An investor is comparing two stocks, A and B. She wants to know if over the long run, there is a significant difference in the return on investment as measured by the percent increase or decrease in the price of the stock from its date of purchase. The investor takes a random sample of $50$annualized daily returns over the past $5$years for each stock. The data are summarized in the table.

a. The investor uses the data to perform a two-sample t test of $H0:\mu A-\mu B=0$versus $Ha:\mu A-\mu B\ne 0,$where $\mu A=$the true mean annualized daily return for Stock A and $\mu B$the true mean annualized daily return for Stock B. The resulting P-value is $0.042$. Interpret this value in context. What conclusion would you make?

b. The investor believes that although the return on investment for Stock A usually exceeds that of Stock B, Stock A represents a riskier investment, where the risk is measured by the price volatility of the stock. The sample variance $S_x^2$is a statistical measure of the price volatility and indicates how much an investment’s actual performance during a specified period varies from its average performance over a longer period. Do the price fluctuations in Stock A significantly exceed those of Stock B, as measured by their variances? State an appropriate set of hypotheses that the investor is interested in testing.

c. To measure this, we will construct a test statistic defined as $F=\frac{largersamplevariance}{smallersamplevariance}$

Calculate the value of the F statistic using the information given in the table. Explain how the value of the statistic provides some evidence for the alternative hypothesis you stated in part (b).

d. Two hundred simulated values of this test statistic, F, were calculated assuming that the two stocks have the same variance in daily price. The results of the simulation are displayed in the following dotplot

Use these simulated values and the test statistic that you calculated in part (c) to determine whether the observed data provide convincing evidence that Stock A is a riskier investment than Stock B. Explain your reasoning.

It is given in the question that:

$$\begin{gathered} {\overline{x}}_1=11.8 \\ {\overline{x}}_2=7.1 \\ {n}_1=50 \\ {n}_2=50 \\ {s}_1=12.9 \\ {s}_2=9.6 \\ P-value=0.042=4.2\% \\ {H}_o:{\mu }_A-{\mu }_B=0 \\ {H}_a:{\mu }_A-{\mu }_B\ne 0 \end{gathered}$$

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.

As we know that,

$$\begin{gathered} {\overline{x}}_1=11.8 \\ {\overline{x}}_2=7.1 \\ {n}_1=50a \\ {n}_2=50 \\ {s}_1=12.9 \\ {s}_2=9.6 \end{gathered}$$

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:

$$\begin{gathered} H_o:{\sigma }_A={\sigma }_B \\ {H}_a:{\sigma }_A>{\sigma }_B \end{gathered}$$

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.

It is given the question that:

$$\begin{gathered} {\overline{x}}_1=11.8 \\ {\overline{x}}_2=7.1 \\ {n}_1=50a \\ {n}_2=50 \\ {s}_1=12.9 \\ {s}_2=9.6 \end{gathered}$$

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:

$$\begin{gathered} F=\frac{largersamplevariance}{smallersamplevariance} \\ =\frac{s_1^2}{s_2^2} \\ =\frac{12.9^2}{9.6^2} \\ =1.8057 \end{gathered}$$

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.

It is given in the question that:

$$\begin{gathered} {\overline{x}}_1=11.8 \\ {\overline{x}}_2=7.1 \\ {n}_1=50a \\ {n}_2=50 \\ {s}_1=12.9 \\ {s}_2=9.6 \end{gathered}$$

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:

role="math" localid="1663669889487" $$\begin{gathered} F=\frac{largersamplevariance}{smallersamplevariance} \\ =\frac{s_1^2}{s_2^2} \\ =\frac{12.9^2}{9.6^2} \\ =1.8057 \end{gathered}$$

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,

$$\begin{gathered} P-value=\frac{6}{200} \\ =0.03 \end{gathered}$$

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

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

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