data A. Data B. Data C. Complete parts A and B for question 1. Please explain st
ID: 3363191 • Letter: D
Question
data A.
Data B.
Data C.
Complete parts A and B for question 1. Please explain steps.
1. You will need your ticker code (show your ticker code on your homework and include a printout of the data) for stock prices for this question. Use your ticker code to obtain the closing prices for the following three time periods to obtain three data sets: March 2, 2016 to March 16, 2016 March 17, 2016 to March 31, 2016 April 1,2016 to April 17, 2016 Data set A Data set B Data set C To ensure equal sample size per group use only the first 8 observations in each data set. Add 0.1 and 0.2 to each one of the closing prices in data sets B and C, respectively Imagine you are testing for the effects of two experimental drugs (data sets B and C) relative to a control group (Data set A) on a physiological variable. a) Conduct an ANOVA on the data set. Show all calculations, table of results, and state your conclusion. b) Use the Bonferroni-Holm (regardless of whether part "a" is significant or not) to examine all pairwise comparisons. Show all you calculations and state your conclusionsExplanation / Answer
a)
SSA=8(16.22875-(16.03))^2……8(15.68625-(16.0354))^2=1.47
SSW=(16.01-(16.22875))^2….(16.06-(15.68625))^2=5.63
MSA=1.47/(3-1)=0.73
MSW=5.63/(24-3)=0.27
F=0.73/0.27=2.74
Since F is less than F crit, we cannot reject the null hypothesis.
b)
BONFERRONI'S METHOD
S^p=E(n-1)*s^2/E(n-1), n=3
3(0.01975536)+3(0.2775)+3(0.5063)
2.41/24
0.1
tcrit is 2.06
t* S^p*1/ni+1/nj
2.06*0.1*1/8+1/8
0.103
Pairwise difference
X1-X2=0.0375
X1-X3=0.5425
X2-X3=0.505
Thus X1-X3 and X2-X3 have significant differences amongst the means
A B C 16.01 16.26 16.97 16.4 17.2 16.22 16.29 16.54 15.72 16.37 16.46 15.05 16.06 15.69 14.77 16.19 15.79 15.35 16.2 15.81 15.35 16.31 15.78 16.06 X1 X2 X3 16.22875 16.19125 15.68625 X 16.03541667 0.019755357 0.277555357 0.5063125 VarianceSSA=8(16.22875-(16.03))^2……8(15.68625-(16.0354))^2=1.47
SSW=(16.01-(16.22875))^2….(16.06-(15.68625))^2=5.63
MSA=1.47/(3-1)=0.73
MSW=5.63/(24-3)=0.27
F=0.73/0.27=2.74
Since F is less than F crit, we cannot reject the null hypothesis.
Anova: Single Factor SUMMARY Groups Count Sum Average Variance A 8 129.83 16.22875 0.019755357 B 8 129.53 16.19125 0.277555357 C 8 125.49 15.68625 0.5063125 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 1.47 2.00 0.73 2.74 0.09 3.47 Within Groups 5.63 21.00 0.27 Total 7.09 23.00b)
BONFERRONI'S METHOD
S^p=E(n-1)*s^2/E(n-1), n=3
3(0.01975536)+3(0.2775)+3(0.5063)
2.41/24
0.1
tcrit is 2.06
t* S^p*1/ni+1/nj
2.06*0.1*1/8+1/8
0.103
Pairwise difference
X1-X2=0.0375
X1-X3=0.5425
X2-X3=0.505
Thus X1-X3 and X2-X3 have significant differences amongst the means
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