Chapter 10, Problem AP3.35CPT

(a)

To determine

To interpret the value in this context and find to what conclusion would you make.

Answer to Problem AP3.35CPT

There is a $4.2\%$ chance of obtaining the sample results or extreme when the true mean annualized daily return for Stock A is the same as the true mean annualized daily return for Stock B.

Explanation of Solution

It is given in the question that:

  $\begin{array}{l}{\overline{x}}_1=11.8\\ {\overline{x}}_2=7.1\\ n_1=50\\ n_2=50\\ s_1=12.9\\ s_2=9.6\\ P\text{-value}=0.042=4.2\%\\ H_0:{\mu }_A-{\mu }_B=0\\ H_a:{\mu }_A-{\mu }_B\ne 0\end{array}$

Since we know that the P-value is the probability of obtaining the sample results or more extreme, when the null hypothesis is true.

Thus, it means that there is a $4.2\%$ chance of obtaining the sample results or extreme when the true mean annualized daily return for Stock A is the same as the true mean annualized daily return for Stock B.

(b)

To determine

To explain do the price fluctuations in stock A significantly increase those of stock B, as measured by their variance or not and state an appropriate set of hypotheses that the investor is interested in testing.

Answer to Problem AP3.35CPT

The hypotheses will be:

  $\begin{array}{l}{H}_0:{\sigma }_A={\sigma }_B\\ H_a:{\sigma }_A>{\sigma }_B\end{array}$

Explanation of Solution

As we know that,

  $\begin{array}{l}{\overline{x}}_1=11.8\\ {\overline{x}}_2=7.1\\ n_1=50\\ n_2=50\\ s_1=12.9\\ s_2=9.6\end{array}$

Then, the standard deviation of Stock A is more than that of Stock B. So, we can say that there is more variations in Stock A than Stock B as the standard deviation of Stock A is more than of Stock B. Thus, it can be that the price fluctuations in stock A significantly increase those of stock B, as measured by their variance or standard deviations. The appropriate set of hypothesis than the investor will be interested in can be:

  $\begin{array}{l}{H}_0:{\sigma }_A={\sigma }_B\\ H_a:{\sigma }_A>{\sigma }_B\end{array}$

Where we want the evidence that the claim, the price fluctuations in stock A significantly increase those of stock B, as measured by their variance, is true. So, the investor will conduct this hypotheses testing.

(c)

To determine

To calculate the value of F statistics using the information given in the table and explain how the value of statistics provide some evidence for the alternative hypothesis you stated in part (b).

Answer to Problem AP3.35CPT

The F value is $1.8057$ .

Explanation of Solution

It is given the question that:

  $\begin{array}{l}{\overline{x}}_1=11.8\\ {\overline{x}}_2=7.1\\ n_1=50\\ n_2=50\\ s_1=12.9\\ s_2=9.6\end{array}$

Now, we have to calculate the value of F statistics. This can be done as:

Firstly, the value of F statistics is the larger sample variance $s_1^2$ divided by the smaller sample variance $s_2^2$ . Then, it will be:

  $\begin{array}{l}F=\frac{\text{Larger sample variance}}{\text{Smaller sample variance}}\\ =\frac{s_1^2}{s_2^2}\\ =\frac{{12.9}^2}{{9.6}^2}\\ =1.8057\end{array}$

Since the F statistics value is $s_1^2$ divided by $s_2^2$ , this indicates that the sample variance of stock A is greater than the sample variance of stock B and this agrees with the alternative hypothesis $H_a:{\sigma }_A>{\sigma }_B$ , as in the part (b).

This implies that there is some evidence for the alternative hypothesis $H_a:{\sigma }_A>{\sigma }_B$ to be true.

(d)

To determine

To explain whether the observed data provide convincing evidence that stock A is a riskier investment than stock B.

Answer to Problem AP3.35CPT

There is convincing evidence that stock A is a riskier investment than stock B.

Explanation of Solution

It is given in the question that:

  $\begin{array}{l}{\overline{x}}_1=11.8\\ {\overline{x}}_2=7.1\\ n_1=50\\ n_2=50\\ s_1=12.9\\ s_2=9.6\end{array}$

And let us that the significance level will be,

  $\alpha =0.05$

Now, the value of F statistics will be as in part (c),

The value of F statistics is the larger sample variance $s_1^2$ divided by the smaller sample variance $s_2^2$ . Then, it will be:

  $\begin{array}{l}F=\frac{\text{Larger sample variance}}{\text{Smaller sample variance}}\\ =\frac{s_1^2}{s_2^2}\\ =\frac{{12.9}^2}{{9.6}^2}\\ =1.8057\end{array}$

Then, the P-value is the probability of obtaining the sample results or more extreme, when the null hypothesis is true.

And we note that six of the $200$ dots in the dot plot lie to the right of $1.8057$ , then we have,

  $\begin{array}{l}P\text{-value}=\frac{6}{200}\\ =0.03\end{array}$

If the P-value is smaller than the significance level, then reject the null hypothesis:

  $P<0.05\Rightarrow \text{Reject }{H}_0$

Thus, we can conclude that there is convincing evidence that stock A is a riskier investment than stock B.