Question

Beta value of a stock. The “beta coefficient” of a stock is a measure of the stock’s volatility (or risk) relative to the market as a whole. Stocks with beta coefficients greater than 1 generally bear the greater risk (more volatility) than the market, whereas stocks with beta coefficients less than 1 are less risky (less volatile) than the overall market (Alexander, Sharpe, and Bailey, Fundamentals of Investments, 2000). A random sample of 15 high-technology stocks was selected at the end of 2009, and the mean and standard deviation of the beta coefficients were calculated: $\overline{\mathbf{x}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{23}$s = .37.

a. Set up the appropriate null and alternative hypotheses to test whether the average high-technology stock is riskier than the market as a whole.

b. Establish the appropriate test statistic and rejection region for the test. Use $\mathit{a}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}$.

c. What assumptions are necessary to ensure the validity of the test?

d. Calculate the test statistic and state your conclusion.

e. What is the approximate p-value associated with this test? Interpret it.

f. Conduct a test to determine if the variance of the stock beta values differs from .15. Use $\mathit{a}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$.

Short answer

a. The null and alternative hypotheses are: $H_0:\mu =1$ Against $H_0:\mu >1$.

b. The test statistic used for the hypothesis test is the t-test, and the rejection region is $t>1.345$.

c. The population from which the random samples are taken should be normally distributed.

d. The value of the test statistic is $t=2.408$.

e. The p-value is 0.0152; there is enough evidence to claim that, on average high-technology stock is riskier than the market as a whole.

f. There is no sufficient evidence to reject the claim that the variance of the stock beta values differs from 0.15.

Step-by-step solution

Given Information

The population variance of the stock beta values is 0.15. It is tested based on the sample variance and the sample size.

Concept

The chi-square test statistic is used to test the significance of the population variance. The degrees of freedom are required to find the critical value.

Conducting the test for variance

f.

The null and alternative hypotheses are:

$H_0:{\sigma }^2=0.15Against{H}_a:{\sigma }^2\ne 0.15$

The test statistic is:

$$\begin{gathered} x^2=\frac{(n-1)s^2}{{\sigma }^2} \\ =\frac{14\times {(0.37)}^2}{0.15} \\ =\frac{14\times 0.1369}{0.15} \\ =12.777 \end{gathered}$$

The test statistic is 12.777.

The chi-square critical value at a 5% level of significance for the two-tailed test is

${x^2}_{0.025,14}=26.119$

A calculator is used to find the critical value.

Since 12.777 is less than 26.119, it fails to reject the null hypothesis.

There is no sufficient evidence to reject the claim that the variance of the stock beta values differs from 0.15.