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When only two treatments are involved, ANOVA andthe Students ttest (Chapter 11) result in the same conclusions. Also, for computed test statistics, F. To demonstrate this relationship use the following example. Fourteen randomly selected students enrolled in a history course were divided into two groups, one consisting of 6 students who took the course in the normal lecture format. The other group of 8 students took the course as a distance course format. At the end of the course, each group was examined with a 50-item test. The following is a ist of the number correct for each of the two groups. Traditional Leczure Distance 96 35 38 35 43 43 48 32 38 41

a-1. Complete the ANOVA table. (Round your SS, MS, and Fvalues to 2 decimal places and pvalue to 4 decimal places.) SourceSS Factors Emor Total MS a2 Use ad 0.01 level of significance. (Round your answer to 2 decimal places.) critical value of F b. Using the ttest from Chapter 11, compute t (Negative smount should be indiceted by e minus sign. Round your answer to 3 decimal places)

c. There is any difference in the mean test scores. HD There is in the mean scores betweer lecture and internet-based formats

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We use the one way ANOVA with replication tool of Excel and get the following ANOVA table SUMMARY Groups Count Sum Average Variance 212 340 35.33 37.6 6.27 42.5 13.43 107.4286 ecture Distance ANOVA Source Factors Error MS P value 176.10 125.33 1176.10 12 16.86 0.0015 10.44 Total 301.13 13 3-2) As from the table above, we can see the test statistic is F = 16.86 Critical value of F is obtained from F-tables or Excel function F.INV.RT using degrees of freedom 1 and 12 and level of significance 0.01 F-crit F.INV.RT 0.01, 1, 12) The critical value of F 9.33 9.330212 b) Using t-test We get the mean and standard deviation for the two samples using Excel tunctions, Average and Stdev.s MeanSS 6 35.33 37.6 8 42.5 107.43 Lecture Distance a 0.01 We use independent Samples T test The null and alternative hypothesis are and μ2 are the population means for scores of students attending lecture or distance education respectively We use following forrnulae df -(n2) SM- t2 df df -6 8-2 12 10.4445 10.4445 SM (35.33-42.5)- (0) 1.7454 に4.106 ] c a0.01 For t-4.106 and df 12, we find the two tailed p-value from Excel function t.dist.2t p-value t.dist.2t(4.106, 12) For two-tailed,we use positive value 1.615 in the tormula) p-value 0.0015 0.0015 <0.01 that is p-value < o Hence, we reject Ho Answer: Reject Ho. There is a difference between the mean scores between lecture and internet based formats

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