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Sum of Squares, Error Mean Squares, Treatment Difference Absolute Value Mean Squares, Error Calculate the value of the test statistic (to 2 decimals). 59.38 TA-TB The p-value is less than 0.01 What is your conclusion? Conclude that not all treatment means are equal b. Calculate the value of Fisher's LSD (to 2 decimals). 3.23 Use Fisher's LSD procedure to test whether there is a significant difference between the means for treatments A and B, treatments A and C, and treatment FA-FC 16 66 10 326.6 -11.77 5.5 Hide Feedback Partially Correct Conclusion FB-FC Significant difference c. Use Fisher's LSD procedure to develop a 95% confidence interval estimate of the difference between the means of treatments A and B (to 2 decimals). E numbers. -21.23 Significant difference Significant difference

Sum of Squares, Error Mean Squares, Treatment Difference Absolute Value Mean Squares, Error Calculate the value of the test statistic (to 2 decimals). 59.38 TA-TB The p-value is less than 0.01 What is your conclusion? Conclude that not all treatment means are equal b. Calculate the value of Fisher's LSD (to 2 decimals). 3.23 Use Fisher's LSD procedure to test whether there is a significant difference between the means for treatments A and B, treatments A and C, and treatment FA-FC 16 66 10 326.6 -11.77 5.5 Hide Feedback Partially Correct Conclusion FB-FC Significant difference c. Use Fisher's LSD procedure to develop a 95% confidence interval estimate of the difference between the means of treatments A and B (to 2 decimals). E numbers. -21.23 Significant difference Significant difference

MATLAB: An Introduction with Applications
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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i only need part c. please do it correct.

 

