Question

Test the significance of the correlation coefficient at α = 0.05, using Table I.

Answer 1

The correlation coefficient is given by the following equation:

`r=\frac{N\sum^{}_{}xy-\sum^{}_{}x\sum^{}_{}y}{\sqrt[]{\lbrack N\sum^{}_{}x^2-(\sum^{}_{}x)^2\rbrack\cdot\lbrack N\sum^{}_{}y^2-(\sum^{}_{}y)^2\rbrack}}`

All sums on the previous equation are already stated in the table:

`\begin{gathered} \sum ^{}_{}x=32 \\ \sum ^{}_{}y=1105 \\ \sum ^{}_{}xy=3405 \\ \sum ^{}_{}x^2=220 \\ \sum ^{}_{}y^2=364525 \end{gathered}`

N is the number of samples. 6 in this case.

Now, replacing the values in the equation:

`r=\frac{6\cdot3405-32\cdot1105}{\sqrt[]{\lbrack6\cdot220-32^2\rbrack\cdot\lbrack6\cdot364525-1105^2\rbrack}}`

Solving the equation, we obtain a correlation coefficient of r = -0.88287.

Then, |r| = 0.88287.

To calculate the critical value for the correlation coefficient, we need to know the degrees of freedom (DF):

`\begin{gathered} DF=N-2=6-2 \\ DF=4 \end{gathered}`

For a level significance of α = 0.05, and DF = 4, the critical correlation value 0.811.

The critical value is 0.811.

Since the correlation factor (0.88287) is higher than the critical value (0.811), we can say that the correlation is significant. There is a significant linear relationship between x and y values.

Finally: |r| = 0.88287 and the Table I critical value is 0.811 so we can say that the correlation is significant. There is a significant linear relationship between x and y values.