Question

Cigarette Tar and Nicotine The table below lists measured amounts (mg) of tar, carbonmonoxide (CO), and nicotine in king size cigarettes of different brands (from Data Set 13“Cigarette Contents” in Appendix B).

a. Is there is sufficient evidence to support a claim of a linear correlation between tar and nicotine?

b. What percentage of the variation in nicotine can be explained by the linear correlation between nicotine and tar?

c. Letting yrepresent the amount of nicotine and letting xrepresent the amount of tar, identify the regression equation.

d. The Raleigh brand king size cigarette is not included in the table, and it has 23 mg of tar. What is the best predicted amount of nicotine? How does the predicted amount compare to the actual amount of 1.3 mg of nicotine?

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

Short answer

a. The value of r is 0.962. There is sufficient evidence to claim that there is a linear correlation between the amount of Tar and Nicotine (mg).

b.The percentage of the variation in nicotine can be explained by the linear correlation between nicotine and tar is 92.5%.

c. The regression equation is\(\hat y = - 0.757 + 0.0920x\).

d. The predicted amount of nicotine is 1.4 mg. Also, the predicted amount of nicotine (1.4 mg) is very close to the actual amount of nicotine (1.3 mg).

Step-by-step solution

Given information

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

Calculate the correlation coefficient

a.

Let x represents the amount of Tar (mg).

Let y represent the amount of Nicotine (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:

x	y	xy	\({x^2}\)	\({y^2}\)
25	1.5	37.5	625	2.25
27	1.7	45.9	729	2.89
20	1.1	22	400	1.21
24	1.6	38.4	576	2.56
20	1.1	22	400	1.21
20	1	20	400	1
21	1.2	25.2	441	1.44
24	1.4	33.6	576	1.96
\(\sum x \)=181	\(\sum y \)=10.6	\(\sum {xy} \)=244.6	\(\sum {{x^2}} \)=4147	\(\sum {{y^2}} \)=14.52

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

\(\begin{aligned} r &= \frac{{n\sum {xy} - \sum x \sum y }}{{\sqrt {n\sum {{x^2}} - {{\left( {\sum x } \right)}^2}} \sqrt {n\sum {{y^2}} - {{\left( {\sum y } \right)}^2}} }}\\ &= \frac{{8\left( {244.6} \right) - \left( {181} \right)\left( {10.6} \right)}}{{\sqrt {8\left( {4147} \right) - {{\left( {181} \right)}^2}} \sqrt {8\left( {14.52} \right) - {{\left( {10.6} \right)}^2}} }}\\ &= 0.962\end{aligned}\)

Therefore, the value of r is equal to 0.962.

Significance of r

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 0.000.

Since the computed value of r equal to 0.962 is less than the larger critical value of 0.707, it can be said that the correlation between the two variables is significant.

Moreover, the p-value is less than 0.05. This also implies that correlation is significant.

Therefore, there is sufficient evidence to claim that there is a linear correlation between the amount of Tar and Nicotine (mg).

Compute the percentage of variation

b.

Thepercentage of the variation in nicotine can be explained by the linear correlation between nicotine and tar is given as,

\(\begin{aligned} {r^2} &= {0.962^2}\\ &= 0.925\end{aligned}\)

Thus, the percentage of the variation in nicotine can be explained by the linear correlation between nicotine and tar is 92.5%.

Compute the regression equation

c.

Let x represents the amount of tar.

Let y represent the amount of nicotine.

The regression equationis given as,

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

The mean value of xis given as,

\(\begin{aligned} \bar x &= \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\\ &= \frac{{25 + 27 + 20 + .... + 24}}{8}\\ &= 22.625\end{aligned}\)

Therefore, the mean value of x is 22.625.

The mean value of yis given as,

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

Therefore, the mean value of y is 1.325.

Thestandard deviation of xis given as,

\(\begin{aligned} {s_x} &= \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ &= \sqrt {\frac{{{{\left( {25 - 22.625} \right)}^2} + {{\left( {27 - 22.625} \right)}^2} + ..... + {{\left( {24 - 22.625} \right)}^2}}}{{8 - 1}}} \\ &= 2.7222\end{aligned}\)

Therefore, the standard deviation of x is 2.7222.

The standard deviation of yis given as,

\(\begin{aligned}{c}{s_y} &= \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({y_i} - \bar y)}^2}} }}{{n - 1}}} \\ &= \sqrt {\frac{{{{\left( {1.5 - 1.325} \right)}^2} + {{\left( {1.7 - 1.325} \right)}^2} + ..... + {{\left( {1.4 - 1.325} \right)}^2}}}{{8 - 1}}} \\ &= 0.2604\end{aligned}\)

Therefore, the standard deviation of y is 0.2604.

The slope of the regression line is given as,

\(\begin{aligned} {b_1} &= r\frac{{{s_Y}}}{{{s_X}}}\\ &= 0.962 \times \frac{{0.2604}}{{2.7222}}\\ &= 0.0920\end{aligned}\)

Therefore, the value of the slope is 0.0920.

Theinterceptis computed as,

\(\begin{aligned} {b_0} &= \bar y - {b_1}\bar x\\ &= 1.325 - \left( {0.0920 \times 22.625} \right)\\ &= - 0.757\end{aligned}\)

Therefore, the value of intercept is -0.757.

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

The regression equationis given as,

\(\begin{aligned} \hat y &= {b_0} + {b_1}x\\ &= - 0.757 + 0.0920x\end{aligned}\)

Thus, the regression equation is \(\hat y = - 0.757 + 0.0920x\).

Predict the amount of nicotine

d.

The amount of tar is 23 mg.

The predicted amount of nicotine is computed as,

\(\begin{aligned} \hat y &= - 0.757 + 0.0920x\\ &= - 0.757 + \left( {0.0920 \times 23} \right)\\ &= 1.359\\ &\approx 1.4\end{aligned}\)

Thus, the predicted amount of nicotine is 1.4 mg.

Therefore, the predicted amount of nicotine (1.4 mg) is very close to the actual amount of nicotine (1.3 mg).