Question

the amount of things that are done to understand a physical educational study was based on the data below. Homework number #1

1. The claimed height (which is in inches) and weights (which is in pounds) of animals is significantly correlated, mean row(p) 0. Data has been collected which is below. Test this claim with statkey at a 5% significance level (alpha) and find the 98% confidence interval (using percentiles) for the true correlation between height and weight. Explain the test results consistent with the confidence interval and why? Note there are two statkey involved.

Height	62	68	66	59	72	69	71	75	67	59
Weight	104	145	168	208	178	167	184	178	119	203

2. What is the Ho and Ha?

3. What is your alpha?

4. What is your p-value?

5. Conclusion?

6. 98% confidence interval?

7. Is it consistent?

8. Can you conclude multiple testing to be a good or bad idea in general and why?

Answer 1

Ʃx =	668
Ʃy =	1654
Ʃxy =	110371
Ʃx² =	44886
Ʃy² =	283812
Sample size, n =	10
x̅ = Ʃx/n = 668/10 =	66.8
y̅ = Ʃy/n = 1654/10 =	165.4
SSxx = Ʃx² - (Ʃx)²/n = 44886 - (668)²/10 =	263.6
SSyy = Ʃy² - (Ʃy)²/n = 283812 - (1654)²/10 =	10240.4
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 110371 - (668)(1654)/10 =	-116.2

Correlation coefficient, r = SSxy/√(SSxx*SSyy) = -116.2/√(263.6*10240.4) = -0.0707

Null and alternative hypothesis:

Ho: ρ = 0 ; Ha: ρ ≠ 0

α = 0.05

Test statistic :  

t = r*√(n-2)/√(1-r²) = -0.0707 *√(10 - 2)/√(1 - -0.0707²) = -0.2005

df = n-2 = 8

p-value = T.DIST.2T(ABS(-0.2005), 8) = 0.8461

Conclusion:

p-value > α , Fail to reject the null hypothesis. There is no correlation between x and y.

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Significance level, α = 0.02

Critical value, t_c = T.INV.2T(0.02, 8) = 2.8965

98% Confidence interval for Correlation:

Lower limit = r - t_c*√((1-r²)/(n-2)) = -1.0922

Upper limit = r + t_c*√((1-r²)/(n-2)) = 0.9508

The confidence interval contain 0. so we fail to reject the null hypothesis.

Yes, it is consistent.

Multiple testing is a good idea. Because give more clear results for the test.