Question

A stockbroker at Critical Securities reported that the mean rate of return on a sample of 10 oil stocks was 12.6% with a standard deviation of 3.9%. The mean rate of return on a sample of 8 utility stocks was 10.9% with a standard deviation of 3.5%. At the 0.05 significance level, can we conclude that there is more variation in the oil stocks?

Answer 1

Answer:

H0: $\sigma^2_1 \leq \sigma^2_2$

H1: $\sigma^2_1 >\sigma^2_2$

$F=\frac{s^2_1}{s^2_2}=\frac{3.9^2}{3.5^2}=1.242$

$p_v =P(F_{9,7}>1.242)=0.396$

Since the $p_v > \alpha$ we have enough evidence to FAIL to reject the null hypothesis. And we can say that we don't have enough evidence to conclude that the variation for the oil stocks it's greater than the variation for the utility stocks at 5% of significance.

Step-by-step explanation:

Data given and notation

$n_1 = 10$ represent the sampe size for the oil stocks

$n_2 =8$ represent the sample size for the utility stocks

$\bar X_1 =12.6$  represent the sample mean for the oil stocks

$\bar X_2 =10.9$  represent the sample mean for the utility stocks

$s_1 = 3.9$ represent the sample deviation for the oil stocks

$s^2_1 = 15.21$ represent the sample variance for the oil stocks

$s_2 = 3.5$ represent the sample deviation for the utility stocks

$s^2_2 = 12.25$ represent the sample variance for the utility stocks

$\alpha=0.05$ represent the significance level provided

Confidence =0.95 or 95%

F test is a statistical test that uses a F Statistic to compare two population variances, with the sample deviations s1 and s2. The F statistic is always  positive number since the variance it's always higher than 0. The statistic is given by:

$F=\frac{s^2_1}{s^2_2}$

Solution to the problem

System of hypothesis

We want to test if the variation for oil stocks it's higher than the variation for utility stocks, so the system of hypothesis are:

H0: $\sigma^2_1 \leq \sigma^2_2$

H1: $\sigma^2_1 >\sigma^2_2$

Calculate the statistic

Now we can calculate the statistic like this:

$F=\frac{s^2_1}{s^2_2}=\frac{3.9^2}{3.5^2}=1.242$

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have $n_1 -1 =10-1=9$ and for the denominator we have $n_2 -1 =8-1=7$ and the F statistic have 9 degrees of freedom for the numerator and 7 for the denominator. And the P value is given by:

P value

$p_v =P(F_{9,7}>1.242)=0.396$

And we can use the following excel code to find the p value:"=1-F.DIST(1.242;9;7;TRUE)"

Conclusion

Since the $p_v > \alpha$ we have enough evidence to FAIL to reject the null hypothesis. And we can say that we don't have enough evidence to conclude that the variation for the oil stocks it's greater than the variation for the utility stocks at 5% of significance.