Problem 1
Problem
The proportions of smokers is calculated: Smokers Total=585975=0.6.
The proportions of non-smokers is calculated: Non-smokers Total=390975=0.4.
These values are used to calculate the expected numbers (Table 2)
The χ2 for the test is calculated using the observed and the expected data:
χ2=(Observed-Expected)2Expected
The χ2 is calculated as the sum of the cells in Table 3:
χ2=3.93 + 0.11 + 1.02 + 5.9 + 0.17 + 1.53 = 12.66
The number of degrees of freedom is: f = (Columns – 1)(Rows – 1) = (3 – 1)(2 – 1)= 2.
The critical value of χ2 at 0.1 significance level is: χ2(2;0.1)critical=4.61.
The null hypothesis is H0: p1 = p2 = p3, all proportions are equal. The alternative hypothesis is H1: p1 ≠ p2 ≠ p3, or at least one proportion is different.
Since χ2>χ2(2;0.1)critical, H1 is true, and at least one proportion is different.
The proportions are presented in Table 4.
The proportions difference is calculated:
| p1 – p2 | = | 0.71 – 0.58 | = 0.1239;
| p2 – p3 | = | 0.58 – 0.57 | = 0.0173.
The pairwise comparison test is calculated as (Anderson, Sweeney & Williams, 515):
CV23=χ2p2(1-p2)n2+p3(1-p3)n3=12.66·0.58(1-0.58)240+0.57(1-0.57)530=0.1367.
The difference in proportions is significant if | p2 – p3 | > CV23 (Anderson, Sweeney & Williams, 516); 0.0173 < 0.1367, and thus the difference in heart and cancer proportions is insignificant.
Multiple-choice questions
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Problem 2
Problem
The null hypothesis is H0: μ1 = μ2 = μ3, all means are equal. The alternative hypothesis is H1: μ1 ≠ μ2 ≠ μ3, or at least one proportion is different.
The calculations for the SStreatment:
Sum(Box 1) = (210 + 230 +190 +180 +190).
Mean (Box 1) = (210 + 230 +190 +180 +190) / 5 = 200.
Variance (Box 1) = ((200 – 210)2 + (200 – 230)2 + (200 – 190)2 + (200 – 180)2 + (200 – 190)2) / (5 – 1) = 400.
GrandMean = Sum of all data / (3 · 5) = 223.2
SStreat=n(xi-x)2=5[(200-223.2)2+(189-223.2)2+(280-223.2)2]=24467.2.
SSerror calculation:
SSerror=(xi-xi)2=(210-200)2+(230-200)2+(190-200)2+(180-200)2+(190-200)2+(195-189.6)2++295-2802+275-2802+290-2802+275-2802+265-2802=2733.2
The block degrees of freedom is: Number of boxes – 1 = 3 – 1 = 2.
The treatment mean square is:
MStreat = SStreat/dftreat = 24467.2 / 2 = 12233.6.
The F-statistics is obtained using formula:
F = MStreat / MSwithin
MSwithin = SSerror/dferror = 2733.2 / 12 = 227.8
F = 12233.6./ 227.8 = 53.7
Fcritical(4;10;0.01) = 6.06.
The test result indicates that the null hypothesis should be rejected; therefore we accept the alternate hypothesis and state that at least one mean is different.
Fisher’s LSD
t(12;0.01) = 3.05 (Anderson, Sweeney & Williams, 976).
LSD = 3.05 ∙227.8∙(15+15+15) = 35.65.
Box 1 – Box 2 = 200 – 189.6 = 10.4 < 35.65.
|Box 1 – Box 3| = |200 – 280| = 80 > 35.65.
|Box 2 – Box 3| = |189.6 – 280| = 90.4 > 35.65.
Therefore, Box 1 and Box 2 means are similar, while Box 3 mean is significantly different from the others.
Multiple-choice questions
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no correct answer (2733.2)
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no correct answer (53.7)
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Problem 3
Problem
The number of cups of coffee sold is the dependent variable (Y), and the temperature is the independent variable (X).
Correlation coefficient calculates as:
r=(xi-x)(yi-y)(xi-x)2(yi-y)2
r=-520068200∙437.5=-0.952.
The correlation coefficient indicates that 95.2% of the coffee sales are related to temperature; the rest 4.8% depend on the other factors.
The regression line coefficients (Table 2):
a=yx2-xxynx2-(x)2
b=nxy-xynx2-(x)2
Multiple-choice questions
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Problem 4
Problem
R2 = SSregression / SSTotal =838.92 / 655.96 = 0.782.
The adjusted coefficient of determination is slightly less than R2, therefore it is 0.734.
Multiple-choice questions
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Works Cited
Anderson, David R, Dennis J. Sweeney, and Thomas A. Williams. Statistics for Business and Economics. Mason, Ohio: Thomson/South-Western, 2005. Print.
