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.)

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.

We define the null and alternative hypotheses as follows:

• Null hypothesis: $H_0: \sigma \geq 5\%$
• Alternative hypothesis: $H_1: \sigma < 5\%$

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: $$\chi^2 = \frac{(25-1)(4.48^2)}{5^2} = 19.2676$$

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: $$\text{Critical Value} = 15.6587$$

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