Suppose a mutual fund qualifies as having moderate risk if the standard deviation of its monthly rate of return is less than 3%. A mutual-fund rating agency randomly selects 27 months and determines the rate of return for a certain fund. The standard deviation of the rate of return is computed to be 2.59%. Is there sufficient evidence to conclude that the fund has moderate risk at the a = 0.05 level of significance? A normal probability plot indicates that the monthly rates of return are normally distributed. (A) Calculate the value of the test statistic. (Round to two decimal places as needed.) (B) Use technology to determine the P-value for the test statistic. (Round to three decimal places as needed.)

Step 1 ：We are given that the standard deviation of the rate of return is 2.59%, the sample size is 27, and we are testing against a null hypothesis that the standard deviation is 3%. We are also given that the data is normally distributed.

Step 2 ：We can use the chi-square test for a single variance to test this hypothesis. The test statistic is given by \((n-1)s^2 / σ^2\), where n is the sample size, s is the sample standard deviation, and σ is the hypothesized population standard deviation.

Step 3 ：We can calculate the test statistic using this formula: \(n = 27\), \(s = 0.0259\), \(sigma = 0.03\), \(test\_statistic = 19.378955555555557\)

Step 4 ：Then, we can calculate the P-value using the chi-square distribution with n-1 degrees of freedom. The P-value is the probability of observing a test statistic as extreme as the one we calculated, assuming the null hypothesis is true. If the P-value is less than the significance level (0.05), we reject the null hypothesis.

Step 5 ：The P-value is calculated as \(p\_value = 0.8200376574369816\)

Step 6 ：The test statistic is approximately 19.38 and the P-value is approximately 0.820. Since the P-value is greater than the significance level of 0.05, we do not reject the null hypothesis. This means that there is not sufficient evidence to conclude that the fund has moderate risk.

Step 7 ：Final Answer: The test statistic is approximately \(\boxed{19.38}\) and the P-value is approximately \(\boxed{0.820}\).