Question

this new method. You are being asked to use your newfound statistical abilities with the testing. 1. To accomplish the test for consistency, they each grade the same test for a sample of students and record the information in the table given below. Using the .10 significance level, is there a difference in the variation (variance) in grading techniques for these two professors? Prof. Swarup 52 67 56 45 Prof. Pikachu 59 60 61 51 56 63 57 65 70 54 64 Use the six step process. 2. After the pandemic is over, the school returns to normal in-class sessions in the fall. One of the statistics professors, now, wants to check if the students retained all the information taught through online teaching. They select three random students, and test them in four different topics. They record the results. Is there a difference in the mean scores among these students at 0.05 level of significance? Use the following ANOVA table to solve the problem Source of Variation ANOVA Table Degrees of Freedom Mean Squares Treatment Error Total Sum of Squares 992 90 1082 9

Answer 1

Tus professors grade the same test for a rample of students where Prof. Smarup grades n = 7 students & Prof. Pikachu grades 28 students. let se be the variance in grades given by Prof. Sumary & 6 ² be the variance in grades given by Prof. Pikachu. We want to test the following hypothesis 1 ine whether there is no difference in the variation in grading techniques for these two professors. If se denotes the sample variance in grades given by Prof. Snarup & de denotes the sample variance in grades given by Brof Pikachu. The the test statistic is given by I 7 and since the mean grade for both the professors is unknown (ie the mean of the population or the whole data) Hence under null hypothesis In F F distribution with noi, nel degrees of freedom) Critical region: We reject the null hypothesis at Lt. Lanel of sigrificance if observed F statistic > Feina stat stot thal, 1 /2in1, n2-1 observed i statistic <- - Faszinalyn (since sur alternative is a both sided alternative)

calculations to find test statistic si - Z) S Z -) where 3,- t ai -tha Here x.. 212... aiz are the observations corresponding to Prof. Snarup. and ank . Iza are the observations (grades of students) given by Prof. Pikachu. :. 3, = = [52+67 +56 +45+ 70+54 +64] = 58.2857 5 - $ [59 +60+61 +51 +56 + 63+57 +65] = 472 - 59 $,?=[52?+672+... 642] - (58.2857)? 5,2 = 24266_ - (58-2857)? 5,2= 69.3486 569.35 $ =[59*+60*+ ... +571+65"] – (59)? S?= 27983 - (59) = 16.75 . Estat = 69.35 24.14 16.75 - Find, Festa Here m=0.1 - F..65; 6,7 = 3.86597 -0.05. in observed Estat > F0.05; 6,7 hence we reject Ho and conclude on the basis of given data that at 1%. laual of significance, there is significant difference in the variation in grading techniques for thes two professors

② Here het d,,d 2, L, denote the mean scores of the 3 students who were tested. We want to test Ho L. = 4, at vs H.: Not Ho ar Ho L;=0&i=161) VSH: Not Ho As x; 20 and Possumption made If the treatments are the 3 students in ANOVA always then the test statistic for Ho against H, is given by E- MS (Treatment) MSE where Ms (Treatment) = S(due to treatment degrees of freedom for treatments. similarly MSE = 55 (due to error) degrees of freedom for error Nour under H. Fu Fafedfe 2,9 where df, is the degrees of freedom for treatments df, . " " " " error here df, = 2 & df = 9 in In F . Critical region: We reject to at level & of significance if F> Fa; df, dhe here & 20.05 e critical value Fo.05; 2,9 = 4.2565 Nowe MS (treatment) 2 992 2496 MSE – 90210 F = 496 = 49.6 hence F> F0.05; 2,9 Hence we conclude on the basis of given data that we reject the null hypothesis at 0.05 level of LO

significance with and we conclude that there is a significant difference in the mean scores of the four three students. ANOVA Table Sources of Variation Sum of Squares Degrees of Mean Squares Freedom | treatment 992 496 49.6 Error 90 10 Total 1082