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Chapter 15: Problem 10

Advertising Campaigns The results of an investigation of product recognition following three advertising campaigns were reported in Example \(11.15 .\) The responses were the percentage adults in 15 different groups who were familiar with the newly advertised product. The normal probability plot indicated that the data were not approximately normal and another method of analysis should be used. Is there a significant difference among the three population distributions from which these samples came? Use an appropriate nonparametric method to answer this question. $$ \begin{array}{lcc} \hline & \text { Campaign } & \\ \hline 1 & 2 & 3 \\ \hline 33 & .28 & 21 \\ .29 & .41 & 30 \\ 21 & .34 & 26 \\ 32 & .39 & .33 \\ .25 & .27 & .31 \\ \hline \end{array} $$

Short Answer

Expert verified
Answer: Yes, there is a significant difference among the three advertising campaigns in terms of product recognition.

Step by step solution

01

Rank the data from all groups combined

Combine all the data from the three campaigns and assign a rank to each value, such that the smallest value gets a rank of 1 and the largest value gets the highest rank. In case of ties, assign the average rank. Data:\(/\) Campaign 1: [33, 29, 21, 32, 25]\(/\) Campaign 2: [28, 41, 34, 39, 27]\(/\) Campaign 3: [21, 30, 26, 33, 31] Ranking:\(/\) Campaign 1: [11.5, 7, 1.5, 10, 4]\(/\) Campaign 2: [6, 15, 13, 14, 5]\(/\) Campaign 3: [1.5, 8, 3, 11.5, 9]
02

Sum the ranks for each group

Next, sum the ranks for each group and calculate the average rank. Sum of ranks:\(/\) Campaign 1: 11.5 + 7 + 1.5 + 10 + 4 = 34\(/\) Campaign 2: 6 + 15 + 13 + 14 + 5 = 53\(/\) Campaign 3: 1.5 + 8 + 3 + 11.5 + 9 = 33 Average rank:\(/\) Campaign 1: 34/5 = 6.8\(/\) Campaign 2: 53/5 = 10.6\(/\) Campaign 3: 33/5 = 6.6
03

Calculate the test statistic

Now, calculate the Kruskal-Wallis test statistic (H) using the formula: $$H = \frac{12}{N(N+1)} \sum_{i=1}^k \frac{R_i^2}{n_i} -3(N+1)$$ Where: - N is the total number of observations, - k is the number of groups, - \(R_i\) is the sum of ranks for group i, - \(n_i\) is the number of observations in group i. In this case, we have: - N = 15 (5 observations per group), - k = 3 (3 campaigns), - \(R_1\) = 34, - \(R_2\) = 53, - \(R_3\) = 33, - \(n_1\) = \(n_2\) = \(n_3\) = 5. Plugging the values into the formula, we get: $$H = \frac{12}{15(16)} \left(\frac{34^2}{5} + \frac{53^2}{5} + \frac{33^2}{5}\right) - 3(16) = 7.837$$
04

Find the critical value and compare it with the test statistic

To determine if there is a significant difference among the groups, we need to compare the calculated test statistic (H) with the critical value from the chi-square distribution table. As we don't have a predefined significance level (alpha), let's use the standard of 0.05. Our degrees of freedom for the chi-square distribution are equal to the number of groups minus 1 (k - 1):$$df = 3 - 1 = 2$$ Looking up the chi-square table for df = 2 and a significance level of 0.05, we find the critical value to be approximately 5.99. Since our calculated test statistic, H = 7.837, is greater than the critical value from the table (5.99), we reject the null hypothesis and conclude that there is a significant difference among the three population distributions from which these samples came.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kruskal-Wallis Test
The Kruskal-Wallis test is a nonparametric method used in statistics to determine if there are statistically significant differences between two or more groups of an independent variable on a continuous or ordinal dependent variable. It's particularly useful when the assumption of normality in the data is violated, as with the given exercise involving three advertising campaigns.

When applying the Kruskal-Wallis test, one ranks all data across all groups, calculates the sum of ranks for each group, and then uses these sums to compute the test statistic. If the calculated statistic exceeds the critical value from the chi-square distribution table at a chosen significance level, typically 0.05, the null hypothesis of equal population distributions is rejected, indicating a significant difference among the groups.
Chi-Square Distribution
The chi-square distribution is a fundamental probability distribution in statistics used in various tests, including the Kruskal-Wallis test. Unlike the normal distribution, it is not symmetric and is defined solely by its degrees of freedom (df), which depend on the number of categories or groups under consideration.

