Question

Suppose a mutual fund qualifies as having moderate risk if the standard deviation of its monthly rate of return is less than 5​%. A​ mutual-fund rating agency randomly selects 28 months and determines the rate of return for a certain fund. The standard deviation of the rate of return is computed to be 4.66​%. Is there sufficient evidence to conclude that the fund has moderate risk at the alpha equals 0.10 level of​ significance? A normal probability plot indicates that the monthly rates of return are normally distributed.

Answer 1

Answer:

$\chi^2 =\frac{28-1}{25} 21.7156 =23.453$

$p_v =P(\chi^2$

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(23.453,27,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that $p_v >\alpha$ so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly lower than 5% so we can't conclude that we have a moderate risk for this case.

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

$n=28$ represent the sample size

$\alpha=0.1$ represent the confidence level  

$s^2 =21.7156$ represent the sample variance obtained

$\sigma^2_0 =25$ represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance specification is lower than 25, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \geq 25$

Alternative hypothesis: $\sigma^2$

Calculate the statistic  

For this test we can use the following statistic:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

$\chi^2 =\frac{28-1}{25} 21.7156 =23.453$

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 27. And since is a right tailed test the p value would be given by:

$p_v =P(\chi^2$

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(23.453,27,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that $p_v >\alpha$ so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly lower than 5% so we can't conclude that we have a moderate risk for this case.