The image contains an educational exercise related to ANOVA (Analysis of Variance) from the General Social Survey. It examines whether there are differences in the typical work week length based on the level of education of the worker.**Content Description:**### Header: ANOVA, CHI-SQUARE AND NONPARAMETRIC TESTS ANOVA: Hypothesis tests and the ANOVA table### Text:The General Social Survey is an annual survey given to about 1500 adults in the United States. Among the questions asked are "What is the highest level of education you've completed?" and "If you're employed full-time, how many hours do you spend working at your job during a typical week?"The exercise aims to determine whether differences exist in work week length based on educational level using a one-way, independent-samples ANOVA test. The groups are categorized by education level and the variable is the number of hours worked.### ANOVA Table:- **Source of Variation** - Treatments (between groups) - Error (within groups) - Total - **Degrees of Freedom** - Treatments: 5 - Error: 1012 - Total: 1017- **Sum of Squares** - Treatments: 1888.24 - Error: 106,506.8 - Total: 108,395.04- **Mean Square** - Treatments: 377.65 - Error: 105.24- **F Statistic:** (to be calculated)### Questions:(b) How many respondents total were examined in the ANOVA? (c) For the ANOVA test, it is assumed that the population variance of hours worked per week is the same for each population. What is an unbiased estimate of this common population variance based on the sample variances? (d) Using the 0.10 level of significance, what is the critical value of the F statistic for the ANOVA test? Round your answer to at least two decimal places. (e) From the survey data, can we conclude that at least one of the populations differs significantly from the others in mean hours worked in a typical week? Use the 0.10 level of significance.### Interface: Buttons to check answers or get explanations.### Footer:© 2021 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center.

Solution: Given information: k= 6 groups n= 1018 total observation Degree of freedom between groups = DFB = 5 Degree of freedom within group = DFW = 1012 Degree of freedom total = DFT = 1017 Mean sum of square of treatments or between group =MSBG= 377.65 Mean sum of square of error or within group = MSW = MSE = 105.24 a) To find the missing value in the anova table that is to find F statistic F=MSBGMSEF= 377.65105.24= 3.588464F= 3.59 The missing value is 3.59 b) To find the total respondents in the anova. The degree of freedom associated with total is DFT = n-1 = 1017 n= 1+1017 = 1018 n= 1018 The total respondents in the anova is 1018 c) The unbiased estiamte of the common population variance based on the sample variance is. The mean square error = MSE is an unbiased estiamte of the common population variance based on the sample variances. Unbiased Estimate = MSE Unbiased Estimate = 105.24 d) The critical value of F statistic at α= 0.10 is Fα, DFB, DFW= F0.10,5,1012= 1.852972 From Excel The critical value is 1.85 e) Decision: The F statistic value is greater than F critical value 3.59 > 1.85 Reject the null hypothesis Conclusion: There is sufficient evidence to conclude that at least one of the population means differs significantly from the others in mean hours. Yes