91影视

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

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:

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.

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).

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%.

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 \(\hat y = - 0.757 + 0.0920x\).

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).