Problem Sets in ANOVA

Statistics and Probability A researcher investigated the number of viral infections people contract as a function of the amount of stress they experienced during a 6-month period. The obtained data: Amount of Stress Negligible Stress Minimal Stress Moderate Stress Severe Stress What are Ho and Ha? Ho: Stress levels experienced during a six-month period does not have a significant impact on viral infection rates. HA: Stress levels do have a significant impact on a person's susceptibility to acquiring viral infections over time. Compute Fobt and complete the ANOVA summary table.

Total Sum of Squares: Xtot = 2+4+6+5+1+. +8+1+3+5+4 = 67 (?Xtot) 2 = 672 = 4489 X2tot = 4+16+36+25+1+. +64+1+9+25+16 = 349 ntot = 16 SStot = ?X2tot -- [(?Xtot) 2/ntot] = 349 -- [4489/16] = 349 -- 280.6 = 68.4 Within Groups Sum of Squares: Xno = 2+1+4+1 = 8 (?Xno) 2 = 82 = 64 X2no = 4+1+16+1 = 22 nno = 4 SSno = ?X2no -- [(?Xno) 2/nno] = 22 -- [64/4] = 6 Xmin = 4+3+2+3 = 12 (?Xmin) 2 = 122 = 144 X2min = 32+16+9+4+9 = 38 nmin = 4 SSmin = ?X2min -- [(?Xmin) 2/nmin] = 38 -- [144/4] = 2 Xmod = 6+5+7+5 = 23 (?Xmod) 2 = 232 = 529 X2mod = 36+25+49+25 = 135 nmod = 4 SSmod = ?X2mod -- [(?Xmod) 2/nmod] = 135 -- [529/4] = 2.75 Xsevere = 5+7+8+4 = 24 (?Xsevere) 2 = 242 = 576 X2severe = 25+49+64+16 = 154 nsevere = 4 SSsevere = ?X2severe -- [(?Xsevere) 2/nsevere] = 154 -- [576/4] = 10 SSwithin = ?X2no -- [(?X no) 2/n no] +.

+ ?X2severe -- [(?Xsevere) 2/nsevere] = 6+2+2.75+10 = 20.75 Between Groups Sum of Squares: SSbetween = [(?Xno) 2/nno)]+ [(?XMmin) 2/nmin)]+ [(?Xmod) 2/nmod)]+ [(?Xsevere) 2/nsevere)] -- [(?Xtot) 2/ntot)] = (64/4)+(144/4)+(529/4)+(576/4) -- (4489/16) = 16+36+132.25+144 -- 280.6 = 47.69 Degrees of Freedom: dftot = ntot -- 1 = 16 -- 1 = 15 dfwithin = ntot -- #groups = 16 -- 4 = 12 dfbetween = #groups -- 1 = 4 -- 1 = 3 Mean Squares: MSbetween = SSbetween/dfbetween = 47.69/3 = 15.90 MSwithin = SSwithin/dfwithin = 20.75/12 = 1.73 F-ratio: F = MSbetween/MSwithin = 15.90/1.73 = 9.19 ANOVA Summary Table: Source SS df MS F Between 47.69 3 15.90 9.19 Within 20.75 12 1.73 Total 68.44 15 C. With ? = .05, what is the Fcrit ? Based on the F.

table, using 3 degrees of freedom for the numerator and 12 degrees of freedom for the denominator, and an ? = .05, then Fcrit = 3.49. D. Report your statistical results. There was a significant main effect for treatment, F (3, 12) = 9.19, p < .010; therefore, the null hypothesis must be rejected. E. Perform the appropriate post hoc comparisons. The formula for Tukey's HSD (honestly significant difference) post-hoc test is: HSD = q*?(MSwithin/ngroup), where k = the number of groups (4) and dfwithin = 12, therefore, q = 4.20. HSD = 4.20*?(1.73/4) = 4.20*0.658 = 2.76.

Based on the HSD value, any difference between group means greater than 2.76 would be considered significant: Mno = 2; Mmin = 3; Mmod = 5.75; Msevere = 6 Mno vs. Mmin = 1, no significant difference since 1 < 2.76 Mno vs. Mmod = 3.75, significant difference since 3.75 > 2.76 Mmin vs. Mmod = 2.75, no significant difference since 2.75 < 2.76 Mmin vs. Msevere = 3.0, significant difference since 3.0 > 2.76 Mmod vs. Msevere = 0.25, no significant difference since 0.25 < 2.76 F. What do you conclude about this study? The one-way ANOVA revealed a significant difference between the groups.

Post-hoc analysis using Tukey's HSD test revealed that the difference between the negligible stress group and the moderate and severe stress groups were significant. In addition, the difference between the minimum and severe stress groups was also significant. This finding suggests that there is significant relationship between stress levels and susceptibility to viral infections, although the direction of causality is unknown. G. Compute the effect size and interpret it.

