Question

One of the most controversial issues in the field of Marketing is the long-run impact that advertising has on brand awareness. That is, while a short-term impact is commonly observed (i.e. increased awareness of the advertised brand within a one-two week time frame of the ad appearing), the longer-term impact is less well understood. To this end, the Marketing Research Firm (MRF) conducted a study to assess the longer-term impact of advertising on his newly launched brand of detergent, clean-so-good (CSG).

The way MRF conducted the study for CSG was as follows:

1. MRF split the population into families without children (FWTC) and families with children (FWC).
2. From each of these two populations MRF selected a random sample of 50 households without replacement from phone company records.
3. Before any advertising occurred, MRM asked each of the households to check off from a provided list of all major brands of detergents (of which CSG is one of them) which brands they were aware of. (The BEFORE data)
4. The results of the BEFORE study indicated that 12 out of 50 FWTC were aware of CSG while 18 out of 50 FWC were aware.
5. For a one-week period CSG ran a set of national advertisements in all major markets.
6. Six months later MRF collected data in an identical fashion to step (iii), but now AFTER the advertising campaign.
7. The results of the AFTER study, obtained from a new random sample (without replacement) from the FWTC and FWC populations indicated that 14 out of 50 FWTC and 28 out of 50 FWC were aware of CSG.

Based on the information, answer the following:

1. State an appropriate null and alternative hypothesis to test that the awareness BEFORE the advertising was the same in the FWTC and FWC groups versus the alternative that it was not the same in the FWTC and FWC group.
2. Test the null versus alternative hypothesis from previous part at the a=0.05 significance level. Be sure to support your conclusion with the appropriate test statistic and p-value.
3. Suppose that you wanted to test the null versus alternative hypothesis that the overall fraction of people before the advertising that were aware of CSG was the same as that after the advertising. What additional information would you need to make this assessment? (Hint: the people before the advertising includes both FWTC and FWC groups. How should we combine the two groups in our test so it would be representative of population?)
4. Test separately at the a=0.05 significance level, for both the FWTC and FWC groups, that the fraction of people who were aware of CSG BEFORE the advertising is the same as the fraction of people AFTER the advertising, versus the alternative that advertising changed the awareness.
5. What can you conclude about the long-run effectiveness of advertising?
6. Do you think that the impact of advertising, as measured here, may be different if an established product (as compared to a new one) was studied? If yes, how? If no, why not?

Answer 1

(a) The hypothesis being tested is:

H₀: µ₁ = µ₂

H₁: µ₁ ≠ µ₂

(b) The output is:

p1	p2	pc
0.24	0.36	0.3	p (as decimal)
12/50	18/50	30/100	p (as fraction)
12.	18.	30.	X
50	50	100	n
	-0.12	difference
	0.	hypothesized difference
	0.0917	std. error
	-1.31	z
	.1904	p-value (two-tailed)

Since the p-value (0.1904) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Therefore, we can support our claim.

(c) We need to get the population proportion for both the groups.

(d) The first hypothesis test for FWTC is:

H₀: µ₁ = µ₂

H₁: µ₁ ≠ µ₂

The output is:

p1	p2	pc
0.24	0.28	0.26	p (as decimal)
12/50	14/50	26/100	p (as fraction)
12.	14.	26.	X
50	50	100	n
	-0.04	difference
	0.	hypothesized difference
	0.0877	std. error
	-0.46	z
	.6484	p-value (two-tailed)

Since the p-value (0.6484) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Therefore, we can support our claim.

The second hypothesis test for FWC is:

H₀: µ₁ = µ₂

H₁: µ₁ ≠ µ₂

The output is:

p1	p2	pc
0.36	0.56	0.46	p (as decimal)
18/50	28/50	46/100	p (as fraction)
18.	28.	46.	X
50	50	100	n
	-0.2	difference
	0.	hypothesized difference
	0.0997	std. error
	-2.01	z
	.0448	p-value (two-tailed)

Since the p-value (0.0448) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we cannot support our claim.

(e) The advertising will change the perspective of people according to the hypothesis testing results.

(f) No, because the results show that both the companies have same impact of advertising.