Analysis of Variance Problem Compatibility Model-Microsoft Word HIT insert Page

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Analysis of Variance Problem Compatibility Model-Microsoft Word HIT insert Page LayoutReferences Malings Renew ew Ram-12 . Ka" Aa,=,E- ·RB@n t q AaabCal AaBbo! Aldal. Ame u.de x, xt A"As@a,,, :·L' Emphaus iesing1 | Tion. 1; 'uege 0 Fent 4. Ancedotal evidence suggests that an individual's performance in mathenmatics is an indicator of academic performance in general The data gives the leaving certificate points (total points for six subjects) in grades in mathematics fora random sample of students registered foran IT course OD OC B OA HD 300 330 325 310 365 195 335 425 S 300 325 350 300 300 315 310 305 360 440 325 310 335 385 350 33 95 355 395 335 340 320 400 Whee OD, grade D in Ondinary level mathematics OA: grade A in Ondinary lived mathematics HID: grade D in Higher level mathematcs et Calculate the ANOVA table for this datd. Test the claim that there is no difference in the average leaving certificate points for the different grades in mathematics H, is rejected a post hoc analysis to see which mean differs a. b. Calculate the 95% individualconfidenceintervals forthedifference maverage points between() OA and OD () OB and OD What can you concue

Explanation / Answer

Hypothesis of the data,

we are given above the table of ANOVA,

The hypotheses of interest in an ANOVA are as follows:

where k = the number of independent comparison groups.

the formulas for calculation for ANOVA,

df

ss

mss

f

p

Between

k-1

SSb

MSB=SSB/k-1

F=MSB/MSE

within

n-k

SSE

MSE=SSE/n-k

total

n-1

SST

the analysis detailed is

Source of Variation

SS

df

MS

F

P-value

F crit

Between Groups

20766.28

4

5191.571

2.892661

0.040947

2.727765

Within Groups

48457.94

27

1794.738

Total

69224.22

31

p=0.04, so we can conclude that the there is statistically significant diffreance between the 5 catogeries.

the post hock analysis is carried out in Tukey test.

The p-value corresponding to the F-statistic of one-way ANOVA is lower than 0.05 which strongly suggests that one or more pairs of treatments are significantly different. You have k=5 treatments, for which we shall apply Tukey's HSD test to each of the 10 pairs to pinpoint which of them exhibits statistially significant difference.

We first establish the critical value of the Tukey-Kramer HSD QQ statistic based on the k=5 treatments and =27 degrees of freedom for the error term, for significance level = 0.05 (p-values) in the Studentzed Range distribution. We obtain these critical values for Q critical =0.05,k=5,=27 Q= 4.1306, respectively.

treatments
pair

Tukey HSD
Q statistic

Tukey HSD
p-value

Tukey HSD
inferfence

A vs B

2.2696

0.5060622

insignificant

A vs C

4.3756

0.0339037

* p<0.05

A vs D

4.0612

0.0557042

insignificant

A vs E

2.8831

0.2755837

insignificant

B vs C

1.8021

0.6866214

insignificant

B vs D

1.6906

0.7296880

insignificant

B vs E

0.4317

0.8999947

insignificant

C vs D

0.0352

0.8999947

insignificant

C vs E

1.4930

0.8059950

insignificant

D vs E

1.3858

0.8474275

insignifican

conclusion:

Group A and Group C - i.e the OD and OB are significantlly diffrance between the each other.

thanks,

df

ss

mss

f

p

Between

k-1

SSb

MSB=SSB/k-1

F=MSB/MSE

within

n-k

SSE

MSE=SSE/n-k

total

n-1

SST

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