Exam 2 CPTS 730 Nov 16, 2017 8) Using the data set anova exam2 nov2017 (in Excel

Question

Exam 2 CPTS 730 Nov 16, 2017 8) Using the data set anova exam2 nov2017 (in Excel format now) please answer the following questions. This data set includes 3 variables: ID, EF (Ejection Fraction), and Group (A,B, or C) and looks like this: EF 1 61.51A 2 65.89 A 3 56.55 A 6 55.18 B 7 63.73 A 1 ID group A) A researcher wants to determine if there are any differences in Ejection Fraction between the 3 groups. Perform a statistical test to answer this question. State the Null and Alternative Hypotheses and perform any diagnostics needed to determine whether a particular approach can be used to answer this question. State your conclusions about this test. (12 Points) The researcher is curious about comparing the individual Groups with each other (A, B and C). Using the model from (A) perform these comparisons among the groups. State the Null and Alternative Hypotheses for these 3 comparisons and state your findings. (8 Points) B)

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 table for anova is

Source of Variation

SS

df

MS

F

P-value

F crit

Between Groups

553.4607

2

276.7303

7.834929

0.001283

3.219942

Within Groups

1483.443

42

35.32008

Total

2036.904

44

a)

p<0.05 , so we can conclude that , we reject the null hypothesis and conclude that , there is significant diffreance btween the groups.

B) we can campaise the individual A,B and C with each other using two sample t-test

A vs B

H0: a = b

H1: a /= b

perform t-test in excel and oupput is ,

A

B

Mean

63.674

61.16333

Variance

22.37398286

46.46245

Observations

15

15

Hypothesized Mean Difference

0

df

25

t Stat

1.171994302

P(T<=t) one-tail

0.126123778

t Critical one-tail

1.708140761

P(T<=t) two-tail

0.252247557

t Critical two-tail

2.059538553

here p>o.o5 , so we conclude that there no significant diffreance between the A and B

B vs C

H0: b = c

H1: b /= c

perform t-test in excel and oupput is ,

t-Test: Two-Sample Assuming Unequal Variances

C

B

Mean

69.53333

61.16333

Variance

37.12381

46.46245

Observations

15

15

Hypothesized Mean Difference

0

df

28

t Stat

3.545713

P(T<=t) one-tail

0.0007

t Critical one-tail

1.701131

P(T<=t) two-tail

0.001399

t Critical two-tail

2.048407

here P<0.05, so reject the null, and conclude that there is significant diffreance between the B and C

A vs C

H0: a = c

H1: a /= cperform t-test in excel and oupput is ,

t-Test: Two-Sample Assuming Unequal Variances

C

A

Mean

69.53333

63.674

Variance

37.12381

22.37398

Observations

15

15

Hypothesized Mean Difference

0

df

26

t Stat

2.942005

P(T<=t) one-tail

0.003386

t Critical one-tail

1.705618

P(T<=t) two-tail

0.006772

t Critical two-tail

2.055529

here P<0.05, so reject the null, and conclude that there is significant diffreance between the A and C

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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