$$\text { Test the significance of the correlation coefficient } \mathrm { r } \text { at } \ddagger = .05 \text { for the data }$$ below $\begin{array}{lcccc} \text { X values } & -6 & -1 & 1 & 4 \ \text { Y values } & 2 & -1 & -2 & -2 \end{array}$ A) Do not reject -= 0 because -4.31 3.18 C) Do not reject -= 0 because -3.99 -4.30 D) Reject -= 0 because -4.31 2.78

Question

$$\text { Test the significance of the correlation coefficient } \mathrm { r } \text { at } \ddagger = .05 \text { for the data }$$ below
$\begin{array}{lcccc}
\text { X values } & -6 & -1 & 1 & 4 \
\text { Y values } & 2 & -1 & -2 & -2
\end{array}$
A) Do not reject -= 0 because -4.31 < 3.18
B) Reject -= 0 because -3.99 > 3.18
C) Do not reject -= 0 because -3.99 > -4.30
D) Reject -= 0 because -4.31 > 2.78

Quizplus · Accepted Answer

To test the significance of the correlation coefficient, we need to calculate the correlation coefficient using the formula:

$$r=\frac{n\sum{XY}-(\sum{X}\sum{Y})}{\sqrt{[n\sum{X^2}-(\sum{X})^2][n\sum{Y^2}-(\sum{Y})^2]}}$$

where $n$ is the number of observations, $\sum{X}$ and $\sum{Y}$ are the sums of $X$ values and $Y$ values respectively, $\sum{XY}$ is the sum of the products of corresponding $X$ and $Y$ values, and $\sum{X^2}$ and $\sum{Y^2}$ are the sums of the squares of $X$ and $Y$ values respectively.

Substituting the given values, we get:

$$r=\frac{4(-12)-8(-1)-2(6)+2(-8)}{\sqrt{[4(37)-(-12)^2][4(9)-(-1)^2]}}=-\frac{4.31}{3.18}=-1.36$$

To test the significance of $r$, we need to calculate the t-statistic given by:

$$t=\frac{r}{\sqrt{\frac{1-r^2}{n-2}}}$$

Substituting the given values, we get:

$$t=-\frac{1.36}{\sqrt{\frac{1-(-1.36)^2}{4-2}}}= -3.99$$

Since the degrees of freedom are $(n-2)=2$, we look up the critical t-value from the t-distribution table with significance level $\ddagger = 0.05$ and degrees of freedom $df=2$ as $t_{crit}= 4.303$. Since $|t|<t_{crit}$, we do not reject the null hypothesis that the correlation coefficient is zero i.e., $\rho=0$.

Therefore, the answer is C) Do not reject $\rho=0$ because $-3.99>-4.303$.

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$\text { Test the significance of the correlation coefficient } \mathrm { r } \text { at } \ddagger = .05 \text { for the data }$