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IncorrectQuestion 11 / 1 ptsA random sample of 40 companies with assets over $10 million was surveyed and asked to indicate their industry and annual computer technology expense. The ANOVA comparing the average computer technology expense among three industries rejected the null hypothesis. The Mean Square Error (MSE) was 195. The following table summarized the results:Based on the comparison between the mean annual computer technology expense for companies in the tax service and food service industries, the 95% confidence interval shows an interval of -5.85 to 14.85 for the difference. This result indicates that _______________.There is no significant difference between the two industry technology expensesThe interval contains a difference of 20.7Companies in the food service industry spend significantly more than companies in the tax service industryCompanies in the food service industry spend significantly less than companies in the tax service industryAssuming the difference is calculated as (Food services expense - Tax services expense), the interval is [-5.85, 14.85]. Because 0 is in this interval, we conclude that the pair of means does not differ. In this case, we can conclude food services technology expense is NOT significantly different than tax services technology expenses.Question 21 / 1 ptsWhen testing for differences between treatment means, a confidence interval is computed with __________________.
The mean square errorThe standard deviationThe sum of squared errorsThe standard error of the meanThe mean square error term in the ANOVA table is an estimate of the common population variance.Question 31 / 1 ptsGiven the following ANOVA table for three treatments each with six observations:What is the computed value of F?7.487.848.84
8.48The computed value of F is the treatment mean square divided by the error mean square. To compute the mean squares, there are three treatments and a total of 18 observations, so k = 3 and n = 18. The treatment degrees of freedom are k - 1 = 3 - 1 = 2 and the error degrees of freedom are n - k = 18 - 3 = 15. The treatment mean square is The mean square error is The computed value of F is Question 41 / 1 ptsAnalysis of variance is used to _____________.Compare nominal dataCompute a t testCompare population proportionsSimultaneously compare several population meansWe use the ANOVA technique to compare several treatment means to see if they are equal.