Question

QUESTION 3 A marketing executive is creating a commercial for a new drug that treats insomnia. Using the same voiceover script, he wants to know which one of three images people prefer to see: 1) a person who cannot sleep at night, 2) a person who is sleeping soundly at night, or 3) a person who is very sleepy during the day. He made four focus groups (comprised of individuals with difficulty sleeping) and showed them all three of the commercials. Each group rated the commercials on a scale of 1-10 (with being least likely to use the new drug and 10 being most likely to use the new drug). The results are presented in the table below Commercial Focus Group 1 Focus Group 2 Focus Group 3 Focus Group 4 Focus Group 5 5.2 6.0 4.6 6.0 1.14 1.22 x2 - 519, G-85 17.2 6.4 207

Answer 1

(a) One-way blocked analysis of variance (ANOVA) is used to test the null hypothesis that multiple population means are all equal, allowing for block effects.

(b) Assumptions:

• The measurement errors are independent, and identically normally distributed with mean 0 and the same variance.
• The population (treatment) effect does not interact with the block effect.
• The blocks may be considered either fixed or random, although they are usually considered random.
• If the blocks are fixed, then all the measurement values are independent, and normally distributed with the same variance.
• If the blocks are random, then all the measurement values are normally distributed with the same variance.
• Measurements from the same random block will be positively correlated.
• Once a particular block has been selected (i.e., the block effect has been accounted for), then observations in that block are independent.
• For a multiple comparisons test of the sample means to be meaningful, the treatment effect is viewed as fixed, so that the populations (treatment groups) in the experiment include all those of interest.

(c) The hypothesis being tested is:

H0: µ1 = µ2 = µ3

Ha: At least one means is not equal

(d)

	Mean	n	Std. Dev
	4.600	5	1.140	Treatment 1
	6.000	5	1.225	Treatment 2
	6.400	5	2.074	Treatment 3
	6.667	3	1.528	Block 1
	6.667	3	1.155	Block 2
	3.333	3	0.577	Block 3
	5.667	3	0.577	Block 4
	6.000	3	1.732	Block 5
	5.667	15	1.633	Total
ANOVA table
Source	SS	df	MS	F	p-value
Treatments	8.93	2	4.467	6.23	.0234
Blocks	22.67	4	5.667	7.91	.0070
Error	5.73	8	0.717
Total	37.33	14

The p-value is 0.0234.

Since the p-value (0.0234) is greater than the significance level (0.01), we fail to reject the null hypothesis.

Therefore, we cannot conclude that the results are significant.

(e) Tukey's JSD would not be needed to be calculated because the results are not significant.

(f) n2 = 8.93/37.33 = 0.24

This is a small effect size.

(g) We cannot conclude that there is a main effect of treatments.