### Hypothesis Testing Using One-Way ANOVA**Objective:** Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and the final conclusion that addresses the original claim.**Significance Level:** 0.05**Data: One-way ANOVA Table**| Source | DF | SS | MS | F | P ||--------|----|--------|-------|------|------|| Size | 3 | 13.500 | 4.500 | 5.17 | 0.011|| Error | 16 | 13.925 | 0.870 | | || Total | 19 | 27.425 | | | |**Hypothesis Setup**- **Null Hypothesis (H0):** Fill this with the statement that indicates no effect or no difference.- **Alternative Hypothesis (H1):** Fill this with the statement that indicates there is an effect or a difference.**Test Statistic and P-value Determination**- **Test Statistic (F):** The F-value from the ANOVA table.- **P-value:** The P-value from the ANOVA table.**Conclusion Criteria**- Compare the P-value to the significance level (0.05).- **Significance Level:** 0.05- Draw a conclusion based on whether the P-value is less than or greater than the significance level.**Conclusion Structure**- If P-value < Significance Level: Reject the null hypothesis (H0).- If P-value ≥ Significance Level: Fail to reject the null hypothesis (H0).Following this structure ensures the clarity and correctness of hypothesis testing using one-way ANOVA for educational purposes.

The following information has been given: One Way Anova Table: Source DF SS MS F P Size 3…

Answered: Identify the null hypothesis,…

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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Salma claimed that the gas price in Blue Park neighborhood (mean of $3.65) is higher than that in Green Spring (mean of $3.59). She obtained the t-value of 2.223 after calculating data from 6 gas stations in Blue Park and 5 gas stations in Green Spring. Use the chart below and determine if Salma's claim is supported! Table 10-3. Critical values of "t" significance level p= 0.05 significance level p= 0.01 TP 1 12.706 63.657 2 4.303 9.925 3 3.182 5.841 2.776 4.604 5 2.571 4.032 6 2.447 3.707 7 2.365 3.499 2.306 3.355 9 2.262 3.250 10 2.228 3.169 11 2.201 3.106 Copy the following into your text box, fill in the blanks, and answer the questions A. The degree of freedom is: B. The t-calculated value is: and the t-critical value is : C. Comparison: t-calc t-crit (answer with ()) D. Is the null hypothesis accepted or rejected? Explain how you determined that! Also, explain what it means to reject or accept the hypothesis! E. Explain if Salma's claim is supported based on your t-value…
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Use a significance level of0.05 to test the claim. One-way ANOVA: Source DF SS MS Size 30 10.00 1.6 0.264 Error 50 6.25 Total 11 80 Null Hypothesis, HO Alternative Hypothesis, H1 Test Statistic, F P-value Significance level Conclusion
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