Question

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

Answer 1

bwt, x	hwt, y	XY	X²	Y²
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.

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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: