According to the manufacturer of M&Ms, 13% of the plain M&Ms in a bag should be brown, 14% yellow, 13% red, 24% blue, 20% orange, and 16% green. A student randomly selected a bag of plain M&Ms. He counted the number of M&Ms of each color and obtained the results shown in the table. Test whether plain M&Ms follow the distribution stated by the manufacturer at the a = 0.05 level of significance. E Click the icon to view the table. Determine the null and alternative hypotheses. Choose the correct answer below. O A. Ho: The distribution of colors is not the same as stated by the manufacturer. H: The distribution of colors is the same as stated by the manufacturer. O B. Hn: The distribution of colors is the same as stated by the manufacturer. H;: The distribution of colors is not the same as stated by the manufacturer. OC. Ho: The distribution of colors is at least as uniform as stated by the manufacturer. H1: The distribution of colors is less uniform than stated by the manufacturer. O D. Hn: The distribution of colors is at most as uniform as stated by the manufacturer. H,: The distribution of colors is more uniform than stated by the manufacturer. Compute the expected counts for each color. (Round to two decimal places as needed.) Color Frequency Expected Count Brown 60 Yellow 63 Red 57 Blue 60 Orange 94 Green 64 What is the test statistic? X6 = (Round to three decimal places as needed.) What is the P-value of the test? P-value = (Round to three decimal places as needed.) Based on the results, do the colors follow the same distribution as stated in the problem? V Ho. There is V evidence that the distribution of colors is V stated by the manufacturer. Observed Distribution of Colors Color Brown Yellow Red Blue Orange Green Frequency Claimed Proportion 60 63 57 60 94 64 0.13 0.14 0.13 0.24 0.20 0.16 Print Done
According to the manufacturer of M&Ms, 13% of the plain M&Ms in a bag should be brown, 14% yellow, 13% red, 24% blue, 20% orange, and 16% green. A student randomly selected a bag of plain M&Ms. He counted the number of M&Ms of each color and obtained the results shown in the table. Test whether plain M&Ms follow the distribution stated by the manufacturer at the a = 0.05 level of significance. E Click the icon to view the table. Determine the null and alternative hypotheses. Choose the correct answer below. O A. Ho: The distribution of colors is not the same as stated by the manufacturer. H: The distribution of colors is the same as stated by the manufacturer. O B. Hn: The distribution of colors is the same as stated by the manufacturer. H;: The distribution of colors is not the same as stated by the manufacturer. OC. Ho: The distribution of colors is at least as uniform as stated by the manufacturer. H1: The distribution of colors is less uniform than stated by the manufacturer. O D. Hn: The distribution of colors is at most as uniform as stated by the manufacturer. H,: The distribution of colors is more uniform than stated by the manufacturer. Compute the expected counts for each color. (Round to two decimal places as needed.) Color Frequency Expected Count Brown 60 Yellow 63 Red 57 Blue 60 Orange 94 Green 64 What is the test statistic? X6 = (Round to three decimal places as needed.) What is the P-value of the test? P-value = (Round to three decimal places as needed.) Based on the results, do the colors follow the same distribution as stated in the problem? V Ho. There is V evidence that the distribution of colors is V stated by the manufacturer. Observed Distribution of Colors Color Brown Yellow Red Blue Orange Green Frequency Claimed Proportion 60 63 57 60 94 64 0.13 0.14 0.13 0.24 0.20 0.16 Print Done
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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![### Testing the Distribution of M&M Colors
**Objective:**
To determine if the distribution of colors in a sample of plain M&Ms matches the distribution stated by the manufacturer using a chi-square test at the α = 0.05 level of significance.
**Manufacturer's Color Distribution:**
According to the manufacturer:
- Brown: 13%
- Yellow: 14%
- Red: 13%
- Blue: 24%
- Orange: 20%
- Green: 16%
**Sample Data:**
A student randomly selects a bag of plain M&Ms and counts the number of each color, obtaining the following results:
- Brown: 60
- Yellow: 63
- Red: 57
- Blue: 60
- Orange: 94
- Green: 64
#### Define the Null and Alternative Hypotheses
- **H₀ (Null Hypothesis):** The distribution of colors is the same as stated by the manufacturer.
- **H₁ (Alternative Hypothesis):** The distribution of colors is not the same as stated by the manufacturer.
#### Select the Hypotheses Pair
- **Option A is correct:**
- **H₀:** The distribution of colors is the same as stated by the manufacturer.
- **H₁:** The distribution of colors is not the same as stated by the manufacturer.
#### Compute the Expected Counts for Each Color
Based on the total number of M&Ms in the sample (60 + 63 + 57 + 60 + 94 + 64 = 398), calculate the expected counts using the manufacturer's distribution percentages.
**Expected Counts (rounded to two decimal places):**
- **Brown:** 0.13 * 398 ≈ 51.74
- **Yellow:** 0.14 * 398 ≈ 55.72
- **Red:** 0.13 * 398 ≈ 51.74
- **Blue:** 0.24 * 398 ≈ 95.52
- **Orange:** 0.20 * 398 ≈ 79.60
- **Green:** 0.16 * 398 ≈ 63.68
#### Calculate the Test Statistic
The chi-square test statistic is calculated as follows:
\[ \chi^{2}_{0} = \sum \frac{(O_i - E_i)^2}{E_i} \]
Where \(O](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1a0802a2-9dd1-43c0-9b3f-3c3accbadbd6%2F4b317612-5a3d-49ce-82a1-ef55f0c831ee%2Fubic3nk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Testing the Distribution of M&M Colors
**Objective:**
To determine if the distribution of colors in a sample of plain M&Ms matches the distribution stated by the manufacturer using a chi-square test at the α = 0.05 level of significance.
**Manufacturer's Color Distribution:**
According to the manufacturer:
- Brown: 13%
- Yellow: 14%
- Red: 13%
- Blue: 24%
- Orange: 20%
- Green: 16%
**Sample Data:**
A student randomly selects a bag of plain M&Ms and counts the number of each color, obtaining the following results:
- Brown: 60
- Yellow: 63
- Red: 57
- Blue: 60
- Orange: 94
- Green: 64
#### Define the Null and Alternative Hypotheses
- **H₀ (Null Hypothesis):** The distribution of colors is the same as stated by the manufacturer.
- **H₁ (Alternative Hypothesis):** The distribution of colors is not the same as stated by the manufacturer.
#### Select the Hypotheses Pair
- **Option A is correct:**
- **H₀:** The distribution of colors is the same as stated by the manufacturer.
- **H₁:** The distribution of colors is not the same as stated by the manufacturer.
#### Compute the Expected Counts for Each Color
Based on the total number of M&Ms in the sample (60 + 63 + 57 + 60 + 94 + 64 = 398), calculate the expected counts using the manufacturer's distribution percentages.
**Expected Counts (rounded to two decimal places):**
- **Brown:** 0.13 * 398 ≈ 51.74
- **Yellow:** 0.14 * 398 ≈ 55.72
- **Red:** 0.13 * 398 ≈ 51.74
- **Blue:** 0.24 * 398 ≈ 95.52
- **Orange:** 0.20 * 398 ≈ 79.60
- **Green:** 0.16 * 398 ≈ 63.68
#### Calculate the Test Statistic
The chi-square test statistic is calculated as follows:
\[ \chi^{2}_{0} = \sum \frac{(O_i - E_i)^2}{E_i} \]
Where \(O
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