please I need help with the bonus 3 point spss you help me w

please I need help with the bonus 3 point ..spss
you help me with the previous part and you forget this part which is the last one ..
thanks

Look at the dats from the \"handedness\" stuly given below. Each score represents the number of errors a subject made during a series of pitch discrimination trials. The psychologist wants to conduct a one-factor ANOVA to determine if there are any significant differences among the groups. Answer questions 6 throgh roup mean 6) Using ambolic.notation, state the null bypothesis that would be tested. (2 pa) 7) In worda, state the nul bypothesis that would be tested (2 ps) 8) Sate the aliemative hypothesis (2 pts) 9) Create a complete ANOVA summary table that includes the source of variance, df, SS, MS, and F test statistic. Make sare you show all th 1op 14and to rTive at 10) Using the information in your summary table, test the mull hypothesis you stated in #6 at the a-05 level. Compare the calculated F statistic to the critical F statistic (make sure you state the degrees of freedom used to find the critical F) Then, make a statistical decision regarding the null hypothesis (3 pts) 11) What conclusion about the study can be reached regarding the test of the mal hypothesis you conducted? (2 pts) Would the researcher need to conduct any follow-up analysis in this study? Explain. (2 pts) 12) BONUS (3 points) Enter this data into SPSS and conduct a one-factor ANOVA. and annotated output to receive the full 3 points. Attach compiete, correct,

Solution

Given n =5 k = 3 N= n*k = 15

Here we set up the hypothesis for the given data

Symbolic notation for null hypothesis is

H_o : to test whether there is significant difference among the groups

Against the alternative hypothesis

H₁ : to test whether there is no significant difference among the groups

Raw Sum of Squares (R.S.S) =(x_ij)² = (6)² +(4)²+………..+(1)² = 102

Grand total (G) = (x_ij) =(6) +(4)+………..+(1) = 30

Correction factor (C.F) = (G)²/ N = (30)²/15 =60

Total sum of squares = R.S.S- C.F = 102-60 = 42

Treatment sum of squares (Tr.S.S) = ((T²)/n) – C.F = 90 – 60 =30

Error sum of squares (E.S.S) = T.S.S – Tr.S.S = 42-30=12

Sources of variation

Degrees of freedom

Sum of squares

Mean sum of squares

F – ratio

Treatment

5-1 = 4

(S_tr )²=30

MSS_tr = 30/4=7.5

F = MSS_tr/ MSS_e                      

Error

15-5=10

(S_e)² =12

MSS_e=12/10 =1.2

F= 7.5/1.2= 6.25F_(4,10)

Total

15-1=14

(S_t)² = 42

F_cal at (4,10) degrees of freedom is 6.25

F_tab at (4,10) degrees of freedom and 0.05 level of significance is 3.48

Therefore F_cal > F_tab

We reject the null hypothesis at 5% level of significance i.e., we accept the alternative hypothesis

Conclusion : there is no significant difference among the groups

Sources of variation	Degrees of freedom	Sum of squares	Mean sum of squares	F – ratio
Treatment	5-1 = 4	(Str )2=30	MSStr = 30/4=7.5	F = MSStr/ MSSe
Error	15-5=10	(Se)2 =12	MSSe=12/10 =1.2	F= 7.5/1.2= 6.25F(4,10)
Total	15-1=14	(St)2 = 42