Answer :
We want to find the Pearson correlation coefficient
[tex]$$
r = \frac{\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i=1}^n (x_i-\bar{x})^2 \, \sum_{i=1}^n (y_i-\bar{y})^2}}
$$[/tex]
and then use the transformation
[tex]$$
t = r\sqrt{\frac{n-2}{1-r^2}},
$$[/tex]
which follows a [tex]$t$[/tex]-distribution with [tex]$n-2$[/tex] degrees of freedom. Finally, we obtain the two-tailed [tex]$p$[/tex]-value as
[tex]$$
p\text{-value} = 2\cdot P\big(T> |t|\big).
$$[/tex]
Below is a step‐by‐step outline of the solution.
──────────────────────────────
Step 1. Compute the Means
There are [tex]$n = 19$[/tex] data pairs. The given values are:
[tex]$$
\begin{array}{c|ccccccccccccccccccc}
x: & 51.7,&45.7,&56.3,&51.4,&49.2,&51.6,&47.1,&53.8,&44.1,&49.7,\\
&49.4,&50.1,&51.2,&44.0,&44.5,&43.5,&50.6,&41.0,&52.7\\[1mm]
y: & 78.2,&68.3,&37.0,&65.6,&42.5,&63.2,&59.1,&67.7,&56.5,&63.1,\\
&61.7,&67.3,&66.2,&62.4,&57.3,&74.0,&45.6,&69.1,&66.6\\
\end{array}
$$[/tex]
Let
[tex]$$
\bar{x} = \frac{\sum_{i=1}^{19} x_i}{19}, \qquad \bar{y} = \frac{\sum_{i=1}^{19} y_i}{19}.
$$[/tex]
After summing the given values we find approximately
[tex]$$
\bar{x} \approx 48.8 \quad \text{and} \quad \bar{y} \approx 61.7.
$$[/tex]
──────────────────────────────
Step 2. Compute the Sums of Squares
For each pair [tex]$(x_i, y_i)$[/tex] compute:
[tex]\[
\begin{array}{c|c|c|c|c}
i & x_i & y_i & x_i - \bar{x} & y_i - \bar{y} \\
\hline
1 & 51.7 & 78.2 & 2.9 & 16.5 \\
2 & 45.7 & 68.3 & -3.1 & 6.6 \\
3 & 56.3 & 37.0 & 7.5 & -24.7 \\
4 & 51.4 & 65.6 & 2.6 & 3.9 \\
5 & 49.2 & 42.5 & 0.4 & -19.2 \\
6 & 51.6 & 63.2 & 2.8 & 1.5 \\
7 & 47.1 & 59.1 & -1.7 & -2.6 \\
8 & 53.8 & 67.7 & 5.0 & 6.0 \\
9 & 44.1 & 56.5 & -4.7 & -5.2 \\
10 & 49.7 & 63.1 & 0.9 & 1.4 \\
11 & 49.4 & 61.7 & 0.6 & 0.0 \\
12 & 50.1 & 67.3 & 1.3 & 5.6 \\
13 & 51.2 & 66.2 & 2.4 & 4.5 \\
14 & 44.0 & 62.4 & -4.8 & 0.7 \\
15 & 44.5 & 57.3 & -4.3 & -4.4 \\
16 & 43.5 & 74.0 & -5.3 & 12.3 \\
17 & 50.6 & 45.6 & 1.8 & -16.1 \\
18 & 41.0 & 69.1 & -7.8 & 7.4 \\
19 & 52.7 & 66.6 & 3.9 & 4.9 \\
\end{array}
\][/tex]
Then compute the following quantities:
- The product [tex]$(x_i-\bar{x})(y_i-\bar{y})$[/tex] and sum over all [tex]$i$[/tex]: denote this sum by [tex]$S_{xy}$[/tex].
- The squared differences [tex]$(x_i-\bar{x})^2$[/tex] summed gives [tex]$S_{xx}$[/tex].
- The squared differences [tex]$(y_i-\bar{y})^2$[/tex] summed gives [tex]$S_{yy}$[/tex].
