Question

You want to test which candy gets you the best ratings on student evaluations (0-10). The first set of students gets no candy, the second gets black licorice, and the third gets skittles. You test n = 4 people in each group. No Black Candy Licorice Skittles 4 8 N = 12 زرا زرا 4 G = 48 5 3 6 2x2 = 238 4 1 6 T = 16 8 24 M= 4 2 6 SS = 2 4 8 a. State the null hypothesis and alternative hypothesis, in both symbols and words. b. Calculate the appropriate ANOVA and fill in the ANOVA Summary table below: SS DF MS F Between Within Total c. What is the Fcrit value you should be using to make your hypothesis testing decision at a = .05? d. Should you reject or fail to reject the null at a = .05? e. So, did candy have an effect on evaluation scores or not? f. Figure out which groups are significantly different from each other using the Tukey HSD follow-ups. g. Should the professor give students candy- if so, what kind(s)? Why?

Answer 1

(a) Ho :M, = 42 = 43 Means not equal Hi : are Ho: There is no of candies evence in the ratings agjeve Hy: Ratings of candies differ from each other.

(6) G2 T.SS = EEye - {X² - G2 and within ss E Tini Eye Ex² - { Tizni ..T.SS 238 (4824 238 - 192 T-ss 46 بیان (Total sum of squares) je wad 24. within s.s= 238 162 + 8zict 24 4 238 - 64 + 16 + 144) or withins.s 238 - 224 9 14 Since Total s.s di Between ss & within s. Between s.s - Total s.. within s.s Betweens.s Les M46-14 32

Su ummary Table SS DF MS f Between 32 2(K-1) 32 = 16 Msb = 16 1.56 MSw It is 10:25 4 9(NK) 14 = 1.56 4 within Total 46 II (N1)

off for (c) from the f- ratio table critical value f for K+ = 31= 2, N-K - 12-3' = i e 2, 9 degrees of freedom and at 0.05 level significa cance is f = 4.26 q g summary ANOVA nell rypothesis d) & & af f-value calculated from summa is greater than critical value if from tables then we reject the at guin level of significance... since calculated = 10.25 > fcritical (4.26) we reject the Null Hypothesis LO Asus