Practice Quiz 1 Solution1

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 , talpha/2= TINV(α, df2) Please, see example in the word document ch.10 notes ANOVA Source of Variation DF SS MS F Between Treatments df1=k-1 SSTR MSTR Within Treatments (Error) df2 =n-k SSE MSE Total n-1

2 F = MSTR MSE Practice Quiz 1 (Solution) 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: C 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: B Solution: The F distribution has k – 1 degrees of freedom in the numerator and n T – k degrees of freedom in the denominator. The numerator degrees of freedom df 1 =4-1=3 The denominator degrees of freedom df 2 =20-4=16 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: C Solution: k=3 n T =5+10+5=20 overall mean is (5*4+10*8+5*9)/20=145/20=7.25 4. In a completely randomized design involving four treatments, the following information is provided.

4 7. An ANOVA procedure is used for data obtained from four populations. Four samples, each comprised of 30 observations, were taken from the four populations. The numerator and denominator (respectively) degrees of freedom for the critical value of F are a. 3 and 30 b. 4 and 30 c. 3 and 119 d. 3 and 116 Answer: D The numerator degrees of freedom df 1 is k – 1=4-1=3 and the denominator degrees of freedom df 2 is n T – k =(4*30-4)=116 n T = n 1 + n 2 + … + n k 8. Which of the following is not a required assumption for the analysis of variance? a. The random variable of interest for each population has a normal probability distribution. b. The variance associated with the random variable must be the same for each population. c. At least 2 populations are under consideration. d. Populations have equal means. Answer: D 9. In an analysis of variance, one estimate of  2 is based upon the differences between the treatment means and the a. means of each sample b. overall sample mean c. sum of observations d. populations have equal means Answer: B 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.

5 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: B Solution: We have 3 treatments A, B, C. So answer is b. For 11,12,13,14,15 we should enter our data in Excel and develop ANOVA table. Please see below. Please look at Excel file “Practice Quiz 2 Solution” in the folder “week 7”. ANOVA: Single Factor SUMMARY Groups Coun t Sum Averag e Varianc e A 4 108 27 32.6666 7 B 4 96 24 13.3333 3 C 4 120 30 56 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 72 2 36 1.05882 4 0.38639 7 8.02151 7 Within Groups 306 9 34 Total 378 11         11. Refer to Exhibit 10-11. The mean square between treatments (MSTR) equals a. 1.872 b. 5.86 c. 34 d. 36 Answer: D 12. Refer to Exhibit 10-11. The mean square within treatments (MSE) equals a. 1.872 b. 5.86

7 Answer: B Solution: df 1 =k-1=3 MSTR=SSTR/(k-1)=180/3=60 k =df 1 +1=3+1=4 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: D Solution: MSE=SSE/n T -k. we know that SSTR+SSE=SST, 180+SSE=480, SSE=480-180=300 MSE=300/ n T -k. But n T -k=df 2 =18-3=15. MSE=300/15=20 ANOVA Source of Variation DF SS MS F Between Treatments 3 (df1) 180 (SSTR) 60 (MSTR) Within Treatments (Error) 15 (df2) 300 (SSE) 20 (MSE) Total 18 480 80 18. Refer to Exhibit 10-13. The test statistic is a. 2.25 b. 6 c. 2.67 d. 3 Answer: D Solution: F=MSTR/MSE=60/20=3 19. Refer to Exhibit 10-13. 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: C Solution: Alpha=0.05

8 p-value we can find using Excel FDIST (3,3,15)=0.063784. So, answer C 95% confidence means alpha=0.05 p-value >0.05 (alpha). Don’t Reject H 0 Exhibit 10-15 The following is part of an ANOVA table that was obtained from data regarding three treatments and a total of 15 observations. Source of Variation DF SS Between Treatments 64 Error (Within Treatments) 96 20. Refer to Exhibit 10-15. The number of degrees of freedom corresponding to between treatments is a. 12 b. 2 c. 3 d. 4 Answer: B Solution: df1=k-1=3-1=2 ANOVA Source of Variation DF SS MS F Between Treatments df1=k-1 SSTR MSTR Within Treatments (Error) df2 =n-k SSE MSE Total n-1 21. Refer to Exhibit 10-15. The number of degrees of freedom corresponding to within treatments is a. 12 b. 2 c. 3 d. 15 Answer: A Solution: df 2 =n-k=15-3=12 ANOVA Source of Variation DF SS MS F Between Treatments df1=k-1 SSTR MSTR Within Treatments (Error) df2 =n-k SSE MSE Total n-1 22. Refer to Exhibit 10-15. The mean square between treatments (MSTR) is a. 36 b. 16 c. 8 d. 32 Answer: D

10 n T = 20 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: D 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: B 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: D 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 d. between .05 and .10 Answer: B 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: A 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.

11 Answer: ANOVA Source of Variation SS df MS F P-value F crit Between Groups 36 2 18 4.5 0.0206 3.35 Within Groups 108 27 4 Total 144 29 b. At 95% confidence, test to see if there is a significant difference among the means. Answer: Using F table (2 numerator and 27 denominator degrees of freedom), for F = 4.5, the p -value is between 0.01 and 0.025 Actual p-value using Excel = 0.0206 Since the p -value is <  =0.05; reject H o and conclude there is a significant difference among the means. (Also, test statistic F = 4.5 > 3.35.) 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: 38.25 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: ANOVA Source of Variation SS df MS F P-value F crit Between Groups 80.25 2 40.125 4.63 0.0414 3.01 Within Groups 78 9 8.67 Total 158.25 11 Using the F table (2 numerator and 9 denominator degrees of freedom) for F = 4.63, the p - value is between 0.025 and 0.05. Actual p -value using Excel = 0.0414 Since the p -value is   = 0.1, reject H o and conclude at least one mean is different from others. (critical F = 3.01)