Step 1

(a)

The value of \(\displaystyle\sum{x}\) is,

\(\displaystyle\sum{x}={13.1}+{23.3}+{31.2}+{38.5}+{51.3}+{20.5}+{22.7}={200.6}\)

Thus, the value of \(\displaystyle\sum{x}\ {i}{s}\ {200.6}\)

The value of \(\displaystyle\sum{y}\) is,

\(\displaystyle\sum{y}={2.78}+{2.18}+{1.48}+{1.03}+{0.75}+{2.38}+{2.2}={12.8}\)

Thus, the value of \(\displaystyle\sum{y}\ {i}{s}\ {12.8}\)

The value of \(\displaystyle\sum{x}^{2}\) is,

\(\displaystyle\sum{x}^{2}={13.1}^{2}+{23.3}^{2}+{31.2}^{2}+{38.5}^{2}+{51.3}^{2}+{20.5}^{2}+{22.7}^{2}\)

\(\displaystyle={171.61}+{542.89}+{973.44}+{1482.25}+{2631.69}+{420.25}+{515.29}\)

\(\displaystyle={6737.42}\) Thus, the value of \(\displaystyle\sum{x}^{2}\ {i}{s}\ {6737.42}\)

Thus, the value of \(\displaystyle\sum{y}^{2}\) is,

\(\displaystyle\sum{y}^{2}={2.78}^{2}+{2.18}^{2}+{1.48}^{2}+{1.03}^{2}+{0.75}^{2}+{2.38}^{2}+{2.2}^{2}\)

\(\displaystyle={7.7284}+{4.7524}+{2.1904}+{1.0609}+{0.5625}+{5.6644}+{4.84}\)

\(\displaystyle={26.7990}\)

Thus, the value of \(\displaystyle\sum{y}^{2}\ {i}{s}\ {26.7990}.\)

The value of \(\displaystyle\sum{x}{y}\) is,

\(\displaystyle\sum{x}{y}={\left({13.1}\times{2.78}\right)}+{\left({23.3}\times{2.18}\right)}+{\left({31.2}\times{1.48}\right)}+{\left({38.5}\times{1.03}\right)}+{\left({51.3}\times{0.75}\right)}+{\left({20.5}\times{2.38}\right)}+{\left({22.7}\times{2.2}\right)}\)

\(\displaystyle={36.418}+{50.794}+{46.176}+{39.655}+{38.475}+{48.790}+{49.940}\)

\(\displaystyle={310.248}\)

Thus, the value of \(\displaystyle\sum{x}{y}\ {i}{s}\ {310.248}.\)

The value of r is,

\(\displaystyle{r}=\frac{{{n}\sum{x}{y}-{\left(\sum{x}\right)}{\left(\sum{y}\right)}}}{\sqrt{{{\left[{n}\sum{x}^{2}-{\left(\sum{x}\right)}^{2}\right]}{\left[\sum{y}^{2}-{\left(\sum{y}\right)}^{2}\right]}}}}\)

\(\displaystyle=\frac{{{7}{\left({310.248}\right)}-{\left({200.6}\times{12.8}\right)}}}{{\sqrt{{{\left[{\left({7}\times{6737.42}\right)}-{\left({200.6}\right)}^{2}\right]}{\left[{\left({7}\times{26.7990}\right)}-{\left({12.8}\right)}^{2}\right]}}}}}\)

\(\displaystyle=\frac{{-{395.944}}}{{405.4729}}\)

\(\displaystyle=-{0.976}\)

Hence, the correlation coefficient r is \(\displaystyle-{0.976}.\)

Step 2

(b)

The hypothesis is,

Null hypothesis:

\(\displaystyle{H}_{{0}}:\rho={0}\)

Alternative hypothesis:

\(\displaystyle{H}_{{1}}:\rho<{0}\)</span>

The test statistic is,

\(\displaystyle{t}=-{r}\sqrt{{\frac{{{n}-{2}}}{{{1}-{r}^{2}}}}}\)

\(\displaystyle=-{0.976}\sqrt{{\frac{{{7}-{2}}}{{{1}-{\left(-{0.976}\right)}^{2}}}}}\)

\(\displaystyle=\frac{{-{2.1824}}}{{0.2178}}\)

\(\displaystyle=-{10.02}\)

Thus, the test statistic is \(\displaystyle-{10.02}.\)

The degrees of freedom is,

\(\displaystyle{d}{f}={n}-{2}\)

\(\displaystyle={7}-{2}\)

\(\displaystyle={5}\)

The degrees of freedom is 5.

Decision rules for correlation coefficient:

If the test statistic is less than the negative critical value, then reject the null hypothesis.

Computation of critical value:

The critical value of t-distribution for 0.01 level of significance at 5 degrees of freedom can be obtained using the excel formula “=T.INV(0.01,5)”. The critical value is \(\displaystyle-{3.36}.\)

Thus, the critical value is \(\displaystyle-{3.36}.\)

Conclusion:

The test statistic is \(\displaystyle-{10.02}\) less than critical value \(\displaystyle-{3.36}.\)

Based on the decision rule, reject the null hypothesis.

