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Modern medical practice tells us not to encourage babies to become too fat. Is there a positive correlation between the weight x of a 1-year-old baby and the weighty of the mature adult (30 years old)? A random sample of medical files produced the following information for 14 females:

x (lb)2125232420152521172426221819y (lb)125125120125130120145130130130130140110115\scriptstyle\begin{matrix} \text{x (lb)} & \text{21} & \text{25} & \text{23} & \text{24} & \text{20} & \text{15} & \text{25} & \text{21} & \text{17} & \text{24} & \text{26} & \text{22} & \text{18} & \text{19}\\ \text{y (lb)} & \text{125} & \text{125} & \text{120} & \text{125} & \text{130} & \text{120} & \text{145} & \text{130} & \text{130} & \text{130} & \text{130} & \text{140} & \text{110} & \text{115}\\ \end{matrix}

Σx=300\Sigma x=300; Σy=1775\Sigma y=1775; Σx2=6572\Sigma x^{2}=6572; Σy2=226,125\Sigma y^{2}=226,125; Σxy=38,220\Sigma x y=38,220 Test the claim that the population correlation coefficient ρ\boldsymbol{\rho} is positive at the 1% level of significance.

Solution

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Step 1
1 of 2

Given:

α=1%=0.01\alpha=1\%=0.01

n=14n=14

Result part c:

r=0.4680r=0.4680

Determine the hypotheses:

H0:ρ=0H_0:\rho=0

H1ρ>0H_1\rho>0

The value of the test statistic is:

t=r(1r2)/(n2)=0.4680(10.46802)/(142)1.835t=\dfrac{r}{\sqrt{(1-r^2)/(n-2)}}=\dfrac{0.4680}{\sqrt{(1-0.4680^2)/(14-2)}}\approx 1.835

Determine the corresponding probability (P-value) using table 6 with df=n2=142=12df=n-2=14-2=12 (one-tail area):

0.025<P<0.0500.025<P<0.050

If the P-value is smaller than the significance level, reject the null hypothesis:

P>0.01 Reject H0P>0.01\Rightarrow \text{ Reject } H_0

There is not sufficient evidence to support the claim that the population correlation coefficient is positive.

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