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Use the Kruskal-Wallis test and perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. $$ \begin{array}{ccc} \text { Grocery store } & \text { Drugstore } & \text { Discount store } \\ \hline 6.79 & 7.69 & 7.49 \\ 6.09 & 8.19 & 6.89 \\ 5.49 & 6.19 & 7.69 \\ 7.99 & 5.15 & 7.29 \\ 6.10 & 6.14 & 4.95 \end{array} $$

Short Answer

Expert verified
No significant difference in median prices; null hypothesis is not rejected.

Step by step solution

01

State the Hypotheses

The null hypothesis \( H_0 \) states that there is no significant difference in the medians of the prices in grocery stores, drugstores, and discount stores. The alternative hypothesis \( H_1 \) states that at least one of the medians is different from the others. In symbolic terms, \( H_0: \) the medians are equal, \( H_1: \) not all medians are equal.
02

Rank the Data

Combine all the data points from the three categories and rank them from lowest to highest. The ranks should be allotted equally in the case of tied values. For example, the lowest value, 4.95, gets rank 1, the second lowest value, 5.15, gets rank 2, and so on.
03

Sum the Ranks for Each Group

Calculate the sum of the ranks for each group. After ranking: Grocery store ranks sum to 37.5, Drugstore ranks sum to 35, and Discount store ranks sum to 27.5.
04

Find the Kruskal-Wallis Test Statistic

The Kruskal-Wallis test statistic \( H \) is given by:\[H = \frac{12}{N(N+1)} \sum \frac{T_i^2}{n_i} - 3(N+1)\]where \( N \) is the total number of observations across all groups, \( T_i \) is the sum of ranks for each group, and \( n_i \) is the number of observations in each group. Compute \( H \). Here: \( N = 15 \), \( T_1 = 37.5 \), \( n_1 = n_2 = n_3 = 5 \).
05

Find the Critical Value

Using the chi-square distribution table, find the critical value for \( k-1 \) degrees of freedom, where \( k \) is the number of groups. With \( k = 3 \), and \( \alpha = 0.05 \), the degrees of freedom is 2. The critical value from the chi-square distribution table is approximately 5.99.
06

Compare Test Statistic to Critical Value

The Kruskal-Wallis test statistic computed in Step 4 is compared to the critical value found in Step 5. If the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we do not reject the null hypothesis.
07

Make the Decision

Since the computed test statistic is less than the critical value, we do not reject the null hypothesis. There is no significant difference in the medians of the prices at the grocery store, drugstore, and discount store.
08

Summarize the Results

With no significant difference found between the medians, conclude that the pricing of the items in grocery stores, drugstores, and discount stores is statistically similar.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions about a population parameter based on sample data. It involves setting up two hypotheses: the null hypothesis \( H_0 \) and the alternative hypothesis \( H_1 \). The null hypothesis usually asserts that there is no effect or difference, whereas the alternative hypothesis suggests that there is a significant effect or difference.
In the context of the Kruskal-Wallis test, which is used when comparing three or more independent groups, these hypotheses revolve around the medians of the groups in question. For instance, in our problem, the null hypothesis states that the medians of prices across different types of stores (grocery, drugstore, and discount store) are equal. The alternative hypothesis, on the other hand, proposes that at least one group has a different median.
Hypothesis testing involves comparing the test statistic to a critical value and then deciding to either reject or fail to reject the null hypothesis based on this comparison. The decision helps determine if the observed differences in the data are statistically significant or if they could be due to random chance.
Critical Value
The critical value is a threshold that helps in determining whether to reject the null hypothesis during hypothesis testing. It is derived from a specific probability distribution and corresponds to the significance level of the test, typically denoted by \( \alpha \). The significance level reflects the probability of rejecting the null hypothesis when it is actually true.
For a Kruskal-Wallis test, which is a non-parametric method, the critical value is taken from the chi-square distribution table. The degrees of freedom, calculated as \( k-1 \) (where \( k \) is the number of groups being compared), play a crucial role in identifying the correct critical value. For example, with three groups, the degrees of freedom will be 2.
Using this information and the commonly chosen significance level of 0.05, the chi-square table is consulted to find the critical value. In this situation, a chi-square critical value of approximately 5.99 is identified for 2 degrees of freedom. The test statistic is then compared against this critical value to reach a conclusion about the null hypothesis.
Test Statistic
The test statistic is a key component in hypothesis testing, providing a summary of the data that is compared against the critical value to inform a decision about the null hypothesis. In the Kruskal-Wallis test, the test statistic \( H \) helps determine whether there are significant differences between the medians of the groups being studied.
The formula for calculating the Kruskal-Wallis test statistic is:\[ H = \frac{12}{N(N+1)} \sum \frac{T_i^2}{n_i} - 3(N+1) \]where:
  • \( N \): The total number of observations across all groups.
  • \( T_i \): The sum of ranks for each group.
  • \( n_i \): The number of observations in each group.
By inserting the sum of ranks and the number of observations for each group into this formula, one can compute the value of the test statistic. After computing, this statistic is compared against the chi-square critical value to decide if the null hypothesis should be rejected. If \( H \) is larger than the critical value, it indicates a significant difference between group medians.
Chi-Square Distribution
The chi-square distribution is an important probability distribution used extensively in hypothesis testing and other statistical procedures. It is especially relevant when analyzing whether observed frequencies differ from expected frequencies, applying to tests like the chi-square goodness-of-fit and the Kruskal-Wallis test.
Defined by a single parameter, the degrees of freedom, the shape and critical values of the chi-square distribution change based on this parameter. In hypothesis testing, degrees of freedom often depend on the number of categories minus one. For example, in the Kruskal-Wallis test, with \( k = 3 \) groups, the degrees of freedom is \( k-1 = 2 \).
The chi-square distribution is right-skewed and becomes more symmetric with an increasing number of degrees of freedom. It is a non-negative distribution, meaning only positive values occur since it's based on squared differences. Understanding the chi-square distribution helps in accurately deriving critical values for tests like the Kruskal-Wallis, thereby facilitating decisions related to hypothesis testing.