Reject the null hypothesis. There is sufficient evidence that \(\displaystyle\rho<{0}.\)</span>

Step 3

c)

The value of \(\displaystyle\overline{{x}}\) is,

\(\displaystyle\overline{{x}}=\frac{{\sum{x}}}{{n}}\)

\(\displaystyle=\frac{200.6}{{7}}\)

\(\displaystyle={28.6571}\)

The value of \(\displaystyle\overline{{y}}\) is,

\(\displaystyle\overline{{y}}=\frac{{\sum{y}}}{{n}}\)

\(\displaystyle=\frac{12.8}{{7}}\)

\(\displaystyle={1.8286}\)

The value of b is,

\(\displaystyle{b}=\frac{{{n}\sum{x}{y}-{\left(\sum{x}\right)}{\left(\sum{y}\right)}}}{{{n}\sum{x}^{2}-{\left(\sum{x}\right)}^{2}}}\)

\(\displaystyle=\frac{{{7}{\left({310.248}\right)}-{\left({200.6}\times{12.8}\right)}}}{{{\left({7}\times{6737.42}\right)}-{\left({200.6}\right)}^{2}}}\)

\(\displaystyle=-\frac{395.944}{{6921.58}}\)

\(\displaystyle=-{0.0572}\)

Thus, the value of b is \(\displaystyle-{0.0572}.\)

The value of a is,

\(\displaystyle{a}=\overline{{y}}-{b}\overline{{x}}\)

\(\displaystyle={1.8286}-{\left(-{0.0572}\times{28.6571}\right)}\)

\(\displaystyle={1.8286}+{1.6392}\)

\(\displaystyle={3.4678}\)

Thus, the value of a is 3.4678.

The value of \(\displaystyle{S}_{{e}}\) is,

\(\displaystyle{S}_{{e}}=\sqrt{{\frac{{\sum{y}^{2}-{a}\sum{y}-{b}\sum{x}{y}}}{{{n}-{2}}}}}\)

\(\displaystyle=\sqrt{{\frac{{{26.7990}-{\left({3.4678}\times{12.8}\right)}-{\left(-{0.0572}\times{310.248}\right)}}}{{{7}-{2}}}}}\)

\(\displaystyle=\sqrt{{\frac{{{0.1573456}}}{{5}}}}\)

\(\displaystyle={0.1774}\)

Hence, the value of \(\displaystyle{S}_{{e}}\) is 0.1776.

(a)

The value of \(\displaystyle\sum{x}\) is,

\(\displaystyle\sum{x}={13.1}+{23.3}+{31.2}+{38.5}+{51.3}+{20.5}+{22.7}={200.6}\)

Thus, the value of \(\displaystyle\sum{x}\ {i}{s}\ {200.6}\)

The value of \(\displaystyle\sum{y}\) is,

\(\displaystyle\sum{y}={2.78}+{2.18}+{1.48}+{1.03}+{0.75}+{2.38}+{2.2}={12.8}\)

Thus, the value of \(\displaystyle\sum{y}\ {i}{s}\ {12.8}\)

The value of \(\displaystyle\sum{x}^{2}\) is,

\(\displaystyle\sum{x}^{2}={13.1}^{2}+{23.3}^{2}+{31.2}^{2}+{38.5}^{2}+{51.3}^{2}+{20.5}^{2}+{22.7}^{2}\)

\(\displaystyle={171.61}+{542.89}+{973.44}+{1482.25}+{2631.69}+{420.25}+{515.29}\)

\(\displaystyle={6737.42}\) Thus, the value of \(\displaystyle\sum{x}^{2}\ {i}{s}\ {6737.42}\)

Thus, the value of \(\displaystyle\sum{y}^{2}\) is,

\(\displaystyle\sum{y}^{2}={2.78}^{2}+{2.18}^{2}+{1.48}^{2}+{1.03}^{2}+{0.75}^{2}+{2.38}^{2}+{2.2}^{2}\)

\(\displaystyle={7.7284}+{4.7524}+{2.1904}+{1.0609}+{0.5625}+{5.6644}+{4.84}\)

\(\displaystyle={26.7990}\)

Thus, the value of \(\displaystyle\sum{y}^{2}\ {i}{s}\ {26.7990}.\)

The value of \(\displaystyle\sum{x}{y}\) is,

\(\displaystyle\sum{x}{y}={\left({13.1}\times{2.78}\right)}+{\left({23.3}\times{2.18}\right)}+{\left({31.2}\times{1.48}\right)}+{\left({38.5}\times{1.03}\right)}+{\left({51.3}\times{0.75}\right)}+{\left({20.5}\times{2.38}\right)}+{\left({22.7}\times{2.2}\right)}\)

\(\displaystyle={36.418}+{50.794}+{46.176}+{39.655}+{38.475}+{48.790}+{49.940}\)

\(\displaystyle={310.248}\)

