**Educational Content on Statistical Analysis: One-Way ANOVA**In this example, we will identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. We will use a significance level of 0.05 to test the claim.### One-way ANOVA Results:Below is a table containing the ANOVA results:| Source | DF | SS | MS | F | P ||--------|-----|-----|------|-----|-------|| Size | 3 | 30 | 10.00| 1.6 | 0.264 || Error | 8 | 50 | 6.25 | | || Total | 11 | 80 | | | |**Explanation of Table Components:**- **Source**: This column indicates the sources of variation (e.g., "Size" and "Error").- **DF (Degrees of Freedom)**: This represents the number of values in the final calculation of a statistic that are free to vary.- **SS (Sum of Squares)**: This depicts the total variation, within and between groups.- **MS (Mean Square)**: This is an average of the sum of squares (SS) divided by their respective DF.- **F**: This is the test statistic derived from dividing the Mean Square between groups by the Mean Square within groups.- **P**: The P-value corresponds to the test statistic, indicating the probability of observing the results assuming that the null hypothesis is true.### Step-by-Step Identification and Calculation:1. **Null Hypothesis, H0**: - This is the hypothesis that there is no difference between the means. Formally: \( H0: \mu_1 = \mu_2 = \mu_3 = \mu_4 \)2. **Alternative Hypothesis, H1**: - This hypothesis states that at least one group mean is different. Formally: \( H1: \text{At least one }\mu_i \text{ differs} \)3. **Test Statistic, F**: - The F-value from the table is 1.6.4. **P-value**: - From the table, the P-value is 0.264.5. **Significance Level**:

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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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Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use a significance level of 0.05 to test the claim. One-way ANOVA: 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 can you show how you worked it out please
The US Department of Energy reported that 48% of homes were heated by natural gas. A random sample of 336 homes in Oregon found that 142 were heated by natural gas. Test the claim that proportion of homes in Oregon that were heated by natural gas is different than what was reported. Use a 10% significance level. Give answer to at least 4 decimal places. What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.) Based on the hypotheses, compute the following:Test Statistic = p-value =
Use a significance level of 0.05 to test the claim. One-way ANOVA: Source DF SS MS Size 3 13.500 4.500 5.17 0.011 Error 16 13.925 0.870 Total 19 27.425 Null Hypothesis, HO Alternative Hypothesis, H1 Test Statistic, F P-value Significance level Conclusion
Think about doing an Independent-Measures ANOVA, Step 3. If the ANOVA Source table looks as follows, what are the MS-between, MS-within, and F-obtained? SOURCE SS df MS Fobt Between 20.22 2 ??? ??? Within 51.33 6 ??? -- Total 71.56 8 -- --
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
Identify the test statistic. Identify the P-value. State the final conclusion that addresses the original claim.
A home improvement company is interested in improving customer satisfaction rate from the 80% currently claimed. The company sponsored a survey of 101 customers and found that 87 customers were satisfied. What is the test statistic z? What is the p-value? Does sufficient evidence exist that the customer satisfaction rate is different than the claim by the company at a significance level of a=0.05? (yes or no)
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Please answer all of the questions thank you
A cold medicine lists 600 milligrams of acetaminophen per fluid ounce as an active ingredient. A test of 75 one-ounce samples of the medicine finds that the mean amount of acetaminophen for the sample is 598 milligrams with a standard deviation of 20 milligrams. Test the claim that the medicine does not contain the required amount of acetaminophen. Use a 0.05 significance level. State the null and alternative hypotheses. O A. Ho: μ = 600 milligrams H₂:μ>600 milligrams OC. Ho: μ = 600 milligrams H₂: μ< 600 milligrams Find the z-score. Z= (Round to two decimal places as needed.) B. Ho: μ = 600 milligrams H₂: μ#600 milligrams O D. Ho: μ#600 milligrams H₂: μ = 600 milligrams M-I
Use a 0.05 significance level to test the claim that pedestrian fatalities are independent of the intoxication of the driver and the intoxication of the pedestrian. Use the traditional method. Be sure to include all the expected values in the chart. List the null and alternative hypotheses, the critical value, the test statistic, Draw the graph and label the test statistic, critical value, and rejection regions. Reject or fail to reject? State the conclusion as outlined in the template or book Pedestrian Intoxicated Expected Value Pedestrian not intoxicated Expected Value Totals Driver Intoxicated 59 79 Driver not Intoxicated 266 581 Totals
There is no difference in the opinion towards stress between private and government facultymembers. for testing of this null hypothesis which statistical tools to be applied and why?

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