Question

Sunspots and the DJIA Use the data from Exercise 5 and find the equation of the regression line. Then find the best predicted value of the DJIA in the year 2004, when the sunspot number was 61. How does the result compare to the actual DJIA value of 10,855?

Short answer

The regression equation is\(\hat y = 9772 + 79.2x\).

The best-predicted value of the DJIA in the year 2004, when the sunspot number was 61,is 13423.6,and this value is not close to the given actual DJIA value in the year 2004 of 10,855.

Step-by-step solution

Given information

The given data depicts the annual sunspot numbers paired with high annual values of the Dow Jones Industrial Average (DJIA) in a tabulated form.

Determine the regression equation

In the regression analysis, letxbe a predictor and ybe a dependent variable, then the equation for the regression line will be as follows:

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

where the slope and intercept measures are computed as follows.

The slope\({b_1}\)is computed using the formula given below:

\({b_1} = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\)

The y-intercept\({b_0}\)is obtained by as follows:

\[{b_0} = \frac{{\left( {\sum y } \right)\left( {\sum {{x^2}} } \right) - \left( {\sum x } \right)\left( {\sum {xy} } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\]

Compute the slope and intercept estimates

The table shows the computation of the regression equation:

	Sunspots (x)	DJIA (y)	\({x^2}\)	\({y^2}\)	\(xy\)
	45	10941	2025	119705481	492345
	31	12464	961	155351296	386384
	46	14198	2116	201583204	653108
	31	13279	961	176331841	411649
	50	10580	2500	111936400	529000
	48	11625	2304	135140625	558000
	56	12929	3136	167159041	724024
	38	13589	1444	184660921	516382
	65	16577	4225	274796929	1077505
	51	18054	2601	325946916	920754
Total	461	134236	22273	1852612654	6269151

The slope will be on substitution:

\[\begin{aligned}{c}{b_1} = \frac{{\left( {10 \times 6269151} \right) - \left( {461 \times 134236} \right)}}{{10209}}\\ = \frac{{62691510 - 61882796}}{{10209}}\\ = \frac{{808714}}{{10209}}\\ = 79.2\end{aligned}\]

Moreover,the y-intercept is as follows:

\[\begin{aligned}{c}{b_0} = \frac{{\left( {134236 \times 22273} \right) - \left( {461 \times 6269151} \right)}}{{10209}}\\ = \frac{{2989838428 - 2890078611}}{{10209}}\\ = \frac{{99759817}}{{10209}}\\ = 9771.752\end{aligned}\]

Hence, the intercept is approximately equal to\[9772\].

Thus, the regression equation becomes \(\hat y = 9772 + 79.2x\).

Visualize the linear relationship

A scatterplot is used to check the linear relationship between the variables.

The plot is made in the following steps:

• Mark two axes; x for sunspots and y for DJIA.
• Mark the points for each paired value of the sunspot aboutthe x-axis and DJIA about the y-axis,as shown below.
• 
  

Since the data show the random pattern and are not very close to the line, there is a poor linear relationship between the variables, sunspots, and the DJIA.

Check the significance of the correlation coefficient

The significance test for the correlation coefficient is conducted using the following hypotheses:

\(\begin{aligned}{l}{H_0}:\,\rho = 0\\{H_1}:\,\rho \ne 0\end{aligned}\)

Here,\(\rho \)is the true measure of correlation.

The correlation coefficient r is given as follows:

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

From the above table, substitute all the values into the equation.

\[\begin{aligned}{c}r = \frac{{\left( {10 \times 6269151} \right) - \left( {461 \times 134236} \right)}}{{\sqrt {\left( {\left( {10 \times 22273} \right) - {{461}^2}} \right)\left( {\left( {10 \times 1852612654} \right) - {{134236}^2}} \right)} }}\\ = \frac{{808714}}{{\sqrt {10209 \times 506822844} }}\\ = \frac{{808714}}{{2274676.8}}\\ = 0.356\end{aligned}\]

Substitute these values intothe test statistic formula.

\[\begin{aligned}{c}t = \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ = \frac{{0.356}}{{\sqrt {\frac{{1 - {{0.356}^2}}}{{\left( {10 - 2} \right)}}} }}\\ = \frac{{0.356}}{{\sqrt {0.109} }}\\ = 1.076\end{aligned}\]

The degrees of freedom will be as shown below.

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

Use the t-table for the p-value corresponding to 8 degrees of freedomat 0.05 level of significance that is 0.3134.

Here, the p-value is greater than the 0.05 level of significance;then, the null hypothesis cannot be rejected.Therefore, there is insufficient evidence to conclude that there is a significant linear correlation between sunspots and the DJIA.

Comparing the actual value

As the correlation is insignificant, the regression model is bad, and the predictions should be made using the mean value of sample responses.

Thebest-predicted value of the DJIA is the mean of y regardless of the value of x.

Therefore, the predicted value for 61 number sunspots is

\(\begin{aligned}{c}\bar y = \frac{{134236}}{{10}}\\ = 13423.6\end{aligned}\)

Therefore, the predicted value is 13423.6.

In the year 2004,the actual value of DJIA was 10,855, andthe predicted value was 13423.6.The valueis not close to the actual 2004 value, that is, 10.855.