A careful calculation gives approximately:
[tex]$$
S_{xy} \approx -190.22,\quad S_{xx} \approx 296.54,\quad S_{yy} \approx 1944.49.
$$[/tex]
──────────────────────────────
Step 3. Compute the Correlation Coefficient
The correlation coefficient is
[tex]$$
r = \frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}.
$$[/tex]
Substitute the values:
[tex]$$
r \approx \frac{-190.22}{\sqrt{296.54\times1944.49}}.
$$[/tex]
First, compute the denominator:
[tex]$$
\sqrt{296.54\times1944.49} \approx \sqrt{576619} \approx 759.41.
$$[/tex]
Thus,
[tex]$$
r \approx \frac{-190.22}{759.41} \approx -0.250.
$$[/tex]
──────────────────────────────
Step 4. Compute the [tex]$t$[/tex]-Statistic
Using the formula
[tex]$$
t = r\sqrt{\frac{n-2}{1-r^2}},
$$[/tex]
with [tex]$n = 19$[/tex] so that [tex]$n-2 = 17$[/tex], first compute [tex]$r^2$[/tex]:
[tex]$$
r^2 \approx (-0.250)^2 = 0.0625.
$$[/tex]
Then,
[tex]$$
1 - r^2 \approx 1 - 0.0625 = 0.9375.
$$[/tex]
The ratio becomes:
[tex]$$
\sqrt{\frac{17}{0.9375}} \approx \sqrt{18.1333} \approx 4.260.
$$[/tex]
Hence, the [tex]$t$[/tex]-statistic is
[tex]$$
t \approx -0.250 \times 4.260 \approx -1.065.
$$[/tex]
──────────────────────────────
Step 5. Find the Two‐Tailed [tex]$p$[/tex]-Value
With [tex]$t \approx -1.065$[/tex] and degrees of freedom [tex]$df = 17$[/tex], we use the [tex]$t$[/tex]-distribution to determine the probability of obtaining a value as extreme as [tex]$|t|$[/tex]. The two-tailed [tex]$p$[/tex]-value is given by:
[tex]$$
p\text{-value} = 2\cdot P(T > |t|).
$$[/tex]
For [tex]$|t| \approx 1.065$[/tex] with [tex]$df = 17$[/tex], the one-tailed probability is roughly [tex]$0.15$[/tex]. Therefore,
[tex]$$
p\text{-value} \approx 2 \times 0.15 = 0.3000.
$$[/tex]
──────────────────────────────
Final Answers
The correlation coefficient is:
[tex]$$
r \approx -0.250,
$$[/tex]
and the two-tailed [tex]$p$[/tex]-value is:
[tex]$$
p\text{-value} \approx 0.3000.
$$[/tex]
Since the significance level is [tex]$\alpha = 0.01$[/tex], the [tex]$p$[/tex]-value is much larger than [tex]$\alpha$[/tex], so there is not enough evidence to conclude there is a statistically significant negative linear correlation between these two variables.
[tex]$$
r = \frac{\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i=1}^n (x_i-\bar{x})^2 \, \sum_{i=1}^n (y_i-\bar{y})^2}}
$$[/tex]
and then use the transformation
[tex]$$
t = r\sqrt{\frac{n-2}{1-r^2}},
$$[/tex]
which follows a [tex]$t$[/tex]-distribution with [tex]$n-2$[/tex] degrees of freedom. Finally, we obtain the two-tailed [tex]$p$[/tex]-value as
[tex]$$
p\text{-value} = 2\cdot P\big(T> |t|\big).
$$[/tex]
Below is a step‐by‐step outline of the solution.
──────────────────────────────
Step 1. Compute the Means
There are [tex]$n = 19$[/tex] data pairs. The given values are:
[tex]$$
\begin{array}{c|ccccccccccccccccccc}
x: & 51.7,&45.7,&56.3,&51.4,&49.2,&51.6,&47.1,&53.8,&44.1,&49.7,\\
&49.4,&50.1,&51.2,&44.0,&44.5,&43.5,&50.6,&41.0,&52.7\\[1mm]
y: & 78.2,&68.3,&37.0,&65.6,&42.5,&63.2,&59.1,&67.7,&56.5,&63.1,\\
&61.7,&67.3,&66.2,&62.4,&57.3,&74.0,&45.6,&69.1,&66.6\\
\end{array}
$$[/tex]
Let
[tex]$$
\bar{x} = \frac{\sum_{i=1}^{19} x_i}{19}, \qquad \bar{y} = \frac{\sum_{i=1}^{19} y_i}{19}.
$$[/tex]
After summing the given values we find approximately
[tex]$$
\bar{x} \approx 48.8 \quad \text{and} \quad \bar{y} \approx 61.7.
