please I need help with the bonus 3 point ..spss you help me with the previous p

ID: 3383280 • Letter: P

Question

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,

Explanation / Answer

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

Ho : 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) =(xij)2 = (6)2 +(4)2+………..+(1)2 = 102

Grand total (G) = (xij) =(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

(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

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

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

Therefore Fcal > Ftab

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