Thus, the value of \(\displaystyle\sum{x}{y}\ {i}{s}\ {310.248}.\)

The value of r is,

\(\displaystyle{r}=\frac{{{n}\sum{x}{y}-{\left(\sum{x}\right)}{\left(\sum{y}\right)}}}{\sqrt{{{\left[{n}\sum{x}^{2}-{\left(\sum{x}\right)}^{2}\right]}{\left[\sum{y}^{2}-{\left(\sum{y}\right)}^{2}\right]}}}}\)

\(\displaystyle=\frac{{{7}{\left({310.248}\right)}-{\left({200.6}\times{12.8}\right)}}}{{\sqrt{{{\left[{\left({7}\times{6737.42}\right)}-{\left({200.6}\right)}^{2}\right]}{\left[{\left({7}\times{26.7990}\right)}-{\left({12.8}\right)}^{2}\right]}}}}}\)

\(\displaystyle=\frac{{-{395.944}}}{{405.4729}}\)

\(\displaystyle=-{0.976}\)

Hence, the correlation coefficient r is \(\displaystyle-{0.976}.\)

Step 2

(b)

The hypothesis is,

Null hypothesis:

\(\displaystyle{H}_{{0}}:\rho={0}\)

Alternative hypothesis:

\(\displaystyle{H}_{{1}}:\rho<{0}\)</span>

The test statistic is,

\(\displaystyle{t}=-{r}\sqrt{{\frac{{{n}-{2}}}{{{1}-{r}^{2}}}}}\)

\(\displaystyle=-{0.976}\sqrt{{\frac{{{7}-{2}}}{{{1}-{\left(-{0.976}\right)}^{2}}}}}\)

\(\displaystyle=\frac{{-{2.1824}}}{{0.2178}}\)

\(\displaystyle=-{10.02}\)

Thus, the test statistic is \(\displaystyle-{10.02}.\)

The degrees of freedom is,

\(\displaystyle{d}{f}={n}-{2}\)

\(\displaystyle={7}-{2}\)

\(\displaystyle={5}\)

The degrees of freedom is 5.

Decision rules for correlation coefficient:

If the test statistic is less than the negative critical value, then reject the null hypothesis.

Computation of critical value:

The critical value of t-distribution for 0.01 level of significance at 5 degrees of freedom can be obtained using the excel formula “=T.INV(0.01,5)”. The critical value is \(\displaystyle-{3.36}.\)

Thus, the critical value is \(\displaystyle-{3.36}.\)

Conclusion:

The test statistic is \(\displaystyle-{10.02}\) less than critical value \(\displaystyle-{3.36}.\)

Based on the decision rule, reject the null hypothesis.

Reject the null hypothesis. There is sufficient evidence that \(\displaystyle\rho<{0}.\)</span>

Step 3

c)

The value of \(\displaystyle\overline{{x}}\) is,

\(\displaystyle\overline{{x}}=\frac{{\sum{x}}}{{n}}\)

\(\displaystyle=\frac{200.6}{{7}}\)

\(\displaystyle={28.6571}\)

The value of \(\displaystyle\overline{{y}}\) is,

\(\displaystyle\overline{{y}}=\frac{{\sum{y}}}{{n}}\)

\(\displaystyle=\frac{12.8}{{7}}\)

\(\displaystyle={1.8286}\)

The value of b is,

\(\displaystyle{b}=\frac{{{n}\sum{x}{y}-{\left(\sum{x}\right)}{\left(\sum{y}\right)}}}{{{n}\sum{x}^{2}-{\left(\sum{x}\right)}^{2}}}\)

\(\displaystyle=\frac{{{7}{\left({310.248}\right)}-{\left({200.6}\times{12.8}\right)}}}{{{\left({7}\times{6737.42}\right)}-{\left({200.6}\right)}^{2}}}\)

\(\displaystyle=-\frac{395.944}{{6921.58}}\)

\(\displaystyle=-{0.0572}\)

Thus, the value of b is \(\displaystyle-{0.0572}.\)

The value of a is,

\(\displaystyle{a}=\overline{{y}}-{b}\overline{{x}}\)

\(\displaystyle={1.8286}-{\left(-{0.0572}\times{28.6571}\right)}\)

\(\displaystyle={1.8286}+{1.6392}\)

\(\displaystyle={3.4678}\)

Thus, the value of a is 3.4678.

The value of \(\displaystyle{S}_{{e}}\) is,

\(\displaystyle{S}_{{e}}=\sqrt{{\frac{{\sum{y}^{2}-{a}\sum{y}-{b}\sum{x}{y}}}{{{n}-{2}}}}}\)

\(\displaystyle=\sqrt{{\frac{{{26.7990}-{\left({3.4678}\times{12.8}\right)}-{\left(-{0.0572}\times{310.248}\right)}}}{{{7}-{2}}}}}\)

\(\displaystyle=\sqrt{{\frac{{{0.1573456}}}{{5}}}}\)

\(\displaystyle={0.1774}\)

Hence, the value of \(\displaystyle{S}_{{e}}\) is 0.1776.