Cigarette Nicotine and Carbon Monoxide Refer to the table of data given in Exercise 1 and use the amounts of nicotine and carbon monoxide (CO).

a. Construct a scatterplot using nicotine for the xscale, or horizontal axis. What does the scatterplot suggest about a linear correlation between amounts of nicotine and carbon monoxide?

b. Find the value of the linear correlation coefficient and determine whether there is sufficient evidence to support a claim of a linear correlation between amounts of nicotine and carbon monoxide.

c. Letting yrepresent the amount of carbon monoxide and letting xrepresent the amount of nicotine, find the regression equation.

d. The Raleigh brand king size cigarette is not included in the table, and it has 1.3 mg of nicotine. What is the best predicted amount of carbon monoxide?

Tar	25	27	20	24	20	20	21	24
CO	18	16	16	16	16	16	14	17
Nicotine	1.5	1.7	1.1	1.6	1.1	1.0	1.2	1.4

The table representing the amount of Cigarette Nicotine and Carbon monoxide is provided.

a.

A plot that shows observations from two variables scaling them on two axes is referred to as a scatterplot.

Steps to sketch a scatterplot:

1. Mark horizontal axis for Nicotine (mg) and vertical axis for Carbon Monoxide (mg).
2. Mark points for each paired value with respect to both axes.
3. The resultant graph is the scatterplot.

It can be observed from the scatter plot that both nicotine and CO increase from left to right, but there is a weak pattern and the points are not very close to a straight-line pattern.

b.

Let x represents the Nicotine (mg).

Let y represent the Carbon monoxide (mg).

The formula for computing the correlation coefficient (r) between the values of Nicotine (mg)and Carbon monoxide (mg)is as follows:

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

The following calculations are done to compute the value of r:

Substituting the above values, the value of r is obtained as,

\(\begin{aligned}{c}r &= \frac{{n\sum {xy} - \sum x \sum y }}{{\sqrt {n\sum {{x^2}} - {{\left( {\sum x } \right)}^2}} \sqrt {n\sum {{y^2}} - {{\left( {\sum y } \right)}^2}} }}\\ &= \frac{{8\left( {171.6} \right) - \left( {129} \right)\left( {10.6} \right)}}{{\sqrt {8\left( {2089} \right) - {{\left( {129} \right)}^2}} \sqrt {8\left( {14.52} \right) - {{\left( {10.6} \right)}^2}} }}\\ &= 0.329\end{aligned}\)

Therefore, the value of r is equal to 0.329.

Here, n=8.

If the value of the correlation coefficient lies between the critical values, then the correlation between the two variables is considered significant else, it is considered insignificant.

The critical values of r for n=8 and\(\alpha = 0.05\)are -0.707 and 0.707.

The corresponding p-value of r is equal to 0.427.

Since the computed value of r equal to 0.329 is less than the larger critical value of 0.707, it can be said that the correlation between the two variables is insignificant.

Moreover, the p-value is greater than 0.05. This also implies that correlation is insignificant.

Therefore, there is not sufficient evidence to claim that there is a linear correlation between the amount of Nicotine (mg) and the amount of Carbon monoxide (mg).

c.

Let x represents the amount of nicotine.

Let y represent the amount of carbon monoxide.

The regression equationis given as,

\(\hat y = {b_0} + {b_1}x\)

The mean value of xis given as,

\(\begin{aligned} \bar x &= \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\\ &= \frac{{1.5 + 1.7 + .... + 1.4}}{8}\\ &= 1.325\end{aligned}\)

Therefore, the mean value of x is 1.325.

The mean value of yis given as,

\(\begin{aligned} \bar y &= \frac{{\sum\limits_{i = 1}^n {{y_i}} }}{n}\\ &= \frac{{18 + 16 + 16 + .... + 17}}{8}\\ &= 16.125\end{aligned}\)

Therefore, the mean value of y is 16.125.

The standard deviation of xis given as,

\(\begin{aligned} {s_x} &= \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ &= \sqrt {\frac{{{{\left( {1.5 - 1.325} \right)}^2} + {{\left( {1.7 - 1.325} \right)}^2} + ..... + {{\left( {1.4 - 1.325} \right)}^2}}}{{8 - 1}}} \\ &= 0.260\end{aligned}\)

Therefore, the standard deviation of x is 0.260.

The standard deviation of yis given as,

\(\begin{aligned} {s_y} &= \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({y_i} - \bar y)}^2}} }}{{n - 1}}} \\ &= \sqrt {\frac{{{{\left( {18 - 16.125} \right)}^2} + {{\left( {16 - 16.125} \right)}^2} + ..... + {{\left( {17 - 16.125} \right)}^2}}}{{8 - 1}}} \\ &= 1.126\end{aligned}\)

Therefore, the standard deviation of y is 1.126.

Theslopeof the regression line is given as,

\(\begin{aligned} {b_1} &= r\frac{{{s_Y}}}{{{s_X}}}\\ &= 0.329 \times \frac{{1.126}}{{0.260}}\\ &= 1.42\end{aligned}\)

Therefore, the value of the slope is 1.42.

Theinterceptis computed as,

\(\begin{aligned} {b_0} &= \bar y - {b_1}\bar x\\ &= 16.125 - \left( {1.42 \times 1.325} \right)\\ &= 14.2\end{aligned}\)

Therefore, the value of the intercept is 14.2.

Substituting the above values, the regression equation is given as,

Theregression equationis given as,

\(\begin{aligned} \hat y &= {b_0} + {b_1}x\\ &= 14.2 + 1.42x\end{aligned}\)

Thus, the regression equation is \(\hat y = 14.2 + 1.42x\).

d.

The amount of nicotine is 1.3 mg.

The predicted amount of carbon monoxide is computed as,

\(\begin{aligned} \hat y &= 14.2 + 1.42x\\ &= 14.2 + \left( {1.42 \times 1.3} \right)\\ &= 16.046\\ &\approx 16\end{aligned}\)

Thus, the predicted amount of carbon monoxide is 16 mg.

Therefore, the predicted amount of carbon monoxide (16 mg) is close to the actual amount of carbon monoxide (15 mg).