Question

Critical Thinking: Is the pain medicine Duragesic effective in reducing pain? Listed below are measures of pain intensity before and after using the drug Duragesic (fontanels) (based on data from Janssen Pharmaceutical Products, L.P.). The data are listed in order by row, and corresponding measures are from the same subject before and after treatment. For example, the first subject had a measure of 1.2 before treatment and a measure of 0.4 after treatment. Each pair of measurements is from one subject, and the intensity of pain was measured using the standard visual analog score. A higher score corresponds to higher pain intensity.

Pain intensity before Duragestic Treatment

1.2	1.3	1.5	1.6	8	3.4	3.5	2.8	2.6	2.2
3	7.1	2.3	2.1	3.4	6.4	5	4.2	2.8	3.9
5.2	6.9	6.9	5	5.5	6	5.5	8.6	9.4	10
7.6

Pain intensity after Duragestic Treatment

0.4	1.4	1.8	2.9	6.0	1.4	0.7	3.9	0.9	1.8
0.9	9.3	8.0	6.8	2.3	0.4	0.7	1.2	4.5	2.0
1.6	2.0	2.0	6.8	6.6	4.1	4.6	2.9	5.4	4.8
4.1

Regression:Use the given data to find the equation of the regression line. Let the response (y) variable be the pain intensity after treatment. What would be the equation of the regression line for a treatment having absolutely no effect?

Short answer

The regression equation with pain intensity after treatment as the response variable is equal to\(\hat y = 1.9103 + 0.2966x\).

The regression equation for the treatment having absolutely no effect is equal to \(\hat y = 2.00 + 0x\).

Step-by-step solution

Given information

The value of the measures of pain intensity before Duragesic treatment and after Duragesic treatment are considered.

Define the variables and the regression line

Let x represent thepain intensity before Duragesic treatment

Let y represent the painintensity afterDuragesic treatment

Here, y is the response variable, and x is the predictor variable

The regression line of y on x has the following expression:

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

\({b_0}\) is the intercept term

\({b_1}\) is the slope coefficient

Show the necessary calculations

The following table shows the required computations to compute the regression line:

x	y	xy	\({x^2}\)
1.2	0.4	0.48	1.44
1.3	1.4	1.82	1.69
1.5	1.8	2.70	2.25
1.6	2.9	4.64	2.56
8.0	6.0	48.00	64.00
3.4	1.4	4.76	11.56
3.5	0.7	2.45	12.25
2.8	3.9	10.92	7.84
2.6	0.9	2.34	6.76
2.2	1.8	3.96	4.84
3.0	0.9	2.70	9.00
7.1	9.3	66.03	50.41
2.3	8.0	18.40	5.29
2.1	6.8	14.28	4.41
3.4	2.3	7.82	11.56
6.4	0.4	2.56	40.96
5.0	0.7	3.50	25.00
4.2	1.2	5.04	17.64
2.8	4.5	12.60	7.84
3.9	2.0	7.80	15.21
5.2	1.6	8.32	27.04
6.9	2.0	13.80	47.61
6.9	2.0	13.80	47.61
5.0	6.8	34.00	25.00
5.5	6.6	36.30	30.25
6.0	4.1	24.60	36.00
5.5	4.6	25.30	30.25
8.6	2.9	24.94	73.96
9.4	5.4	50.76	88.36
10.0	4.8	48.00	100.00
7.6	4.1	31.16	57.76
\(\sum {x = } \)144.9	\(\sum {y = } \)102.2	\(\sum {xy} = \)533.78	\(\sum {{x^2}} = \)866.35

Calculate the slope of the regression line

The slopeof the regression lineis computed as follows:

\(\begin{aligned} {b_1} &= \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ &= \frac{{31\left( {533.78} \right) - \left( {144.9} \right)\left( {102.2} \right)}}{{31\left( {866.35} \right) - {{\left( {144.9} \right)}^2}}}\\ &= 0.2966\end{aligned}\)

Therefore, the value of the slope coefficient is 0.2966.

Calculate the intercept of the regression line

The interceptis computed as follows:

\(\begin{aligned} {b_0} &= \frac{{\left( {\sum y } \right)\left( {\sum {{x^2}} } \right) - \left( {\sum x } \right)\left( {\sum {xy} } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ &= \frac{{\left( {102.2} \right)\left( {866.35} \right) - \left( {144.9} \right)\left( {533.78} \right)}}{{31\left( {866.35} \right) - {{\left( {144.9} \right)}^2}}}\\ &= 1.9103\end{aligned}\)

Therefore, the value of the intercept is 1.9103.

Regression equation

Theregression equationis given as follows:

\(\begin{aligned} \hat y &= {b_0} + {b_1}x\\ &= 1.9103 + 0.2966x\end{aligned}\)

Thus, the regression equation is \(\hat y = 1.9103 + 0.2966x\).

Regression equation when the treatment has no effect

If Duragesic treatment has no effect, then the values of the pain intensities before and after the treatment will be equal.

In the given data, the following two pairs of data have the same pain intensities before and after the treatment:

x	y	xy	\({x^2}\)
6.9	2.0	13.80	47.61
6.9	2.0	13.80	47.61
\(\sum x \)=13.8	\(\sum y \)=4	\(\sum {xy} \)=27.60	\(\sum {{x^2}} \)=95.22

Now, the modified slope coefficient is equal to:

\(\begin{aligned}{c}{b_1} &= \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ &= \frac{{2\left( {27.60} \right) - \left( {13.8} \right)\left( 4 \right)}}{{2\left( {95.22} \right) - {{\left( {13.8} \right)}^2}}}\\ &= 0\end{aligned}\)

The modified intercept term is equal to:

\(\begin{aligned} {b_0} &= \frac{{\left( {\sum y } \right)\left( {\sum {{x^2}} } \right) - \left( {\sum x } \right)\left( {\sum {xy} } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ &= \frac{{\left( 4 \right)\left( {95.22} \right) - \left( {13.8} \right)\left( {27.60} \right)}}{{4\left( {95.22} \right) - {{\left( {13.8} \right)}^2}}}\\ &= 2.00\end{aligned}\)

Thus, the regression equation when the treatment has no effect is as follows:

\(\hat y = 2.00 + 0x\)