Question

A marketing research group shows three different television advertisements for the same product to each of sixty people to see if the ads are equally effective or not. Each participant is asked which of the three advertisements they believe is most effective. The following table summarizes the counts of how many people found each advertisement to be the most effective.

Advertisement 1 2 3 Count 16 30 14

Use 5% level of significance to show and test if there is equal proportions of people believing each advertisement to be most effective.

Answer 1

Answer:-

Given That:-

A marketing research group shows three different television advertisements for the same product to each of sixty people to see if the ads are equally effective or not. Each participant is asked which of the three advertisements they believe is most effective. The following table summarizes the counts of how many people found each advertisement to be the most effective.

Advertisement 1 2 3 Count 16 30 14

Use 5% level of significance to show and test if there is equal proportions of people believing each advertisement to be most effective.

Given,

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: The proportion of different adevertisement is equal or p1 = p2 = p3

Alternative hypothesis: At least one of the proportions in the null hypothesis is false.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.

Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = k - 1 = 3 - 1
D.F = 2
(Ei) = n * pi

	Observed	Expected
Advertisement 1	16	20.00	0.8
Advertisement 2	30	20.00	5.00
Advertisement 3	14	20.00	1.80
Total	60	60.00	7.60

X2 = 7.60

where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, Ei is the expected frequency count for level i, Oi is the observed frequency count for level i, and X2 is the chi-square test statistic.

The P-value is the probability that a chi-square statistic having 2 degrees of freedom is more extreme than 7.60.

We use the Chi-Square Distribution Calculator to find P(X2 > 7.60) = 0.022

Interpret results. Since the P-value (0.022) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that proportions of people believing each advertisement to be most effective is equal.

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