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The scatterplot generated on Minitab is given.

a.

A scatterplot is a two-dimensional graph thatrepresents a pair of values for two variables.

Here,the scatterplot represents an overall upward pattern, which means the values of one variable are expected to increase with the other.

Due to this pattern and moderately close observations, it can be expected that there exists a linear correlation between the two variables.

b.

The observations obtained from the scatterplot are as follows:

The formula for the correlation coefficient is shown below:

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

The valuesare in the table below:

Substitute the values to obtain the correlation coefficient.

\(\begin{aligned} r &= \frac{{10\left( {136} \right) - \left( {28} \right)\left( {28} \right)}}{{\sqrt {10\left( {142} \right) - {{\left( {28} \right)}^2}} \sqrt {10{{\left( {142} \right)}^2} - {{\left( {28} \right)}^2}} }}\\ &= 0.906\end{aligned}\)

Thus, the correlation coefficient is 0.906.

Let\(\rho \)be the true correlation coefficient.

Form the hypotheses as shown:

\(\begin{array}{l}{{\rm{H}}_{\rm{o}}}:\rho = 0\\{{\rm{{\rm H}}}_{\rm{a}}}:\rho \ne 0\end{array}\)

The samplesize is10(n).

The test statistic is computed as follows:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{0.906}}{{\sqrt {\frac{{1 - {{0.906}^2}}}{{10 - 2}}} }}\\ &= 6.054\end{aligned}\)

Thus, the test statistic is 6.054.

The degree of freedom is computedbelow:

\(\begin{aligned} df &= n - 2\\ &= 10 - 2\\ &= 8\end{aligned}\)

The p-value is computedfrom the t-distribution table.

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T > t} \right)\\ &= 2P\left( {T > 6.054} \right)\\ &= 2\left( {1 - P\left( {T < 6.054} \right)} \right)\\ &= 0.0003\end{aligned}\)

As thep-value is lesser than 0.05, the null hypothesis is rejected.

Therefore, there is sufficient evidence to prove theexistence of a linear correlation between thetwo variables.

c.

The data without coordinate (10,10) is

The scatterplot hence formed is shown below:

Thus, there appears to be an association between the two variables.

The values are in the table below:

Substitute the values to obtain the correlation coefficient:

\(\begin{aligned} r &= \frac{{9\left( {36} \right) - \left( {18} \right)\left( {18} \right)}}{{\sqrt {9\left( {42} \right) - {{\left( {18} \right)}^2}} \sqrt {9{{\left( {42} \right)}^2} - {{\left( {18} \right)}^2}} }}\\ &= 0\end{aligned}\)

Thus, the correlation coefficient is 0.

Let\(\rho \)denote the actual correlation coefficient.

The hypotheses areformulatedas shown below

\(\begin{array}{l}{{\rm{H}}_{\rm{o}}}:\rho = 0\\{{\rm{{\rm H}}}_{\rm{a}}}:\rho \ne 0\end{array}\)

The samplesize is9(n).

The test statistic is computed as follows:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{0}{{\sqrt {\frac{{1 - {0^2}}}{{9 - 2}}} }}\\ &= 0\end{aligned}\)

Thus, the test statistic is 0.

The degree of freedom is computedbelow:

\(\begin{aligned} df &= n - 2\\ &= 9 - 2\\ &= 7\end{aligned}\)

The p-value is computed from the t-distribution table.

\(\begin{aligned} p - value &= 2P\left( {t > 0} \right)\\ &= 2\left( {1 - P\left( {t < 0} \right)} \right)\\ &= 1\end{aligned}\)

As the p-value is greater than 0.05, the null hypothesis fails to be rejected.

Therefore, there is not sufficient evidence to support the claim that the variables have a linear correlation between them.

The result changes to a large extent as one single paired observation is removed from the data. The correlation measure changes from 0.906 to 0 as one pair is removed from the data.