparisons of means
95% family-wise confidence level
Fit: aov(formula = Sales ~ Design, data = tt)
$Design
diff lwr upr p adj
B-A 17.6 14.52 20.68 0
C-A 11.4 8.32 14.48 0
C-B -6.2 -9.28 -3.12 0
The HSD shows that the differences in mean at each difference level is significant.
(c) The 95% confidence interval is given below:
> confint(sales_aov)
2.5 % 97.5 %
(Intercept) 12.82 16.4
DesignB 15.08 20.1
DesignC 8.88 13.9
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