Question

In: Math

1.) The file cats.csv contains a data set consisting of the body weight (in kilograms) and...

1.) The file cats.csv contains a data set consisting of the body weight (in kilograms) and heart weight (in grams) for 12 cats. Test at the 5% significance level that a positive linear relationship exists between the body weight of cat and their mean heart weight. Provides all parts of the test including hypotheses, test statistic, p-value, decision, and interpretation.

2.) The file cats.csv contains a data set consisting of the body weight (in kilograms) and heart weight (in grams) for 12 cats. Construct and interpret a 90% confidence interval for β1

cats.csv file is

bwt hwt
2.6 9.8
3.8 16.3
3.7 16.2
3.4 16.3
2 7.7
3.8 13.3
2.5 10.5
2.1 8.7
2.1 7.3
3.1 13.7
3.6 14.4
3.2 13.4

Solutions

Expert Solution

bwt, x hwt, y XY
2.6 9.8 25.48 6.76 96.04
3.8 16.3 61.94 14.44 265.69
3.7 16.2 59.94 13.69 262.44
3.4 16.3 55.42 11.56 265.69
2 7.7 15.4 4 59.29
3.8 13.3 50.54 14.44 176.89
2.5 10.5 26.25 6.25 110.25
2.1 8.7 18.27 4.41 75.69
2.1 7.3 15.33 4.41 53.29
3.1 13.7 42.47 9.61 187.69
3.6 14.4 51.84 12.96 207.36
3.2 13.4 42.88 10.24 179.56
Ʃx = Ʃy = Ʃxy = Ʃx² = Ʃy² =
35.9 147.6 465.76 112.77 1939.88
Sample size, n = 12
x̅ = Ʃx/n = 35.9/12 = 2.99166667
y̅ = Ʃy/n = 147.6/12 = 12.3
SSxx = Ʃx² - (Ʃx)²/n = 112.77 - (35.9)²/12 = 5.36916667
SSyy = Ʃy² - (Ʃy)²/n = 1939.88 - (147.6)²/12 = 124.4
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 465.76 - (35.9)(147.6)/12 = 24.19

1) Null and alternative hypothesis:

Ho: ρ = 0 ; Ha: ρ > 0

α = 0.05

Correlation coefficient, r = SSxy/√(SSxx*SSyy) = 24.19/√(5.36917*124.4) = 0.9360

Test statistic :  

t = r*√(n-2)/√(1-r²) = 0.936 *√(12 - 2)/√(1 - 0.936²) = 8.4082

df = n-2 = 10

p-value = T.DIST.RT(8.4082, 10) = 0.0000

Conclusion:

p-value < α, Reject the null hypothesis.

There is enough evidence to conclude that at the 5% significance level a positive linear relationship exists between the body weight of cat and their mean heart weight.

---

2) Slope, b = SSxy/SSxx = 24.19/5.36917 = 4.5054

Significance level, α = 0.10

Critical value, t_c = T.INV.2T(0.1, 10) = 1.8125  

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 124.4 - (24.19)²/5.36917 = 15.41547

Standard error, se = √(SSE/(n-2)) = √(15.41547/(12-2)) = 1.2416

90% confidence interval for β1:


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