Chi square is used to test __________________ dependent variables. 

Is there a difference among the 3 treatments in the table below?  Test using α = .05 Complete the ANOVA table that follows as an aid (Recommend doing so on a separate sheet of paper, or separate document), then answer the questions that follow. Treatment A Treatment B Treatment C 2 3 7 ∑X2 = 461 2 3 7 G = 75 3 4 8 N = 15 3 5 8 k = 3 5 5 10 SS1 = 6 SS2 = 4 SS3 = 6 Source           Sums of Squares           df                Mean Square           FBetween         SSbn                                dfbn             MSbn                       FWithin             SSwn                               dfwn             MSwn Total                SStot       SStot =______      SSwn = ______        SSbn = ______

A normal distribution has a mean of µ = 60 with σ = 5. If a vertical line is drawn through the distribution at X = 65, what proportion of the scores are on the left side of the line?

[example continued…] Is there a difference among the 3 treatments in the table below?  Test using α = .05  Treatment A Treatment B Treatment C 2 3 7 ∑X2 = 461 2 3 7 G = 75 3 4 8 N = 15 3 5 8 k = 3 5 5 10 SS1 = 6 SS2 = 4 SS3 = 6   Does the type of treatment make a difference?

Assume a normal distribution. What is p(z > +2.50)?

[example continued…] Is there a difference among the 3 treatments in the table below?  Test using α = .05  Treatment A Treatment B Treatment C 2 3 7 ∑X2 = 461 2 3 7 G = 75 3 4 8 N = 15 3 5 8 k = 3 5 5 10 SS1 = 6 SS2 = 4 SS3 = 6   What is the value of the critical F boundary?   (from the tables)

[example continued…] Below is data for two variables, X and Y.  Answer the questions that follow.  X Y X-MX Y-MY (X-MX)2 (Y-MY)2 (X-MX)(Y-MY) 3 2 -2 -1 4 1 2 4 2 -1 -1 1 1 ___ 6 4 1 ___ 1 1 1 7 4 2 1 ___ 1 2   r = ____

The average YouTube user spends 40 minutes per session, using YouTube (μ = 40), and σ is unknown. Test whether ARC students who are YouTube users (n = 16, M = 38, SS = 960) differ from the average YouTube user on number of minutes spent using YouTube per session. α = .05. What is the alternative hypothesis (H1)

[example continued…] For the following data: [µ = 80, σ = 24, n = 36, M = 85], zM = ______ [note: zM is the z score for the sample mean]

Does training in meditation make a difference for student success?  10 students received training (M1= 88, SS1= 240) and 5 students did not (M2= 82, SS2= 150). Test using α = .05   Which is the null hypothesis, H0?