Questions: 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 25 months and determines the rate of return for a certain fund. The standard deviation of the rate of return is computed to be 4.48%. Is there sufficient evidence to conclude that the fund has moderate risk at the α=0.10 level of significance? A normal probability plot indicates that the monthly rates of return are normally distributed. What are the correct hypotheses for this test? The null hypothesis is H0: 0.05. The alternative hypothesis is H1: ≠ 0.05 Calculate the value of the test statistic. χ0^2= (Round to two decimal places as needed.)

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 25 months and determines the rate of return for a certain fund. The standard deviation of the rate of return is computed to be 4.48%. Is there sufficient evidence to conclude that the fund has moderate risk at the α=0.10 level of significance? A normal probability plot indicates that the monthly rates of return are normally distributed. What are the correct hypotheses for this test? The null hypothesis is H0: 0.05. The alternative hypothesis is H1: ≠ 0.05 Calculate the value of the test statistic. χ0^2= (Round to two decimal places as needed.)
Transcript text: 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 25 months and determines the rate of return for a certain fund. The standard deviation of the rate of return is computed to be $4.48 \%$. Is there sufficient evidence to conclude that the fund has moderate risk at the $\boldsymbol{\alpha}=0.10$ level of significance? A normal probability plot indicates that the monthly rates of return are normally distributed. What are the correct hypotheses for this test? The null hypothesis is $\mathrm{H}_{0}$ : $\square$ $\square$ 0.05 . The alternative hypothesis is $\mathrm{H}_{1}$ : o $\square$ 0.05 Calculate the value of the test statistic. $\chi_{0}^{2}=$ $\square$ (Round to two decimal places as needed.)
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Solution

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Solution Steps

To determine if there is sufficient evidence to conclude that the fund has moderate risk, we need to perform a hypothesis test for the standard deviation. The hypotheses are:

  • Null hypothesis (\(H_0\)): The standard deviation of the monthly rate of return is 5% or more.
  • Alternative hypothesis (\(H_1\)): The standard deviation of the monthly rate of return is less than 5%.

We will use the chi-square test for the standard deviation. The test statistic is calculated using the formula: \[ \chi^2 = \frac{(n-1)s^2}{\sigma^2} \] where \(n\) is the sample size, \(s\) is the sample standard deviation, and \(\sigma\) is the population standard deviation under the null hypothesis.

Step 1: State the Hypotheses

We define the null and alternative hypotheses as follows:

  • Null hypothesis: \( H_0: \sigma \geq 5\% \)
  • Alternative hypothesis: \( H_1: \sigma < 5\% \)
Step 2: Calculate the Test Statistic

Using the formula for the chi-square test statistic: \[ \chi^2 = \frac{(n-1)s^2}{\sigma^2} \] we substitute the values:

  • \( n = 25 \)
  • \( s = 4.48 \)
  • \( \sigma = 5 \)

Calculating the test statistic: \[ \chi^2 = \frac{(25-1)(4.48^2)}{5^2} = 19.2676 \]

Step 3: Determine the Critical Value

For a significance level of \( \alpha = 0.10 \) and degrees of freedom \( df = n - 1 = 24 \), the critical value from the chi-square distribution is: \[ \text{Critical Value} = 15.6587 \]

Step 4: Compare Test Statistic and Critical Value

We compare the test statistic with the critical value:

  • Test statistic: \( \chi^2 = 19.2676 \)
  • Critical value: \( 15.6587 \)

Since \( 19.2676 > 15.6587 \), we fail to reject the null hypothesis.

Final Answer

There is insufficient evidence to conclude that the fund has moderate risk at the \( \alpha = 0.10 \) level of significance. Thus, the final conclusion is: \[ \boxed{\text{Fail to reject } H_0} \]

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