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

Confidence Interval Construct a 95% confidence interval estimate of the mean sunspot number. Write a brief statement interpreting the confidence interval.

Short answer

The 95% confidence interval for the mean sunspot number is equal to (9.62, 62.21).

There is 95% confidence that the true value of the population mean of the sunspot numbers will lie between 9.62 and 62.21.

Step-by-step solution

Given information

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

Important computations

The mean of the sample sunspot numbers is computed below:

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

The sample standard deviation is computed as shown below:

\(\begin{aligned} s &= \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ &= \sqrt {\frac{{{{(2.9 - 35.91)}^2} + {{(3.1 - 35.91)}^2} + ...... + {{(79.3 - 35.91)}^2}}}{{8 - 1}}} \\ &= 31.45\end{aligned}\)

The sample size (n) is equal to 8.

Level of significance

The following formula is used to compute the level of significance:

\(\begin{aligned} {\rm{Confidence}}\;{\rm{Level}} &= 95\% \\100\left( {1 - \alpha } \right) &= 95\\1 - \alpha &= 0.95\\ &= 0.05\end{aligned}\)

Therefore,

\(\begin{aligned} \frac{\alpha }{2} &= \frac{{0.05}}{2}\\ &= 0.025\end{aligned}\)

t-multiplier

The degrees of freedom are computed below:

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

The value of the t-multiplier at \(\frac{\alpha }{2} = 0.025\) and degrees of freedom equal to 7 is obtained as 2.365.

Confidence interval

Substitute the values obtained above to calculate the value of margin of error (E) as shown:

\(\begin{aligned} E &= {t_{\frac{\alpha }{2}}}\frac{s}{{\sqrt n }}\\ &= \left( {2.365} \right)\frac{{31.45}}{{\sqrt 8 }}\\ &= 26.297\end{aligned}\)

Thus, the confidence interval becomes:

\(\begin{aligned} CI &= \left( {\bar x - E,\bar x + E} \right)\\ &= \left( {35.91 - 26.297,35.91 + 26.297} \right)\\ &= \left( {9.613,62.207} \right)\\ &\approx \left( {9.62,62.21} \right)\end{aligned}\)

Therefore, the 95% confidence interval for the mean sunspot number is equal to (9.62, 62.21).

Interpretation of the confidence interval

There is 95% confidence that the true value of the population mean of the sunspot numbers will lie between 9.62 and 62.21.