Question

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

Hypothesis Test The mean sunspot number for the past three centuries is 49.7. Use a 0.05 significance level to test the claim that the eight listed sunspot numbers are from a population with a mean equal to 49.7.

Short answer

It can be concluded that the sunspot numbers are from a population with a mean of 49.7.

Step-by-step solution

Given information

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

The researcher wants to test the claim that the eight listed sunspot numbers are from a population with a mean of 49.7, at the significance level of 0.05.

Identify the test and frame the statistical hypotheses

The t-test is used to check if the population mean is equal to the given hypothesized value as the population standard deviation is unknown.

The null hypothesis is as follows:

\({H_0}:\mu = 49.7\)

The alternative hypothesis is as follows:

\({H_1}:\mu \ne 49.7\)

The test is two-tailed.

If the absolute value of the test statistic value is greater than the critical value, the null hypothesis is rejected.

Statistics and parameters

Let\(\bar x\)denote the sample mean.

Let s denote the sample standard deviation.

The value of\(\bar x\) is computed as follows:

\(\)

\(\begin{aligned} \bar x &= \frac{{\sum x }}{n}\\ &= \frac{{7.5 + 2.9 + ..... + 79.3}}{8}\\ &= 35.91\end{aligned}\)

The sample standard deviation is computedbelow:

\(\begin{aligned} s &= \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ &= \sqrt {\frac{{{{(7.5 - 35.91)}^2} + {{(2.9 - 35.91)}^2} + ...... + {{(79.3 - 35.91)}^2}}}{{8 - 1}}} \\ &= 31.45\end{aligned}\)

Calculate the test statistic, critical value, and p-value

The test statistic value is computed below:

\(\begin{aligned} t &= \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\;\; \sim {t_{\left( {n - 1} \right)}}\\ &= \frac{{35.91 - 49.7}}{{\frac{{31.45}}{{\sqrt 8 }}}}\\ &= - 1.240\end{aligned}\)

Thus, the value of t is –1.240.

The degrees of freedom for t are computed below:

\(\begin{aligned} df &= n - 1\\ &= 8 - 1\\ &= 7\end{aligned}\)

Using the t-table, the critical value of t for df =7,\(\alpha = 0.05\)and a two-tailed test is 2.3646.

The corresponding p-value is obtained below:

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T > |t|} \right)\\ &= 2P\left( {T > | - 1.240|} \right)\\ &= 0.2549\end{aligned}\)

Since the absolute value of tis less than the critical value, the null hypothesis fails to reject.

Further, since the p-value is greater than 0.05, the null hypothesis fails to reject.

Conclusion

There is not sufficient evidence to reject the claim that the sunspot numbers are from a population with a mean of 49.7.