**Title: Analyzing M&M Color Distribution**The color distribution of plain M&M's is claimed to be 18% green, 22% orange, 18% yellow, 22% blue, 10% red, and 10% brown. This exercise involves testing the claim at the 0.05 significance level. The task is to complete the table by calculating the Expected Frequency and Residual for each color category. All answers should be rounded to three decimal places.**Table: M&M's Color Distribution**| Category | Observed Frequency | Expected Frequency | Residual ||----------|--------------------|--------------------|----------|| Green | 19 | | || Orange | 12 | | || Yellow | 11 | | || Blue | 45 | | || Red | 11 | | || Brown | 9 | | |**Key:**- **Null Hypothesis (\(H_0\)):** The proportions are as claimed: \( p_{green} = 0.18 \); \( p_{orange} = 0.22 \); \( p_{yellow} = 0.18 \); \( p_{blue} = 0.22 \); \( p_{red} = 0.1 \); \( p_{brown} = 0.1 \).- **Alternative Hypothesis (\(H_1\)):** At least one proportion is different.**Instructions:**- Calculate the Expected Frequency for each color based on the total observed frequencies.- Compute the Residuals (difference between Observed and Expected Frequencies).**Note:**A dropdown option is provided to make an original claim based on the data analysis.**Graph/Diagram Explanation:**In the absence of graphs or diagrams, the focus is on calculating the Expected Frequencies and Residuals using statistical methods to determine if the color distribution differs significantly from the claimed proportions. **Chi-Square Distribution Analysis****Description:**This section is designed to help you input critical values and evaluate a chi-square test based on a given significance level and degrees of freedom. The probability graph depicted is a generic representation of the Chi-Square (χ²) distribution, illustrating how probabilities are distributed for various χ² values. **Graph Explanation:**- **Title:** X² - Distribution- **Y-axis:** Probability- **X-axis:** X- The graph illustrates a typical right-skewed chi-square distribution, which rises quickly to a peak and then tails off gradually.**Instructions:**- Enter the values for the chi-square test using the following inputs: - **Critical Value:** Enter the critical chi-square value based on the alpha level (α) and degrees of freedom (df) provided. - **Test Statistic:** Enter the calculated test statistic, rounding to three decimal places. - **P-value:** Enter the p-value from your chi-square calculation, rounding to four decimal places. **Decision:**- Use the dropdown menu to select your decision regarding the hypothesis (e.g., reject or fail to reject).**Conclusion:**- Using the dropdown, conclude whether the data supports the hypothesized color distribution of plain M&Ms: 18% green, 22% orange, 18% yellow, 22% blue, 10% red, and 10% brown.**Note:**- Ensure all calculations are accurate and rounded as specified. The graph is for illustration only, and actual data may differ.

Name: **Title: Analyzing M&M Color Distribution**The color distribution of
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Description: **Title: Analyzing M&M Color Distribution**The color distribution of plain M&M's is claimed to be 18% green, 22% orange, 18% yellow, 22% blue, 10% red, and 10% brown. This exercise involves testing the claim at the 0.05 significance level. The task

The original claim is null hypothesis. Thus, the original claim is H0. The expected frequencies and residuals are, Category Observed frequency Expected frequency Residual Green 19 107×0.18=19.26 19-19.26=-0.26 Orange 12 107×0.22=23.54 12-23.54=-11.54 Yellow 11 107×0.18=19.26 11-19.26=-8.26 Blue 45 107×0.22=23.54 45-23.54=21.46 Red 11 107×0.1=10.7 11-10.7=0.3 Brown 9 107×0.1=10.7 9-10.7=-1.7 Total 107 The degrees of freedom is, df=k-1=6-1=5 The degrees of freedom is 5. The significance level is 0.05. Computation of critical-value: The critical value of χ2-distribution for 0.05 level of significance at 5 degrees of freedom can be obtained using the excel formula “=CHISQ.INV.RT(0.05,5)”. The critical value is 11.071. Thus, the critical value is χ0.05,52=11.071. The test statistic is, χ2=Sum of O-E2E=19-19.26219.26+12-23.54223.54+11-19.26219.26+45-23.54223.54+11-10.7210.7+9-10.7210.7=0.004+5.657+3.542+19.564+0.008+0.270=29.045 Thus, the test statistic is 29.045. Computation of p-value: The p-value of χ2-distribution at 5 degrees of freedom can be obtained using the excel formula “=CHISQ.DIST.RT(29.045,5)”. The p-value is 0.0000. Decision rule: If p-value≤α, then reject the null hypothesis. Otherwise, do not reject the null hypothesis. Conclusion: The p-value (0.00) is less than the level of significance (0.05). Based on decision rule, reject the null hypothesis. Thus, it cannot be concluded that the claim that the color distribution of plain M&Ms is 18% green, 22% orange, 18% yellow, 22% blue, 10% red and 10% brown.

Math

Statistics

The color distribution of plain M&Ms are claimed to be 18% green, 22% orange, 18% yellow, 22% blue, 10% red, and 10% brown. Test the claim at the 0.05 significance level. Complete the table. Round all answers to three decimal places.

The color distribution of plain M&Ms are claimed to be 18% green, 22% orange, 18% yellow, 22% blue, 10% red, and 10% brown. Test the claim at the 0.05 significance level. Complete the table. Round all answers to three decimal places.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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