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- Answer letter a with the red words. Provide a brief explanationQUESTION 3 Significance level (a) aims to control which type(s) of error? O Type I error O Type Il error O Type I and Type II error O Significance level is not related to controlling for error. QUESTION 4 forror?The leader of two postpartum women’s support groups is interested in the depression levels of the women in her groups. She administers the Center for Epidemiologic Studies Depression Scale (CES-D) screening test to the members of her groups. The CES-D is a 20-question self-test that measures depressive feelings and behaviors during the previous week. The mean depression level from the screening test for the 10 women in the first group is μ₁ = 16; the mean depression level for the 14 women in the second group is μ₂ = 10. Without calculating the weighted mean for the combined group, you know that the weighted mean is:
- which of these comparison between an ANOVA and t test is correct? A. a t test provides more flexibilty in research studies than an ANOVA B. An ANOVA examines wheater mean differences exists between conditions wheres t test does not C. An ANOVA can be used to compare three or more conditions wheres a t test cannot D. A t test can be used to compare two conditions wheres an ANOVA cannot.58please help
- This research was conducted with Korean adolescents in the Los Angeles area. Would you be willing to generalize the results of this study to Korean adolescents who live in other regions of the country? Why or why not.Considering that k= 4, n= 58, SSE= 84.75, and SSR= 725 fill in the ANOVA table and state the results of testing the stated Null Hypothesis below DF SS MS Significance Regression 0.455 Residual From the table F= 2.55 Total Họ. by = 0 the result of testing the Null Hypothesis isHello! I need answer for letter F only using Anderson-Darling test to support the normal distribution of the data.
- In an experiment designed to test the output levels of three different treatments, the following results were obtained: SST = 450, SSTR : 120, nT 19. Set up the ANOVA table and test for any significant difference between the mean output levels of the three treatments. Use a = 0.05. Do not round your intermediate calculations. Source Sum Degrees Mean Square F p-value (to 2 decimals) of Variation of Squares of Freedom (to 2 decimals) (to 4 decimals) Treatments 120 Error Total 450 The p-value is - Select your answer - What is your conclusion? - Select your answer -For a fixed alpha level of significance used for a hypothesis test, the critical value for a chi-square statistic decreases as the size of the sample decreases. a. True b. FalseMy answer is as follows: T critical = 2.70 T value= 7.775 P< a (0.01) Therefore we reject the null. Is this correct?