Need help with ANOVA and Chi Squared please!

You may need to use the appropriate technology to answer this question. In a completely randomized design, seven experimental units were used for each of the ve levels of the factor. Complete the following ANOVA table. (Round your values for MSE and Fto two decimal places, and your pvalue to four decimal places.) Treatments 3|]0 I I I I I I I Error | | | | | Total 440 |:| A company manufactures printers and fax machines at plants located in Atlanta, Dallas, and Seattle. To measure how much employees at these plants know about quality management, a random sample of 6 employees was selected from each plant and the employees selected were given a quality awareness examination. The examination scores for these 18 employees are shown in the following table. The sample means, sample variances, and sample standard deviations for each group are also provided. Managers want to use these data to test the hypothesis that the mean examination score is the same for all three plants. Set up the ANOVA table for these data. (Round your values for MSE and F to two decimal places, and your p-value to four decimal places.) Source Sum Degrees Mean F p-value of Variation of Squares of Freedom Square Treatments Error Total Test for any significant difference in the mean examination score for the three plants. Use a = 0.05. State the null and alternative hypotheses. O Ho: Not all the population means are equal. Ha: H1 = H2 = H3 O Ho: H1 = H2 = H3 Ha: Not all the population means are equal. O H: At least two of the population means are equal. H.: At least two of the population means are different. OH: H F H 2 F H3 H: H1 = H2 = #3 O Ho: H1 = H2 = H3 Hai H1 * H2 # H3Find the value of the test statistic. (Round your answer to two decimal places.) Find the p-value. (Round your answer to four decimal places.) p-value = State your conclusion. O Do not reject Ho. There is sufficient evidence to conclude that the means for the three plants are not equal. O Reject Ho. There is not sufficient evidence to conclude that the means for the three plants are not equal. O Do not reject Ho. There is not sufficient evidence to conclude that the means for the three plants are not equal. O Reject H . There is sufficient evidence to conclude that the means for the three plants are not equal.Treatment Treatment Treatment A B C 32 43 33 29 43 36 31 45 36 27 45 35 31 49 40 Sample mean 30 45 36 Sample variance 4.00 6.00 6.50 (a) At the a = 0.05 level of significance, can we reject the null hypothesis that the means of the three treatments are equal? State the null and alternative hypotheses. O Ho: HA # HBF HC HA: HA = HB = PC O H: Not all the population means are equal. H: HA = HB = HC OH: HA = HB = HC H: Not all the population means are equal. O Ho: HA = HB = HC Hai HAF HB # HCFind the value of the test statistic. (Round your answer to two decimal places.) E Find the pvalue. (Round your answer to three decimal places.) State your conclusion. 0 Reject H0. There is not sufcient evidence to conclude that the means of the three treatments are not equal. 0 Do not reject H0. There is not sufcient evidence to conclude that the means of the three treatments are not equal. 0 Do not reject H0. There is sufcient evidence to conclude that the means of the three treatments are not equal. 0 Reject H0. There is sufcient evidence to conclude that the means of the three treatments are not equal. (b) Use Fisher's LSD procedure to test whether there is a signicant difference between the means for treatments A and B, treatments A and C, and treatments B and C. Use a = 0.05. Find the value of LSD. (Round your answer to two decimal places.) Find the pairwise absolute difference between sample means for each pair of treatments. F'A _;B' 2 \\:| FA _ 23' = \\:| Fa _ Ec' = :l Which treatment means differ significantly? (Select all that apply.) O There is a significant difference between the means for treatments A and B. O There is a significant difference between the means for treatments A and C. O There is a significant difference between the means for treatments B and C. O There are no significant differences. (c) Use Fisher's LSD procedure to develop a 95% confidence interval estimate of the difference between the means of treatments A and B. (Use X - X. Round your answers to two decimal places.) to1 2 3 4 6.8 8.9 11.0 9.9 8.2 7.7 10.2 12.6 5.5 9.6 9.6 11.9 7.8 10.4 10.2 10.7 8.6 9.3 9.0 11.2 7.5 9.9 8.2 11.5 (a) At the a = 0.05 level of significance, what is the difference, if any, in the population mean times among the four machines? State the null and alternative hypotheses. Ha: H1 = H2 = M3 = H4 O H : At least two of the population means are equal. H.: At least two of the population means are different. OH: H = Hz = M3 = H4 O H: Not all the population means are equal. Ha: H1 = H2 = M3 = H4 OH: H1 = H2 = M3 = H4 Ha: Not all the population means are equal.Find the value of the test statistic. (Round your answer to two decimal places.) Find the p-value. (Round your answer to three decimal places.) p-value = State your conclusion. O Reject H . There is sufficient evidence to conclude that the mean time between breakdowns is not the same for the four machines. O Do not reject Ho. There is sufficient evidence to conclude that the mean time between breakdowns is not the same for the four machines. O Do not reject H . There is not sufficient evidence to conclude that the mean time between breakdowns is not the same for the four machines. O Reject Ho. There is not sufficient evidence to conclude that the mean time between breakdowns is not the same for the four machines. (b) Use Fisher's LSD procedure to test for the equality of the means for machines 2 and 4. Use a 0.05 level of significance. Find the value of LSD. (Round your answer to two decimal places.) LSD = Find the pairwise absolute difference between sample means for machines 2 and 4. * 2 - X 4 = What conclusion can you draw after carrying out this test? O There is a significant difference between the means for machines 2 and 4. O There is not a significant difference between the means for machines 2 and 4