document.write( "Question 1169144: Suppose a mutual fund qualifies as having moderate risk if the standard deviation of its monthly rate of return is less than 5โ€‹%. Aโ€‹ 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 โ€‹3.73%. Is there sufficient evidence to conclude that the fund has moderate risk at the a=0.10 level ofโ€‹ significance? A normal probability plot indicates that the monthly rates of return are normally distributed.
\n" ); document.write( "o=0.05
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\n" ); document.write( "i need to find x to the second power
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Algebra.Com's Answer #851449 by CPhill(1959)\"\" \"About 
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Let $\sigma$ be the true standard deviation of the monthly rate of return for the mutual fund.
\n" ); document.write( "The mutual fund qualifies as having moderate risk if $\sigma < 0.05$ (or 5%).\r
\n" ); document.write( "\n" ); document.write( "We are given a sample of $n = 24$ months, and the sample standard deviation is $s = 0.0373$ (or 3.73%).
\n" ); document.write( "We want to test the hypothesis that the true standard deviation is less than 0.05 at a significance level of $\alpha = 0.10$.\r
\n" ); document.write( "\n" ); document.write( "The null hypothesis is $H_0: \sigma \ge 0.05$.
\n" ); document.write( "The alternative hypothesis is $H_a: \sigma < 0.05$.\r
\n" ); document.write( "\n" ); document.write( "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:\r
\n" ); document.write( "\n" ); document.write( "$\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}$\r
\n" ); document.write( "\n" ); document.write( "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, $\sigma_0 = 0.05$.\r
\n" ); document.write( "\n" ); document.write( "Plugging in the values:
\n" ); document.write( "$\chi^2 = \frac{(24-1)(0.0373)^2}{(0.05)^2}$
\n" ); document.write( "$\chi^2 = \frac{23 \times (0.00139129)}{0.0025}$
\n" ); document.write( "$\chi^2 = \frac{0.03200}{0.0025}$
\n" ); document.write( "$\chi^2 = 12.8$\r
\n" ); document.write( "\n" ); document.write( "The degrees of freedom for the chi-square distribution are $df = n - 1 = 24 - 1 = 23$.\r
\n" ); document.write( "\n" ); document.write( "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 $\chi^2_{0.10, 23}$.\r
\n" ); document.write( "\n" ); document.write( "From the chi-square distribution table, the critical value $\chi^2_{0.10, 23} \approx 15.659$.\r
\n" ); document.write( "\n" ); document.write( "Now we compare the test statistic to the critical value:
\n" ); document.write( "Test statistic $\chi^2 = 12.8$
\n" ); document.write( "Critical value $\chi^2_{0.10, 23} \approx 15.659$\r
\n" ); document.write( "\n" ); document.write( "Since the test statistic ($12.8$) is less than the critical value ($15.659$), we reject the null hypothesis.\r
\n" ); document.write( "\n" ); document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "Final Answer: The final answer is $\boxed{12.8}$
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