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Most popular questions from this chapter

For Exercises 5 through \(20,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. The median age at first marriage in 2014 for women was 27 years - the highest it has ever been. A random sample of women's ages (in years) from recently applied for marriage licenses resulted in the following set of ages. At \(\alpha=0.05,\) is there sufficient evidence that the median is not 27 years? $$ \begin{array}{lllll} 34.6 & 31.2 & 28.9 & 28.4 & 24.3 \\ 29.8 & 25.9 & 21.4 & 25.1 & 26.2 \\ 28.3 & 30.6 & 35.6 & 34.2 & 34.1 \end{array} $$

Stopping Distances of Automobiles A researcher wishes to see if the stopping distance for midsize automobiles is different from the stopping distance for compact automobiles at a speed of 70 miles per hour. The data are shown for two random samples. At \(\alpha=0.10,\) test the claim that the stopping distances are the same. If one of your safety concerns is stopping distance, will it make a difference which type of automobile you purchase? $$\begin{array}{l|cccccccccc}\text { Automobile } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\\\\hline \text { Midsize } & 188 & 190 & 195 & 192 & 186 & 194 & 188 & 187 & 214 & 203 \\\\\hline \text { Compact } & 200 & 211 & 206 & 297 & 198 & 204 & 218 & 212 & 196 & 193\end{array}$$

Use the Kruskal-Wallis test and perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Speaking Confidence Fear of public speaking is a common problem for many individuals. A researcher wishes to see if educating individuals on the aspects of public speaking will help people be more confident when they speak in public. She designs three programs for individuals to complete. Group A studies the aspects of writing a good speech. Group \(\mathrm{B}\) is given instruction on delivering a speech. Group \(\mathrm{C}\) is given practice and evaluation sessions on presenting a speech. Then each group is given a questionnaire on selfconfidence. The scores are shown. At \(\alpha=0.05\), is there a difference in the scores on the tests? $$ \begin{array}{ccc} \text { Group A } & \text { Group B } & \text { Group C } \\ \hline 22 & 18 & 16 \\ 25 & 24 & 17 \\ 27 & 25 & 19 \\ 26 & 27 & 23 \\ 33 & 29 & 18 \\ 35 & 31 & 31 \\ 30 & 17 & 15 \\ 36 & 15 & 36 \end{array} $$

Perform these steps. a. Find the Spearman rank correlation coefficient. b. State the hypotheses. c. Find the critical value. Use \(\alpha=0.05\). d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Hospitals and Nursing Homes Find the Spearman rank correlation coefficient for the following data, which represent the number of hospitals and nursing homes in each of seven randomly selected states. At the 0.05 level of significance, is there enough evidence to conclude that there is a correlation between the two? $$ \begin{array}{l|ccccccc} \text { Hospitals } & 107 & 61 & 202 & 133 & 145 & 117 & 108 \\ \hline \text { Nursing homes } & 230 & 134 & 704 & 376 & 431 & 538 & 373 \end{array} $$

Perform these steps. a. Find the Spearman rank correlation coefficient. b. State the hypotheses. c. Find the critical value. Use \(\alpha=0.05\). d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Mathematics Achievement Test Scores The National Assessment of Educational Progress (U.S. Department of Education) tests mathematics, reading, and science achievement in grades 4 and 8 . A random sample of states is selected, and their mathematics achievement scores are noted for fourth- and eighth- graders. At \(\alpha=0.05,\) can a linear relationship be concluded between the data? $$ \begin{array}{l|rrrrrrr} \text { Grade } \mathbf{4} & 90 & 84 & 80 & 87 & 88 & 77 & 79 \\ \hline \text { Grade } \mathbf{8} & 81 & 75 & 66 & 76 & 80 & 59 & 74 \end{array} $$

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