$$[/tex]
──────────────────────────────
Step 2. Compute the Sums of Squares
For each pair [tex]$(x_i, y_i)$[/tex] compute:
[tex]\[
\begin{array}{c|c|c|c|c}
i & x_i & y_i & x_i - \bar{x} & y_i - \bar{y} \\
\hline
1 & 51.7 & 78.2 & 2.9 & 16.5 \\
2 & 45.7 & 68.3 & -3.1 & 6.6 \\
3 & 56.3 & 37.0 & 7.5 & -24.7 \\
4 & 51.4 & 65.6 & 2.6 & 3.9 \\
5 & 49.2 & 42.5 & 0.4 & -19.2 \\
6 & 51.6 & 63.2 & 2.8 & 1.5 \\
7 & 47.1 & 59.1 & -1.7 & -2.6 \\
8 & 53.8 & 67.7 & 5.0 & 6.0 \\
9 & 44.1 & 56.5 & -4.7 & -5.2 \\
10 & 49.7 & 63.1 & 0.9 & 1.4 \\
11 & 49.4 & 61.7 & 0.6 & 0.0 \\
12 & 50.1 & 67.3 & 1.3 & 5.6 \\
13 & 51.2 & 66.2 & 2.4 & 4.5 \\
14 & 44.0 & 62.4 & -4.8 & 0.7 \\
15 & 44.5 & 57.3 & -4.3 & -4.4 \\
16 & 43.5 & 74.0 & -5.3 & 12.3 \\
17 & 50.6 & 45.6 & 1.8 & -16.1 \\
18 & 41.0 & 69.1 & -7.8 & 7.4 \\
19 & 52.7 & 66.6 & 3.9 & 4.9 \\
\end{array}
\][/tex]
Then compute the following quantities:
- The product [tex]$(x_i-\bar{x})(y_i-\bar{y})$[/tex] and sum over all [tex]$i$[/tex]: denote this sum by [tex]$S_{xy}$[/tex].
- The squared differences [tex]$(x_i-\bar{x})^2$[/tex] summed gives [tex]$S_{xx}$[/tex].
- The squared differences [tex]$(y_i-\bar{y})^2$[/tex] summed gives [tex]$S_{yy}$[/tex].
A careful calculation gives approximately:
[tex]$$
S_{xy} \approx -190.22,\quad S_{xx} \approx 296.54,\quad S_{yy} \approx 1944.49.
$$[/tex]
──────────────────────────────
Step 3. Compute the Correlation Coefficient
The correlation coefficient is
[tex]$$
r = \frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}.
$$[/tex]
Substitute the values:
[tex]$$
r \approx \frac{-190.22}{\sqrt{296.54\times1944.49}}.
$$[/tex]
First, compute the denominator:
[tex]$$
\sqrt{296.54\times1944.49} \approx \sqrt{576619} \approx 759.41.
$$[/tex]
Thus,
[tex]$$
r \approx \frac{-190.22}{759.41} \approx -0.250.
$$[/tex]
──────────────────────────────
Step 4. Compute the [tex]$t$[/tex]-Statistic
Using the formula
[tex]$$
t = r\sqrt{\frac{n-2}{1-r^2}},
$$[/tex]
with [tex]$n = 19$[/tex] so that [tex]$n-2 = 17$[/tex], first compute [tex]$r^2$[/tex]:
[tex]$$
r^2 \approx (-0.250)^2 = 0.0625.
$$[/tex]
Then,
[tex]$$
1 - r^2 \approx 1 - 0.0625 = 0.9375.
$$[/tex]
The ratio becomes:
[tex]$$
\sqrt{\frac{17}{0.9375}} \approx \sqrt{18.1333} \approx 4.260.
$$[/tex]
Hence, the [tex]$t$[/tex]-statistic is
[tex]$$
t \approx -0.250 \times 4.260 \approx -1.065.
$$[/tex]
──────────────────────────────
Step 5. Find the Two‐Tailed [tex]$p$[/tex]-Value
With [tex]$t \approx -1.065$[/tex] and degrees of freedom [tex]$df = 17$[/tex], we use the [tex]$t$[/tex]-distribution to determine the probability of obtaining a value as extreme as [tex]$|t|$[/tex]. The two-tailed [tex]$p$[/tex]-value is given by:
[tex]$$
p\text{-value} = 2\cdot P(T > |t|).
$$[/tex]
For [tex]$|t| \approx 1.065$[/tex] with [tex]$df = 17$[/tex], the one-tailed probability is roughly [tex]$0.15$[/tex]. Therefore,
[tex]$$
p\text{-value} \approx 2 \times 0.15 = 0.3000.
$$[/tex]
──────────────────────────────
Final Answers
The correlation coefficient is:
[tex]$$
r \approx -0.250,
$$[/tex]
and the two-tailed [tex]$p$[/tex]-value is:
[tex]$$
p\text{-value} \approx 0.3000.
$$[/tex]
Since the significance level is [tex]$\alpha = 0.01$[/tex], the [tex]$p$[/tex]-value is much larger than [tex]$\alpha$[/tex], so there is not enough evidence to conclude there is a statistically significant negative linear correlation between these two variables.
