91Ó°ÊÓ

It is given in the question that:

$$\begin{gathered} {\stackrel{¯}{x}}_1=11.8 \\ {\stackrel{¯}{x}}_2=7.1 \\ {n}_1=50 \\ {n}_2=50 \\ {s}_1=12.9 \\ {s}_2=9.6 \\ P-value=0.042=4.2\% \\ {H}_o:{\mathrm{μ}}_A-{\mathrm{μ}}_B=0 \\ {H}_a:{\mathrm{μ}}_A-{\mathrm{μ}}_B≠0 \end{gathered}$$

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,

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

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:

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

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:

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

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 $H_a:{\mathrm{σ}}_A>{\mathrm{σ}}_B$as in the part (b)

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

It is given in the question that:

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

And let us that the significance level will be,

$\mathrm{Î\pm }=0.05$

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:

$P<0.05⇒Reject{H}_0$

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