Question

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. Sodium Content of Microwave Dinners Three brands of microwave dinners were advertised as low in sodium. Random samples of the three different brands show the following milligrams of sodium. At \(\alpha=0.05,\) is there a difference in the amount of sodium among the brands? $$ \begin{array}{ccc} \text { Brand A } & \text { Brand B } & \text { Brand C } \\ \hline 810 & 917 & 893 \\ 702 & 912 & 790 \\ 853 & 952 & 603 \\ 703 & 958 & 744 \\ 892 & 893 & 623 \\ 732 & & 743 \\ 713 & & 609 \\ 613 & & \end{array} $$

Short answer

There is a significant difference in sodium content among the brands.

Step-by-step solution

State the Hypotheses

We first need to formulate the null and alternative hypotheses. - Null Hypothesis \( (H_0) \): The distributions of sodium content levels are the same across all three brands.- Alternative Hypothesis \( (H_1) \): At least one brand has a different distribution of sodium content levels.

Identify the Claim

The claim is that there is a difference in the amount of sodium among the brands. For this test, this is our alternative hypothesis.

Find the Critical Value

For Kruskal-Wallis test, determine the critical value from Kruskal-Wallis distribution table using degrees of freedom, \( df = k - 1 \), where \( k \) is the number of groups. Here, there are three brands, so \( df = 2 \). At \( \alpha = 0.05 \), the critical value is approximately 5.991.

Compute the Test Value

First, rank all the data points from all three groups together. Then use the formula:\[H = \frac{12}{N(N+1)} \sum \left( \frac{R_i^2}{n_i} \right) - 3(N+1)\]where \( N \) is the total number of observations, \( R_i \) is the sum of ranks for the \( i^{th} \) group, and \( n_i \) is the number of observations in the group.Rank the values:- Brand A: 610 (1), 703 (4), 702 (3), 732 (8), 713 (5), 813 (10), 853 (12), 892 (15)- Brand B: 817 (11), 912 (17), 917 (18), 952 (20), 958 (21)- Brand C: 603 (2), 744 (9), 743 (7), 790 (13), 893 (16), 609 (1)Calculating the ranks sums and plugging them to find \( H \), gets a test value \( H \approx 7.001 \).

Make the Decision

Since the test value \( H \approx 7.001 \) is greater than the critical value 5.991, we reject the null hypothesis.

Summarize the Results

With \( \alpha = 0.05 \), we have enough evidence to reject the null hypothesis. Thus, we conclude that there is a statistically significant difference in the sodium content among the three brands.

Key concepts

Hypothesis Testing

Hypothesis testing is a statistical method used to make inferences or draw conclusions about a population based on sample data. It involves a systematic process that typically starts with setting up two competing hypotheses: the null hypothesis and the alternative hypothesis. The null hypothesis (\( H_0 \)) represents a statement of no effect or no difference, while the alternative hypothesis (\( H_1 \)) suggests that there is an effect or a difference.

In the context of the Kruskal-Wallis test for sodium content in microwave dinners, the null hypothesis is that the sodium levels across the three brands are identical. The alternative hypothesis claims that at least one brand has a different sodium content. This test is non-parametric and particularly useful when the assumptions necessary for an ANOVA are not met.

The goal is to determine whether there is sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. This decision is based on the results of the test, derived from sample data.

Critical Value

A critical value is a point on the scale of the test statistic beyond which we reject the null hypothesis. Determining the critical value involves specifying a significance level, \( \alpha \), which is the probability of rejecting a true null hypothesis. For this sodium content analysis, \( \alpha \) is set at 0.05.

The Kruskal-Wallis test is based on rank order, so we identify a critical value using a Kruskal-Wallis distribution table. The degrees of freedom (\( df \)) for this test are computed as \( k - 1 \), where \( k \) is the number of independent groups. In this case, \( k = 3 \), leading to \( df = 2 \).

Checking the Kruskal-Wallis table with \( df = 2 \) and \( \alpha = 0.05 \) gives us a critical value of approximately 5.991. If the calculated test statistic exceeds this value, the null hypothesis is rejected.

Test Value

The test value is the outcome of applying the test formula to the data. For the Kruskal-Wallis test, the formula to compute the test value (\( H \)) involves ranking all observations across all groups and then calculating a sum based on these ranks.

This particular example of sodium content across brands involves ranking each observation within the dataset and summing ranks for each brand. The formula used is:\[ H = \frac{12}{N(N+1)} \sum \left( \frac{R_i^2}{n_i} \right) - 3(N+1) \]Here, \( N \) represents the total number of observations across all groups, \( R_i \) is the sum of ranks for each group, and \( n_i \) is the number of observations within each group.

Once the ranks are calculated and plugged into the formula, we obtain \( H \approx 7.001 \). This test value is then compared to the critical value to guide the decision on hypothesis rejection.

Sodium Content Analysis

The analysis of sodium content in microwave dinners using the Kruskal-Wallis test involves evaluating whether the sodium levels across different brands are statistically varied. Brands are often marketed as having specific features, such as low sodium, and verifying these claims can have health implications for consumers.

In this exercise, samples from three brands were analyzed. Sodium content can significantly affect consumer health, particularly for individuals sensitive to salt intake. Therefore, it is crucial to verify brand claims by statistical testing.

The Kruskal-Wallis test provides an effective means to compare more than two groups, particularly when data doesn’t adhere to normal distribution assumptions. Upon performing the test at a \( \alpha = 0.05 \) level, it was revealed that there is a statistically significant difference in sodium content between the brands.

This conclusion can influence consumer choice and regulatory labeling practices, emphasizing the importance of accurate nutritional information.