Answer :
We are given a sample of 17 paired observations, so the degrees of freedom for the correlation test is
[tex]$$
df = n - 2 = 17 - 2 = 15.
$$[/tex]
Because we are testing a two‐tailed hypothesis at a significance level of [tex]$\alpha = 0.05$[/tex], the critical value of the [tex]$t$[/tex] statistic comes from
[tex]$$
t_{\alpha/2, 15} = t_{0.975,15}.
$$[/tex]
A standard [tex]$t$[/tex]–table (or calculator) shows that
[tex]$$
t_{0.975,15} \approx 2.131.
$$[/tex]
There is a relationship between the [tex]$t$[/tex] statistic and the sample Pearson correlation coefficient [tex]$r$[/tex] given by
[tex]$$
t = r \sqrt{\frac{df}{1-r^2}}.
$$[/tex]
Hence, the critical correlation coefficient (in absolute value) is the [tex]$r$[/tex] value that would give [tex]$t_{critical}$[/tex] when the equality holds. Solving for [tex]$r$[/tex], we have
[tex]$$
r_c = \sqrt{\frac{t_{critical}^2}{t_{critical}^2 + df}}.
$$[/tex]
Plug in the numbers:
[tex]$$
r_c = \sqrt{\frac{(2.131)^2}{(2.131)^2 + 15}}.
$$[/tex]
First, calculate [tex]$(2.131)^2$[/tex]:
[tex]$$
(2.131)^2 \approx 4.543.
$$[/tex]
Then add the degrees of freedom:
[tex]$$
4.543 + 15 = 19.543.
$$[/tex]
Thus,
[tex]$$
r_c = \sqrt{\frac{4.543}{19.543}} \approx \sqrt{0.2325} \approx 0.482.
$$[/tex]
This is the critical correlation value. In other words, if the absolute value of the sample correlation coefficient is greater than approximately 0.482, it is significant at the [tex]$\alpha=0.05$[/tex] level.
Next, we compute the Pearson correlation coefficient, [tex]$r$[/tex], from the given data:
[tex]\[
\begin{array}{|r|r|}
\hline
x & y \\ \hline
26.8 & 64.2 \\
32.8 & 59.1 \\
45.6 & 52.6 \\
37.7 & 58.8 \\
37.6 & 54.9 \\
27.4 & 61.9 \\
31.7 & 62.3 \\
30.0 & 59.9 \\
29.8 & 58.0 \\
30.9 & 64.8 \\
10.0 & 82.6 \\
35.3 & 51.7 \\
42.8 & 53.0 \\
18.0 & 69.9 \\
27.6 & 58.3 \\
57.4 & 35.3 \\
48.2 & 53.1 \\
\hline
\end{array}
\][/tex]
Going through the calculation (or using an appropriate calculator/software), one finds that
[tex]$$
r \approx -0.927.
$$[/tex]
This means that there is a very strong negative linear relationship between [tex]$x$[/tex] and [tex]$y$[/tex]. Notice that
[tex]$$
|r| = 0.927 > 0.482 = r_c.
$$[/tex]
Since the absolute value of the sample correlation coefficient exceeds the critical value, we conclude that there is a statistically significant correlation between the two variables at the [tex]$\alpha=0.05$[/tex] level.
To summarize:
1. The critical value for the correlation coefficient is
[tex]$$
r_c \approx 0.482.
$$[/tex]
2. The sample correlation coefficient is
[tex]$$
r \approx -0.927.
$$[/tex]
3. Final conclusion: There is sufficient sample evidence to support the claim that there is a statistically significant correlation between the two variables.
[tex]$$
df = n - 2 = 17 - 2 = 15.
$$[/tex]
Because we are testing a two‐tailed hypothesis at a significance level of [tex]$\alpha = 0.05$[/tex], the critical value of the [tex]$t$[/tex] statistic comes from
[tex]$$
t_{\alpha/2, 15} = t_{0.975,15}.
$$[/tex]
A standard [tex]$t$[/tex]–table (or calculator) shows that
[tex]$$
t_{0.975,15} \approx 2.131.
$$[/tex]
There is a relationship between the [tex]$t$[/tex] statistic and the sample Pearson correlation coefficient [tex]$r$[/tex] given by
[tex]$$
t = r \sqrt{\frac{df}{1-r^2}}.
$$[/tex]
Hence, the critical correlation coefficient (in absolute value) is the [tex]$r$[/tex] value that would give [tex]$t_{critical}$[/tex] when the equality holds. Solving for [tex]$r$[/tex], we have
[tex]$$
r_c = \sqrt{\frac{t_{critical}^2}{t_{critical}^2 + df}}.
$$[/tex]
Plug in the numbers:
[tex]$$
r_c = \sqrt{\frac{(2.131)^2}{(2.131)^2 + 15}}.
$$[/tex]
First, calculate [tex]$(2.131)^2$[/tex]:
[tex]$$
(2.131)^2 \approx 4.543.
$$[/tex]
Then add the degrees of freedom:
[tex]$$
4.543 + 15 = 19.543.
$$[/tex]
Thus,
[tex]$$
r_c = \sqrt{\frac{4.543}{19.543}} \approx \sqrt{0.2325} \approx 0.482.
$$[/tex]
This is the critical correlation value. In other words, if the absolute value of the sample correlation coefficient is greater than approximately 0.482, it is significant at the [tex]$\alpha=0.05$[/tex] level.
Next, we compute the Pearson correlation coefficient, [tex]$r$[/tex], from the given data:
[tex]\[
\begin{array}{|r|r|}
\hline
x & y \\ \hline
26.8 & 64.2 \\
32.8 & 59.1 \\
45.6 & 52.6 \\
37.7 & 58.8 \\
37.6 & 54.9 \\
27.4 & 61.9 \\
31.7 & 62.3 \\
30.0 & 59.9 \\
29.8 & 58.0 \\
30.9 & 64.8 \\
10.0 & 82.6 \\
35.3 & 51.7 \\
42.8 & 53.0 \\
18.0 & 69.9 \\
27.6 & 58.3 \\
57.4 & 35.3 \\
48.2 & 53.1 \\
\hline
\end{array}
\][/tex]
Going through the calculation (or using an appropriate calculator/software), one finds that
[tex]$$
r \approx -0.927.
$$[/tex]
This means that there is a very strong negative linear relationship between [tex]$x$[/tex] and [tex]$y$[/tex]. Notice that
[tex]$$
|r| = 0.927 > 0.482 = r_c.
$$[/tex]
Since the absolute value of the sample correlation coefficient exceeds the critical value, we conclude that there is a statistically significant correlation between the two variables at the [tex]$\alpha=0.05$[/tex] level.
To summarize:
1. The critical value for the correlation coefficient is
[tex]$$
r_c \approx 0.482.
$$[/tex]
2. The sample correlation coefficient is
[tex]$$
r \approx -0.927.
$$[/tex]
3. Final conclusion: There is sufficient sample evidence to support the claim that there is a statistically significant correlation between the two variables.
