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Stocks and Sunspots. Listed below are annual high values of the Dow Jones Industrial Average (DJIA) and annual mean sunspot numbers for eight recent years. Use the data for Exercises 1–5. A sunspot number is a measure of sunspots or groups of sunspots on the surface of the sun. The DJIA is a commonly used index that is a weighted mean calculated from different stock values.

DJIA

14,198

13,338

10,606

11,625

12,929

13,589

16,577

18,054

Sunspot

Number

7.5

2.9

3.1

16.5

55.7

57.6

64.7

79.3

Correlation Use a 0.05 significance level to test for a linear correlation between the DJIA values and the sunspot numbers. Is the result as you expected? Should anyone consider investing in stocks based on sunspot numbers?

Short Answer

Expert verified

The value of r is 0.731.

Since the test statistic value is greater than the critical value, the null hypothesis is rejected. There is a significant linear correlation between the sunspot numbers and the stock indices.

The results are not what was expected.

Furthermore, no one should examine the sunspot count before investing because the correlation between them is meaningless.

Step by step solution

01

Given information

Data are given on two variables, “DJIA” and “Sunspot Number”.

02

Hypotheses

The null hypothesis is as follows:

There is no linear correlation between the variables “DJIA” and “Sunspot Number”.

The alternative hypothesis is as follows:

There is a linear correlation between the variables “DJIA” and “Sunspot Number”.

03

Correlation coefficient

Let x denote the DJIA.

Let y denote the sunspot numbers.

The formula for computing the correlation coefficient (r) between the values of DJIA and sunspot numbers 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

\({x^2}\)

\({y^2}\)

14198

7.5

106485

201583204

56.25

13338

2.9

38680.2

177902244

8.41

10606

3.1

32878.6

112487236

9.61

11625

16.5

191812.5

135140625

272.25

12929

55.7

720145.3

167159041

3102.49

13589

57.6

782726.4

184660921

3317.76

16577

64.7

1072532

274796929

4186.09

18054

79.3

1431682

325946916

6288.49

\(\sum x \)=110916

\(\sum y \)=287.3

\(\sum {xy} \)=4376942

\(\sum {{x^2}} \)=1579677116

\(\sum {{y^2}} \)=17241.35

Substitute the above values to find r:

\(\begin{aligned} r &= \frac{{n\sum {xy} - \sum x \sum y }}{{\sqrt {n\sum {{x^2}} - {{\left( {\sum x } \right)}^2}} \sqrt {n\sum {{y^2}} - {{\left( {\sum y } \right)}^2}} }}\\ &= \frac{{8\left( {4376942} \right) - \left( {110916} \right)\left( {287.3} \right)}}{{\sqrt {8\left( {1579677116} \right) - {{\left( {110916} \right)}^2}} \sqrt {8\left( {17241.35} \right) - {{\left( {287.3} \right)}^2}} }}\\ &= 0.731\end{aligned}\)

Therefore, the value of r is 0.731.

04

Test statistic

The test statistic is computed as follows:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\;\;\; \sim {t_{\left( {n - 2} \right)}}\\ &= \frac{{0.731}}{{\sqrt {\frac{{1 - {{0.731}^2}}}{{8 - 2}}} }}\\ &= 2.624\end{aligned}\)

Here, n=8.

The degrees of freedom of t are computed as follows:

\(\begin{aligned} df &= n - 2\\ &= 8 - 2\\ &= 6\end{aligned}\)

Using the t-table, the critical value of t for \(\alpha = 0.05\)and df = 6 is 2.4469.

The corresponding p-value is obtained as shown:

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T > t} \right)\\ &= 2P\left( {T > 2.624} \right)\\ &= 0.039\end{aligned}\)

Since the test statistic value is greater than the critical value and the p-value is less than 0.05, the null hypothesis is rejected.

Therefore, the linear correlation between the sunspot numbers and DJIA is significant.

05

Discrepancy from reality

In reality, there is no correlation between stock indices and sunspot numbers.

But, according to the obtained value of r, the correlation comes out to be significant.

Thus, the results differ from what was expected.

Moreover, nobody should consider the sunspot numbers before investing as, in the real sense, the correlation between them does not hold any meaning.

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Most popular questions from this chapter

In Exercises 5–8, use a significance level of A = 0.05 and refer to theaccompanying displays.Garbage Data Set 31 “Garbage Weight” in Appendix B includes weights of garbage discarded in one week from 62 different households. The paired weights of paper and glass were used to obtain the XLSTAT results shown here. Is there sufficient evidence to support the claim that there is a linear correlation between weights of discarded paper and glass?

Confidence Intervals for a Regression Coefficients A confidence interval for the regression coefficient b1 is expressed

\(\begin{array}{l}{b_1} - E < {\beta _1} < {b_1} + E\\\end{array}\)

Where

\(E = {t_{\frac{\alpha }{2}}}{s_{{b_1}}}\)

The critical t score is found using n –(k+1) degrees of freedom, where k, n, and sb1 are described in Exercise 17. Using the sample data from Example 1, n = 153 and k = 2, so df = 150 and the critical t scores are \( \pm \)1.976 for a 95% confidence level. Use the sample data for Example 1, the Stat diskdisplay in Example 1 on page 513, and the Stat Crunchdisplay in Exercise 17 to construct 95% confidence interval estimates of \({\beta _1}\) (the coefficient for the variable representing height) and\({\beta _2}\) (the coefficient for the variable representing waist circumference). Does either confidence interval include 0, suggesting that the variable be eliminated from the regression equation?

let the predictor variable x be the first variable given. Use the given data to find the regression equation and the best predicted value of the response variable. Be sure to follow the prediction procedure summarized in Figure 10-5 on page 493. Use a 0.05 significance level.

For 30 recent Academy Award ceremonies, ages of Best Supporting Actors (x) and ages of Best Supporting Actresses (y) are recorded. The 30 paired ages yield\(\bar x = 52.1\)years,\(\bar y = 37.3\)years, r= 0.076, P-value = 0.691, and

\(\hat y = 34.4 + 0.0547x\). Find the best predicted value of\(\hat y\)(age of Best Supporting Actress) in 1982, when the age of the Best Supporting Actor (x) was 46 years.

Different hotels on Las Vegas Boulevard (“the strip”) in Las Vegas are randomly selected, and their ratings and prices were obtained from Travelocity. Using technology, with xrepresenting the ratings and yrepresenting price, we find that the regression equation has a slope of 130 and a y-intercept of -368.

a. What is the equation of the regression line?

b. What does the symbol\(\hat y\)represent?

Notation Twenty different statistics students are randomly selected. For each of them, their body temperature (°C) is measured and their head circumference (cm) is measured.

a. For this sample of paired data, what does r represent, and what does \(\rho \)represent?

b. Without doing any research or calculations, estimate the value of r.

c. Does r change if the body temperatures are converted to Fahrenheit degrees

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