The General Social Survey is an annual survey given to a random selection of abo

Question

The General Social Survey is an annual survey given to a random selection of about 1500 adults in the United States. Among the many 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?" We're interested in examining whether or not there are differences in the typical work week length based on the level of education of the worker. We perform a one-way, independent-samples ANOVA test, with the "groups" in the ANOVA being categories of highest educational degree of the worker ("less than high school," "bachelor's degree," etc.) and the variable being the number of hour:s written down by the worker in answering the survey question Suppose that the results of our ANOVA are given in the ANOVA table below. Complete the missing cell of the ANOVA table (rounding your answer to two decimal places), and then answer the questions below the table Source of Degrees ofSum of Freedom Mean Square F statistic Squares Variation Treatments Groups) Error (Between 2 348.44 174.22 (Within 972 94134.1 96.85 Groups) Total 94482.54 How many groups (categories of highest degree earned) of respondents were examined in the ANOVA? For the ANOVA test, it is assumed that the population variance of hours worked per week is the same for each population of workers represented. What is arn unbiased estimate of this common population variance based on the sample variances? 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 place:s From the survey data, can we conclude that at least one of the populations differs significantly from the others in Yes mean hours worked in a typical week? Use the 0.10 level of significance No

Explanation / Answer

Ans:

1)Total df=N-1=974

Total number of sujects observed,N=974+1=975

2)

common population variance=94482.54/975=96.91

common sample variance=94482.54=97.00

3)

critical F value=FINV(0.1,2,972)=2.308

4)

F-statistic=174.22/96.85=1.799

As,F statistic is less than critical F value,we fail to reject null hypothesis and we can not conclude that at least one of the populations differes significantly different from the others.

(No)

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