# 13.3 Practice: Analysis of Variance (ANOVA) ## Given Data: - **Sum of Squares, Treatment**: \( 653.3 \) - **Sum of Squares, Error**: \( 66 \) - **Mean Squares, Treatment**: \( 326.6 \) - **Mean Squares, Error**: \( 5.5 \) ### Task a) Calculate the value of the test statistic (to 2 decimals): \[ F_{\text{Statistic}} = 59.38 \] - **p-value**: less than \( 0.01 \) ### Conclusion: Based on the given p-value, the conclusion is: **Conclude that not all treatment means are equal**. ### Task b) Calculate the value of Fisher's LSD (to 2 decimals): \[ \text{Fisher's LSD} = 3.23 \] Use Fisher's LSD procedure to test whether there is a significant difference between the means for treatments A and B, treatments A and C, and treatments B and C. Use \( \alpha = 0.05 \). #### Difference Analysis Table: | Difference | Absolute Value | Conclusion | |--------------------------|----------------|-----------------------| | \( \overline{x}_A - \overline{x}_B \) | 16 | Significant difference | | \( \overline{x}_A - \overline{x}_C \) | 6 | Significant difference | | \( \overline{x}_B - \overline{x}_C \) | 10 | Significant difference | ### Task c) Use Fisher's LSD procedure to develop a 95% confidence interval estimate of the difference between the means of treatments A and B (to 2 decimals). Enter negative values as negative numbers. \[ \text{95% Confidence Interval} = [-21.23, -11.77] \] The confidence interval indicates that the difference between the means of treatments A and B is significantly different and falls within the range of \(-21.23\) to \(-11.77\). This practice exercise demonstrates the application of the Analysis of Variance (ANOVA) and Fisher's Least Significant Difference (LSD) procedure in determining the significance of differences between treatment means.
Transcribed Image Text:# 13.3 Practice: Analysis of Variance (ANOVA) ## Given Data: - **Sum of Squares, Treatment**: \( 653.3 \) - **Sum of Squares, Error**: \( 66 \) - **Mean Squares, Treatment**: \( 326.6 \) - **Mean Squares, Error**: \( 5.5 \) ### Task a) Calculate the value of the test statistic (to 2 decimals): \[ F_{\text{Statistic}} = 59.38 \] - **p-value**: less than \( 0.01 \) ### Conclusion: Based on the given p-value, the conclusion is: **Conclude that not all treatment means are equal**. ### Task b) Calculate the value of Fisher's LSD (to 2 decimals): \[ \text{Fisher's LSD} = 3.23 \] Use Fisher's LSD procedure to test whether there is a significant difference between the means for treatments A and B, treatments A and C, and treatments B and C. Use \( \alpha = 0.05 \). #### Difference Analysis Table: | Difference | Absolute Value | Conclusion | |--------------------------|----------------|-----------------------| | \( \overline{x}_A - \overline{x}_B \) | 16 | Significant difference | | \( \overline{x}_A - \overline{x}_C \) | 6 | Significant difference | | \( \overline{x}_B - \overline{x}_C \) | 10 | Significant difference | ### Task c) Use Fisher's LSD procedure to develop a 95% confidence interval estimate of the difference between the means of treatments A and B (to 2 decimals). Enter negative values as negative numbers. \[ \text{95% Confidence Interval} = [-21.23, -11.77] \] The confidence interval indicates that the difference between the means of treatments A and B is significantly different and falls within the range of \(-21.23\) to \(-11.77\). This practice exercise demonstrates the application of the Analysis of Variance (ANOVA) and Fisher's Least Significant Difference (LSD) procedure in determining the significance of differences between treatment means.
### 13.3 Practice #### Exercise: The following data are from a completely randomized design. | | Treatment A | Treatment B | Treatment C | |---------------|-------------|-------------|-------------| | Observations | 32 | 45 | 33 | | | 30 | 44 | 36 | | | 30 | 45 | 35 | | | 26 | 47 | 36 | | | 32 | 49 | 40 | - Sample mean: | 30 | 46 | 36 | - Sample variance: | 6.00 | 4.00 | 6.50 | a. At the α = 0.05 level of significance, can we reject the null hypothesis that the means of the three treatments are equal? Compute the values below (to 1 decimal, if necessary). - **Sum of Squares, Treatment:** 653.3 ✔️ - **Sum of Squares, Error:** 66 ✔️ - **Mean Squares, Treatment:** 326.6 ✔️ - **Mean Squares, Error:** 5.5 ✔️ Calculate the value of the test statistic (to 2 decimals). - **Test Statistic (F-value):** 59.38 ✔️ The **p-value** is: - **Less than 0.01** ✔️ What is your conclusion? - **Conclusion:** Reject the null hypothesis and conclude that not all treatment means are equal. ✔️ b. Calculate the value of Fisher's LSD (to 2 decimals). - **Fisher's LSD:** 2.39 ✔️
Transcribed Image Text:### 13.3 Practice #### Exercise: The following data are from a completely randomized design. | | Treatment A | Treatment B | Treatment C | |---------------|-------------|-------------|-------------| | Observations | 32 | 45 | 33 | | | 30 | 44 | 36 | | | 30 | 45 | 35 | | | 26 | 47 | 36 | | | 32 | 49 | 40 | - Sample mean: | 30 | 46 | 36 | - Sample variance: | 6.00 | 4.00 | 6.50 | a. At the α = 0.05 level of significance, can we reject the null hypothesis that the means of the three treatments are equal? Compute the values below (to 1 decimal, if necessary). - **Sum of Squares, Treatment:** 653.3 ✔️ - **Sum of Squares, Error:** 66 ✔️ - **Mean Squares, Treatment:** 326.6 ✔️ - **Mean Squares, Error:** 5.5 ✔️ Calculate the value of the test statistic (to 2 decimals). - **Test Statistic (F-value):** 59.38 ✔️ The **p-value** is: - **Less than 0.01** ✔️ What is your conclusion? - **Conclusion:** Reject the null hypothesis and conclude that not all treatment means are equal. ✔️ b. Calculate the value of Fisher's LSD (to 2 decimals). - **Fisher's LSD:** 2.39 ✔️
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