Question

For each of the questions below calculate (show all work):

1) Null and alternative hypothesis
2) Cutoff score (α, df_BG, df_E)
3) Sum of squares between- and within-groups (SS_BG, SS_E)
4) Mean square between- and within-groups (MS_BG, MS_E)
5) F_obt
6)Decision to retain or reject the null and write-up

Carl wants to know if vodka (5, 10, or 15 shots) affects the number of times (out of 10 throws) people hit a three-pointer with a basketball. He is predicting that the groups will differ significantly (α = .05). (Round to the hundredths place)

5 shots	10 shots	15 shots
4	3	4
4	2	2
5	3	3
4	4	1

Answer 1

(1)

H0:Null Hypothesis: ( the groups will not differ significantly)

HA: Alternative Hypothesis: (At least one mean is different from other 2 means) ( the groups will differ significantly) (Claim)

(2)

From the given Table, the following Table is calculated:

Shots	N	Mean	Std. Dev
5 Shots	4	4.25	0.5
10 Shots	4	3	0.8165
15 Shots	4	2.5	1.291

:From the abovw Table, ANOVA Table is calculated as follows:

Source of Variation	Degrees of Freedom	Sum of Squares	Mean Square	F Stat	P Value
Between Groups	2	6.5	6.5/2=3.25	3.25/0.8611=3.7742	0.0645
Witnin Groups	9	7.7501	7.7501/9=0.8611
Total	11	14.2501

Cutoff scores:

(i) = 0.05

(ii) dfBG = 2

(iii) dfE = 9

(3)

(i) Sum of squares between--groups (SS_BG) =6.5

(ii) Sum of squares within-groups (SS_E) = 7.7501

(4)

(i) Mean squares between--groups (MS_BG) = 3.25

(ii) Mean squares within-groups (MS_E) = 0.8611

(5)

Fobt =3.7742

(6)

Since p - value = 0.0645 is greater than = 0.05, the difference is not significant. Fail to reject null hypothesis.

Conclusion:

The data support the claim that the groups will differ significantly.