2 = [SSbetween -- (k -- 1)(MSwithin)]/(SStot + MSwithin), where k equals number of groups 2 = [47.69 -- (4 -- 1)(1.73)]/(68.4+1.73) = 42.5/70.13 = 0.606 = 60.6% This finding suggests that 60.6% of the variability in viral infections between groups can be explained by stress levels. H. Estimate the value of µ that is likely to be found in the severe stress condition. µ = ?X/N = 24/4 = 6 Question 2 In a study in which k = 3, n = 21, X1 = 45.3, X2 = 16.9, X3 = 8.2 compute the following squares. Source Sum of Squares df Mean Square F Between 2 73.66 1.54 Within 18 47.94 Total 20 A. Complete the ANOVA summary table.

Degrees of Freedom: dftot = ntot -- 1 = 21 -- 1 = 20 dfwithin = ntot -- #groups = 21 -- k = 21 -- 3 = 18 dfbetween = #groups -- 1 = k -- 1 = 3 -- 1 = 2 Mean Squares: MSbetween = SSbetween/dfbetween = 147.32/2 = 73.66 MSwithin = SSwithin/dfwithin = 862.99/18 = 47.94 F-ratio: F = MSbetween/MSwithin = 73.66/47.94 = 1.54 B. With ? = .05, what do you conclude about Fobt ? Using the F. statistic table, with a 2 in the numerator and 18 in the denominator, Fcrit is 3.55 for ? Of 0.05. Since Fobt < Fcrit, the difference between the groups is nonsignificant. C. Report your results in the correct format.

The treatment did not have a significant effect, F (2, 18) = 1.54, p > .10; therefore, the null hypothesis cannot be rejected. D. Perform the appropriate post hoc comparisons. What do you conclude about this relationship? Given HSD = q*?(MSwithin/ngroup), where k = 3 and dfwithin = 18, q = 3.61 for ? = .05. HSD = 3.61*?(47.94/7) = 3.61*2.62 = 9.46. Based on the HSD value, any difference between group means greater than 9.46 would be considered significant. M1 = 45.3; M2 = 16.9; M3 = 8.2 M1 vs. M2 = 28.4, significant difference since 28.4 > 9.46 M1 vs. M3 = 37.1, significant difference since 37.1 > 9.46 M2 vs.

M3 = 8.7, no significant difference since 8.7 < 9.46 Based on the post-hoc analysis there is a significant difference between the group 1 mean and the other group means, but no difference between the means for group 2 and 3. E. What is the effect size in this study, and what does this tell you about the influence of the independent variable? 2 = [SSbetween -- (k -- 1)(MSwithin)]/(SStot + MSwithin) 2 = [147.32 -- (3 -- 1)(47.94)]/(1010.31+47.94) = 95.88/1058.25 = 0.0906 = 9.06% This finding suggests the independent variable is responsible for 9.06% of the variability of the dependent variable. The independent variable is therefore a poor predictor of X.

Question 3 A researcher investigated the effect of volume of background noise on participant's accuracy rates while performing a difficult task. He tested three groups of randomly selected students and obtained the following means and sums of squares: Low Volume Moderate Volume High Volume X 61.5 65.5 48.25 n 4 5 7 Source Sum of Squares df Mean Square F Between Groups 2 6.92 Within Groups 13 47.09 Total 15 A. Complete the ANOVA.

Degrees of Freedom: dftot = ntot -- 1 = 16 -- 1 = 15 dfwithin = ntot -- #groups = 16 -- 3 = 13 dfbetween = #groups -- 1 = 3 -- 1 = 2 Mean Squares: MSbetween = SSbetween/dfbetween = 652.16/2 = 326.08 MSwithin = SSwithin/dfwithin = 612.16/13 = 47.09 F-ratio: F = MSbetween/MSwithin = 326.08/47.09 = 6.92 B. At ? = .05, what of Fcrit? Using ? = .05, a numerator of 2, and a denominator of 13, the Fcrit value equals 3.81. C. Report the statistical results in the proper format. There was a significant main effect for treatment, F (2, 13) = 6.92, p < .010; therefore, the null hypothesis must be rejected. D. Perform the appropriate post hoc tests.

Using the Tukey-Kramer post-hoc test for unequal group sample sizes: HSD = q*?[(MSwithin/2)(1/n1+1/n2)], where k = 3, dfwithin = 13, q = 3.73, for ? = .05. HSDlow vs. mod = 3.73*?[(47.09/2)(1/4+1/5)] = 12.14 HSDlow vs. high = 3.73*?[(47.09/2)(1/4+1/7)] = 11.34 HSDmod vs. high = 3.73*?[(47.09/2)(1/5+1/7)] = 10.60 Mlow = 61.5; Mmod = 65.5; Mhigh = 48.25 Mlow vs. Mmod = 4.0, no significant difference since 4.0 < 12.14 Mlow vs. Mhigh = 13.25, significant difference since 13.25 > 11.34 Mmod vs. Mhigh = 17.25, significant difference since 17.25 > 10.60 E.

What do you conclude about this study? Based on the post-hoc analysis there is a significant performance accuracy difference between the group exposed to high volume background noise and participants exposed to low and moderate levels of background noise. F. Compute the effect size and interpret it. 2 = [SSbetween -- (k -- 1)(MSwithin)]/(SStot + MSwithin), where k equals number of groups 2.