1. Compute the sum of squares due to treatment, SST, and the sum of squares due to error, SSE. 2. Compute the mean square due to treatment, MST, and the mean square due to error, MSE 3. Compute the F-test statistic.
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1. Compute the sum of squares due to treatment, SST, and the sum of squares due to error, SSE.
2. Compute the
3. Compute the F-test statistic.
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- The average French person drinks approximately 30 gallons of soda a year. Some French legislators are recommending instituting a sales tax on soda to raise revenue and fight obesity. Will a sales tax reduce the mean amount of soda consumption? Suppose that a sales tax on soda will be added in a random sample of communities to measure the impact on soda consumption. State the null and alternative hypothese Suppose sample results give a p-value of 0.2. State your general conclusion (reject the null or fail to reject the null). Then state your conclusion based on the context of this situation. Now suppose sample results give a p-value of 0.018. State your general conclusion (reject the null or fail to reject the null). Then state your conclusion based on the context of this situation.The height of women ages 20-29 is normally distributed, with a mean of 63.9 inches. Assume o = 2.5 inches. Are you more likely to randomly select 1 woman with a height less than 65.6 inches or are you more likely to select a sample of 16 women with a mean height less than 65.6 inches? Explain. Click the icon to view page 1 of the standard normal table. Click the icon to view page 2 of the standard normal table. What is the probability of randomly selecting 1 woman with a height less than 65.6 inches? (Round to four decimal places as needed.) What is the probability of selecting a sample of 16 women with a mean height less than 65.6 inches? (Round to four decimal places as needed.) Are you more likely to randomly select 1 woman with a height less than 65.6 inches or are you more likely to select a sample of 16 women with a mean height less than 65.6 inches? Choose the correct answer below. A. It is more likely to select 1 woman with a height less than 65.6 inches because the probability…The average retirement age in America is 63 years old. Do small business owners retire at a different average age? The data below shows the results of a survey of small business owners who have recently retired. Assume that the distribution of the population is normal. 65, 59, 58, 62, 60, 62, 58, 69, 59, 62, 71, 68, 76, 57 What can be concluded at the the a = 0.05 level of significance level of significance? a. For this study, we should use t-test for a population mean b. The null and alternative hypotheses would be: Ho: H₁: O # O O 63 c. The test statistic to d. The p-value = e. The p-value is ? a f. Based on this, we should Select an answer the null hypothesis. g. Thus, the final conclusion is that ... Add Work 63 > Next Question O The data suggest the population mean retirement age for small business owners is not significantly different from 63 at a = 0.05, so there is insufficient evidence to conclude that the population mean retirement age for small business owners is different…
- I need help pleaseIs narcissism a more common personality trait today than it was a few decades ago? It is known that the mean population score on the Narcissistic Personality Inventory (NPI) for students attending University of South Alabama around 20 years ago was μ= 15 (Twenge, 2010). Interested in the narcissism levels of students in the year 2020, a researcher administers the NPI to a random sample of 25 University of Alabama sophomores this Spring term. The mean NPI score from the researcher’s sample of sophomores is M = 16.5, with s = 3.4. 1. Find the obtained (i.e., computed) test statistic for a sample (n=25) with a mean of 16.5 2. Make a statistical decision about the null. Will you reject or fail to reject the null based on your sample data? 3. Justify your decision about the null.Can u answerthis complete please I will rate you
- 21WB_MATH_1016_WBON Ashley Ramos & Homework: Unit 7 Homework Save Score: 0.83 of 5 pts 6 of 14 (11 complete) HW Score: 49.17%, 24.58 of 50 pts 7.2.31-T Question Help A random sample of 75 eighth grade students' scores on a national mathematics assessment test has a mean score of 293. This test result prompts a state school administrator to declare that the mean score for the state's eighth graders on this exam is more than 285. Assume that the population standard deviation is 32. At a = 0.14, is there enough evidence to support the administrator's claim? Complete parts (a) through (e). (a) Write the claim mathematically and identify Ho and Ha. Choose the correct answer below. O A. Ho: H2285 (claim) O B. Ho: H= 285 (claim) Ha:H> 285 Ψc. H : μ# 285 Ha:H 285 (claim) OD. Ho:H= 285 Ο Ε. H : μ 285 Hai H> 285 (claim) (b) Find the standardized test statistic z, and its corresponding area. z = 2.17 (Round to two decimal places as needed.) (c) Find the P-value. P-value = (Round to three decimal…What are the Class 1 error rate and the Class 0 error rate on the test data?An experiment was conducted on rodents to determine if the drug Haldol influenced the amount of "rough play." Five groups were included in the experiment. Subsequent behavior in terms of mean "rough play" behaviors was recorded. The output and box plot below, display the mean differences from baseline (i.e., after the drug was administered compared to before drug administration). What are the null and alternative hypotheses of this test? Summary of Roughplay differences Scd. Dev. Haldol dese Mean Freq. -.63158419 13.290618 19 Vehicle | 0.025 mg i 0.05 mg/ 1 0.1 mg/k -12.754389 13.783537 0.2 mg/k I 18 1.1481501 20.447041 5.0499951 16.035879 20 19 -20.S5 16.431064 20 Total -5.6631968 1B.558994 96 Analysis of Variance Frob F (p-value) Source SS df KS F Test- Statistic 8999.32832 4 2249.33203 9.63 0.0000 Between groups Within groups 91 260.682623 23722.1178 Tozal 32721.4461 95 344.436275 chif (4) 4.1710 Prob>chiz = 0.303 = Bartlett's test for equal variances: Roeghpiay diferences -50 Ho: 0…
- A researcher believes that the mean earnings of top-paid actors, athletes and musicians are the same. The earnings (in millions of dollars) for several randomly selected people from each category are shown in the table below. It has been confirmed that the population variance for each group is equal and that earnings follow a normal distribution for top-pain actors, athletes and musicians. Use a 5% significance to test the claim. Job Actor Actor Actor Actor Actor Actor Actor Actor Actor Actor Actor Actor Actor Actor Athlete Athlete Athlete Athlete Athlete Athlete Athlete Athlete Athlete Athlete Musician Musician Musician Musician Musician Musician Musician Musician Musician Musician Musician Musician Musician Musician Musician Earnings 31 34 33 26 21 33 30 36 27 27 34 37 36 23 30 39 46 41 36 50 30 44 43 44 46 54 65 30 39 52 37 41 45 60 60 31 49 45 45 What is the factor variable? Top-Paid Job Type What is the response variable? Earnings Test Statistic: 15.2043 X P-Value: 0.0000✓ O…Can you write it in text please so I can copy and paste it!! Thank youI need handwritten and correct table can be in photo otherwise work in handwritten