For example, in the Kruskal-Wallis test, the degrees of freedom are the number of groups minus one. You look up the critical value for the chi-square distribution in statistical tables, commonly found in the appendices of statistics textbooks or software. If the Kruskal-Wallis test statistic exceeds this critical chi-square value, the null hypothesis is rejected, implying a significant difference between the groups.
Rank Sum Test
Rank sum tests, such as the Wilcoxon rank-sum test or the Mann-Whitney U test, are nonparametric alternatives to the t-test for comparing two groups. These tests rank all observations from low to high and analyze the sums of these ranks to assess whether the two groups differ statistically.

In contrast, the Kruskal-Wallis test, which was applied in the exercise, is an extension of these rank sum tests to more than two groups. It considers the sum of ranks within each group to evaluate the collective difference across all groups. Rank sum tests are particularly useful when dealing with skewed distributions or ordinal data.
Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions about the properties of populations based on sample data. It begins with an assumption, known as the null hypothesis, which is a statement of no effect or no difference. In the context of the Kruskal-Wallis test, the null hypothesis asserts that there is no difference in the median responses measured across the groups.

The alternative hypothesis counters the null, suggesting that at least one group is different. Researchers then use a test statistic to determine whether the observed data are consistent with the null hypothesis or if there is evidence to support the alternative hypothesis. If the test statistic falls into the critical region defined by the significance level (alpha), the null hypothesis is rejected.

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Most popular questions from this chapter

To compare the effects of three toxic chemicals, \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C},\) on the skin of rats, 2 -centimeter-side squares of skin were treated with the chemicals and then scored from 0 to 10 depending on the degree of irritation. Three adjacent 2-centimeter-side squares were marked on the backs of eight rats, and each of the three chemicals was applied to each rat. Thus, the experiment was blocked on rats to eliminate the variation in skin sensitivity from rat to rat. a. Do the data provide sufficient evidence to indicate a difference in the toxic effects of the three chemicals? Test using the Friedman \(F_{r}\) -test with \(\alpha=.05 .\) b. Find the approximate \(p\) -value for the test and interpret it.

Word-Association Experiments A comparison of reaction times for two different stimuli in a word-association experiment produced the accompanying results when applied to a random sample of 16 people: Do the data present sufficient evidence to indicate a difference in mean reaction times for the two stimuli? Use the Wilcoxon rank sum test and explain your conclusions.

Use the Wilcoxon rank sum test to determine whether population 1 lies to the left of population 2 by (1) stating the null and alternative hypotheses to be tested, (2) calculating the values of \(T_{1}\) and \(T_{l}^{*},(3)\) finding the rejection region for \(\alpha=.05,\) and (4) stating your conclusions. $$ \begin{array}{l|ccccc} \text { Sample } 1 & 6 & 7 & 3 & 1 & \\ \hline \text { Sample } 2 & 4 & 4 & 9 & 2 & 7 \end{array} $$

Use the Wilcoxon rank sum test to determine whether population 1 lies to the left of population 2 by (1) stating the null and alternative hypotheses to be tested, (2) calculating the values of \(T_{1}\) and \(T_{l}^{*},(3)\) finding the rejection region for \(\alpha=.05,\) and (4) stating your conclusions. $$ \begin{array}{l|lllll} \text { Sample } 1 & 1 & 3 & 2 & 3 & 5 \\ \hline \text { Sample } 2 & 4 & 7 & 6 & 8 & 6 \end{array} $$

An experiment was conducted to study the relationship between the ratings of a tobacco leaf grader and the moisture content of the tobacco leaves. Twelve leaves were rated by the grader on a scale of \(1-10\), and corresponding readings of moisture content were made. $$\begin{array}{ccc}\hline \text { Leaf } & \text { Grader's Rating } & \text { Moisture Content } \\\\\hline 1 & 9 & .22 \\\2 & 6 & .16 \\\3 &7 & .17 \\\4 & 7 & .14 \\\5 & 5 & .12 \\\6 & 8 & .19 \\\7 & 2 & .10 \\\8 & 6 & .12 \\\9 & 1 & .05 \\\10 & 10 & .20 \\\11 & 9 & .16 \\\12 & 3 & .09 \\\\\hline\end{array}$$ a. Calculate \(r_{s}\) b. Do the data provide sufficient evidence to indicate an association between the grader's ratings and the moisture contents of the leaves?
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