Question

A marketing specialist wanted to determine the effectiveness of various advertising media in selling a new cleaning product. He requested 160 prospective buyers to randomly listen to , watch or read 4 types of advertisements, then asked them to choose which one convinced them to buy the product. based on their choices in the table below, conduct a goodness-of fit chi square test at 0.05 level of significance using the 5 steps hypothesis tesing.

is there a significant difference in their preference of advertising media?

Radio=30

TV=55

Newspaper=27

Word of mouth=48

Answer 1

Step 1: Making assumptions:

(i) The sampling method is Simple Random Sampling (SRS).

(ii) The variable under study is categorical

(iii) The expected value of the number of sample observations in each level of the variable is at least 5.

Step 2: Stating the research and null hypotheses and selecting alpha

H0: Null Hypothesis: There is no significant difference in various advertising media in selling a new cleaning product.

HA: Alternative Hypothesis: There is significant difference in various advertising media in selling a new cleaning product (Claim)

The level of significance = =0.05

Step 3: Selecting the Sampling Distribution and specifying the Test Statistic.

the Sampling Distribution is Distribution.

The Test Statistic is given by:

Step 4: Computing the Test Statistic and corresponding P - Value:

Advertising Media	Observed (O)	Expected (E)	(O - E)2/E
Radio	30	40	2.5
TV	55	40	3.125
News paper	27	40	4.225
Word of mouth	48	40	3.6
Total = =			13.95

ndf = 4 - 1 =3

By Technology, P - Value = 0.0030

Step 5: Drawing a conclusion:
Since P - Value = 0.0030 is less than =0.05, the difference is significant. Reject null hypothesis.

Conclusion:

The data support the claim that there is significant difference in various advertising media in selling a new cleaning product.