# Solved (Free): For a sample of eight bears, researchers measured the distances around the bears’ chests and weighed the bears

#### ByDr. Raju Chaudhari

Mar 15, 2021

For a sample of eight bears, researchers measured the distances around the bears' chests and weighed the bears. Minitab was used to find that the value of the linear correlation coefficient is r=0.744. Using $\alpha=0.05$, determine if there is a linear correlation between chest size and weight. What proportion of the variation in weight can be explained by the linear relationship between weight and chest size?

### Solution

Given that $n = 8$, $r=0.744$, $\alpha =0.05$.

State the hypothesis testing problem

The hypothesis testing problem is $H_0: \rho = 0$ against $H_a: \rho \neq 0$.

Define the test statistic

The test statistic for testing above hypothesis is

 \begin{aligned} t&=\dfrac{r\sqrt{n-2}}{\sqrt{1-r^2}}\\ &=\frac{0.744\sqrt{8 -2}}{\sqrt{1-0.744^2}}\\ &=2.727 \end{aligned}

The test statistic $t$ follows Students' $t$ distribution with $n-2=8-2 =6$ degrees of freedom.

The level of significance is $\alpha = 0.05$.

Determine the critical values
For the specified value of $\alpha$ determine the critical region.

 \begin{aligned} P(t < t_{1-\alpha/2,n-2} \text{ or } t > t_{\alpha/2,n-2}) = \alpha. \end{aligned}

The critical values are $t_{\alpha/2,n-2}=-2.447$ and $t_{1-\alpha/2,n-2}=2.447$.

Decision

As the observed value of test statistic $t$ falls inside the critical region, we reject the null hypothesis.
We conclude that there is a linear correlation between chest size and weight.

$R^2= 0.5535$ That is $55.3536$ percent of the variation is weight can be explained by the linear relationship between weight and chestsize.