Practice Quiz 1

1 ECON 3300 Ch.10 Analysis of Variance (ANOVA) In the analysis of variance, the null hypothesis states that sample means, obtained from different samples, come from the same distribution. The alternative hypothesis states that there is significant difference among sample means, which indicates that the samples come from different distributions. Formulas: The numerator degrees of freedom df 1 is k – 1, and the denominator degrees of freedom df 2 is n T – k n T = n 1 + n 2 + … + n k Example MSTR when we have k=3 MSTR = n 1 ( x 1 −´ x ) 2 + n 2 ( x 2 −´ x ) 2 + n 3 ( x 3 −´ x ) 2 k − 1 MSE = ∑ j = 1 k ( n j − 1 ) s j 2 n T − k F = MSTR MSE The null and alternative hypotheses along with the rejection rule are restated as follows: H o : μ 1 = μ 2 = μ 3 H a : Not all population means are equal Reject Ho if p-value ≤ α (The test provides sufficient evidence to conclude that the means of the three populations are not equal) p-value=FDIST(F, numerator degrees of freedom df 1 , denominator degrees of freedom df 2 ) Interval Estimate The interval estimate for a single population is given by x ± t α / 2 s √ n Where s = √ MSE , t alpha/2= TINV(α, df2) Please, see example in the word document ch.10 notes

2 Practice Quiz 1 1. The required condition for using an ANOVA procedure on data from several populations is that the a. the selected samples are dependent on each other b. sampled populations are all uniform c. sampled populations have equal variances d. sampled populations have equal means Answer: 2. An ANOVA procedure is used for data that was obtained from four sample groups each comprised of five observations. The degrees of freedom for the critical value of F are a. 3 and 20 b. 3 and 16 c. 4 and 17 d. 3 and 19 Answer: 3. In a completely randomized design involving three treatments, the following information is provided: Treatment 1 Treatment 2 Treatment 3 Sample Size 5 10 5 Sample Mean 4 8 9 The overall mean for all the treatments is a. 7.00 b. 6.67 c. 7.25 d. 4.89 Answer: 4. In a completely randomized design involving four treatments, the following information is provided. Treatment 1 Treatment 2 Treatment 3 Treatment 4 Sample Size 50 18 15 17 Sample Mean 32 38 42 48 The overall mean (the grand mean) for all treatments is a. 40.0 b. 37.3 c. 48.0 d. 37.0 Answer:

4 c. sum of observations d. populations have equal means Answer: Exhibit 10-11 To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assigned to the 3 treatments. You are given the results below. Treatment Observation A 20 30 25 33 B 22 26 20 28 C 40 30 28 22 10. Refer to Exhibit 10-11. The null hypothesis for this ANOVA problem is a.  1 =  2 b.  1 =  2 =  3 c.  1 =  2 =  3 =  4 d.  1 =  2 = ... =  12 Answer: 11. Refer to Exhibit 10-11. The mean square between treatments (MSTR) equals a. 1.872 b. 5.86 c. 34 d. 36 Answer: 12. Refer to Exhibit 10-11. The mean square within treatments (MSE) equals a. 1.872 b. 5.86 c. 34 d. 36 Answer: 13. Refer to Exhibit 10-11. The test statistic to test the null hypothesis equals a. 0.944 b. 1.059 c. 3.13 d. 19.231 Answer: 14. Refer to Exhibit 10-11. The null hypothesis is to be tested at the 1% level of significance. The p - value is

5 a. greater than 0.1 b. between 0.1 to 0.05 c. between 0.05 to 0.025 d. between 0.025 to 0.01 Answer: 15. Refer to Exhibit 10-11. The null hypothesis a. should be rejected b. should not be rejected c. should be revised d. None of these alternatives is correct. Answer: Exhibit 10-13 Part of an ANOVA table is shown below. ANOVA Source of Variation DF SS MS F Between Treatments 3 180 Within Treatments (Error) Total 18 480 16. Refer to Exhibit 10-13. The mean square between treatments (MSTR) is a. 20 b. 60 c. 300 d. 15 Answer: ANOVA Source of Variation DF SS MS F Between Treatments 3 180 60 Within Treatments (Error) 15 300 Total 18 480 17. Refer to Exhibit 10-13. The mean square within treatments (MSE) is a. 60 b. 15 c. 300 d. 20 Answer: 18. Refer to Exhibit 10-13. The test statistic is

7 24. Refer to Exhibit 10-15. If at 95% confidence, we want to determine whether or not the means of the populations are equal, the p -value is a. between 0.01 to 0.025 b. between 0.025 to 0.05 c. between 0.05 to 0.1 d. greater than 0.1 Answer: 25. Refer to Exhibit 10-15. The conclusion of the test is that the means a. are equal b. may be equal c. are not equal d. None of these alternatives is correct. Answer: Exhibit 10-16 26. Refer to Exhibit 10-16. The mean square between treatments (MSTR) equals a. 400 b. 500 c. 1,687.5 d. 2,250 Answer: 27. Refer to Exhibit 10-16. The mean square within treatments (MSE) equals a. 400 b. 500 c. 1,687.5 d. 2,250 Answer: 28. Refer to Exhibit 10-16. The test statistic to test the null hypothesis equals a. 0.22 b. 0.84 c. 4.22 d. 4.5 Answer: 29. Refer to Exhibit 10-16. The null hypothesis is to be tested at the 5% level of significance. The p - value is a. less than .01 b. between .01 and .025 c. between .025 and .05

8 d. between .05 and .10 Answer: 30. Refer to Exhibit 10-16. The null hypothesis a. should be rejected b. should not be rejected c. was designed incorrectly d. None of these alternatives is correct. Answer: 31. ANOVA Source of Variation SS df MS F Between Groups ? 2 ? 4.5 Within Groups ? 27 4 Total ? ? a. Fill in all the blanks in the above ANOVA table. Answer: b. At 95% confidence, test to see if there is a significant difference among the means. 32. Random samples were selected from three populations. The data obtained are shown below. Treatment 1 Treatment 2 Treatment 3 45 31 39 41 34 37 37 35 38 40 40 42 a. Compute the overall sample mean x . Answer : b. At 90% confidence, test to see if there is a significant difference in the means of the three populations. Show the complete ANOVA table. Answer: