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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

AIDS Research Scientists have shown that a newly developed vaccine can shield rhesus monkeys from infection by the SIV virus, a virus closely related to the HIV virus which affects humans. In their work, researchers gave each of \(n=6\) rhesus monkeys five inoculations with the SIV vaccine and one week after the last vaccination, each monkey received an injection of live SIV. Two of the six vaccinated monkeys showed no evidence of SIV infection for as long as a year and a half after the SIV injection. \({ }^{5}\) Scientists were able to isolate the SIV virus from the other four vaccinated monkeys, although these animals showed no sign of the disease. Does this information contain sufficient evidence to indicate that the vaccine is effective in protecting monkeys from SIV? Use \(\alpha=.10 .\)

Use the large-sample approximation to the Wilcoxon signed-rank test with the information from Exercises \(5-6,\) reproduced below. Calculate the \(p\) -value for the test and draw conclusions with \(\alpha=.05 .\) Compare your results with the results in Exercises 5-6. Test for a difference in the two distributions when \(n=30\) and \(T^{+}=249\)

Competitive Running Is the number of years of competitive running experience related to a runner's distance running performance? The data on nine runners, obtained from a study by Scott Powers and colleagues, are shown in the table: $$\begin{array}{crc}\hline & \text { Years of Competitive } & \text { 10-Kilometer Finish } \\\\\text { Runner } & \text { Running } & \text { Time (minutes) } \\\\\hline 1 & 9 & 33.15 \\\2 & 13 & 33.33 \\\3 & 5 & 33.50 \\\4 & 7 & 33.55 \\\5 & 12 & 33.73 \\\6 & 6 & 33.86 \\\7 & 4 &33.90 \\\8 & 5 & 34.15 \\\9 & 3 & 34.90 \\\\\hline\end{array}$$ a. Calculate the rank correlation coefficient between years of competitive running and a runner's finish time in the 10 -kilometer race. b. Do the data provide evidence to indicate a significant rank correlation between the two variables? Test using \(\alpha=.05\)

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.

In Exercise 10 (Section 15.2), you used the sign test to determine whether the data provided sufficient evidence to indicate that assessor A tends to give higher assessments than assessor \(\mathrm{B}\), using the data shown in the table. $$ \begin{array}{ccc} \hline \text { Property } & \text { Assessor A } & \text { Assessor B } \\ \hline 1 & 276.3 & 275.1 \\ 2 & 288.4 & 286.8 \\ 3 & 280.2 & 277.3 \\ 4 & 294.7 & 290.6 \\ 5 & 268.7 & 269.1 \\ 6 & 282.8 & 281.0 \\ 7 & 276.1 & 275.3 \\ 8 & 279.0 & 279.1 \\ \hline \end{array} $$ a. Use the Wilcoxon signed-rank test for a paired experiment to test the null hypothesis that there is no difference in the distributions of property assessments between assessors \(A\) and \(B\) against the onetailed alternative, using a value of \(\alpha\) near. \(05 .\) b. Compare the conclusion of the test in part a with the conclusions derived from the results of the \(t\) -test and the sign test in Exercises \(10(\mathrm{a})\) and \(10(\mathrm{~b})\) (Section 15.2). Explain why these test conclusions are (or are not) consistent.

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