Recent research studies suggest that having certain aromas o

Recent research studies suggest that having certain aromas or fragrances present in a work environment will enhance the productivity levels of the workers. In one such study, subjects were put in environments with different aromas present and asked to try to solve as many anagrams (word jumbles) as possible in a given amount of time. Suppose that four different aromas were compared in one such study. These aroma treatments were: Lemon fragrance, Floral fragrance, Fried food aroma, and No aroma (the control group). Further suppose that 12 persons of similar intelligence participated in such a study, with three being assigned at random to each of four aroma treatments. The subjects were put in a room with the given aroma for a half hour of anagram solving. The table below shows the number of anagrams each person solved.

Aroma

# Anagrams solved

Lemon

11

10

12

Floral

11

14

11

Fried food

5

5

8

None

8

7

6

a) Give the mean number of anagrams solved for each method, along with the overall mean for all groups combined.

b) Complete the ANOVA table below, showing your calculations.

                        Source             df                    SS                  MS                F

               Between Aromas       

            Within Aroma groups

                        Total

Note: You may use software to compute the standard deviations for each group.

c) Carry out the F-test at = .05, by completing all the hypothesis steps.

d) [Software] Carry out Tukey’s procedure by making simultaneous 95% confidence intervals to compare all pairs of treatments. State a conclusion.

e) Run the analyses in Software of choice to check your answers to Parts (a)-(c). Attach the output.

Solution

a)

Variable    Mean StDev
Lemon     11.000 1.000
Floral       12.00   1.73
Fried food 6.00   1.73
None        7.000 1.000
Grand Mean     9.000

b)

Analysis of Variance

Source DF Seq SS Contribution Adj SS Adj MS F-Value P-Value
Factor   3   78.00        82.98%   78.00 26.000    13.00    0.002
Error    8   16.00        17.02%   16.00   2.000
Total   11   94.00       100.00%

c)

Method

Null hypothesis         All means are equal
Alternative hypothesis At least one mean is different
Significance level      = 0.05

Equal variances were assumed for the analysis.

Factor Information

Factor Levels Values
Factor       4 Lemon, Floral, Fried food, None

p-value is less than 0.05, reject null hypothesis of equal mean. At least one mean is different from others.

d)

Tukey Pairwise Comparisons

Grouping Information Using the Tukey Method and 95% Confidence

Factor      N    Mean Grouping
Floral      3   12.00 A
Lemon       3 11.000 A
None        3   7.000    B
Fried food 3    6.00    B

Means that do not share a letter are significantly different. Group Fried food and None and significantly different from Floral and Lemon.

e) I used Minitab.