SOLUTION: 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
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Question 1169144: 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 24 months and determines the rate of return for a certain fund. The standard deviation of the rate of return is computed to be 3.73%. Is there sufficient evidence to conclude that the fund has moderate risk at the a=0.10 level of significance? A normal probability plot indicates that the monthly rates of return are normally distributed.
o=0.05
0<0.05
i need to find x to the second power
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let $\sigma$ be the true standard deviation of the monthly rate of return for the mutual fund.
The mutual fund qualifies as having moderate risk if $\sigma < 0.05$ (or 5%).
We are given a sample of $n = 24$ months, and the sample standard deviation is $s = 0.0373$ (or 3.73%).
We want to test the hypothesis that the true standard deviation is less than 0.05 at a significance level of $\alpha = 0.10$.
The null hypothesis is $H_0: \sigma \ge 0.05$.
The alternative hypothesis is $H_a: \sigma < 0.05$.
Since the population is assumed to be normally distributed and we are testing a hypothesis about the population standard deviation, we will use the chi-square distribution. The test statistic for the standard deviation is:
$\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}$
where $n$ is the sample size, $s$ is the sample standard deviation, and $\sigma_0$ is the hypothesized standard deviation under the null hypothesis. In this case, we use the upper bound of the null hypothesis, $\sigma_0 = 0.05$.
Plugging in the values:
$\chi^2 = \frac{(24-1)(0.0373)^2}{(0.05)^2}$
$\chi^2 = \frac{23 \times (0.00139129)}{0.0025}$
$\chi^2 = \frac{0.03200}{0.0025}$
$\chi^2 = 12.8$
The degrees of freedom for the chi-square distribution are $df = n - 1 = 24 - 1 = 23$.
Since this is a left-tailed test ($H_a: \sigma < 0.05$), we need to find the critical value $\chi^2_{\alpha, df}$ such that the area to the left of it is $\alpha = 0.10$ with $df = 23$. We look up the chi-square distribution table or use a statistical calculator for $\chi^2_{0.10, 23}$.
From the chi-square distribution table, the critical value $\chi^2_{0.10, 23} \approx 15.659$.
Now we compare the test statistic to the critical value:
Test statistic $\chi^2 = 12.8$
Critical value $\chi^2_{0.10, 23} \approx 15.659$
Since the test statistic ($12.8$) is less than the critical value ($15.659$), we reject the null hypothesis.
There is sufficient evidence at the $\alpha = 0.10$ level of significance to conclude that the standard deviation of the monthly rate of return is less than 5%. Therefore, there is sufficient evidence to conclude that the fund has moderate risk.
The question asks to find $x^2$. In the context of the chi-square test, $x^2$ usually refers to the chi-square test statistic.
Final Answer: The final answer is $\boxed{12.8}$
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