A simple random sample of 26 precipitation amounts has a standard deviation of 0.18. Find the test statistic, using a 0.01 significance level to test a claim that the standard deviation of all precipitation amounts is equal to 0.25.

A) 13.478
B) 18.000
C) 12.960
D) 48.225

1 answer

To find the test statistic for the claim regarding the standard deviation of all precipitation amounts, we will use the Chi-Squared test for variance. The formula for the test statistic \( \chi^2 \) is:

\[ \chi^2 = \frac{(n - 1) s^2}{\sigma_0^2} \]

where:

  • \( n \) = sample size,
  • \( s \) = sample standard deviation,
  • \( \sigma_0 \) = claimed population standard deviation.

From the problem, we have:

  • \( n = 26 \),
  • \( s = 0.18 \),
  • \( \sigma_0 = 0.25 \).

Step 1: Calculate the sample variance

First, we calculate \( s^2 \):

\[ s^2 = (0.18)^2 = 0.0324 \]

Next, we calculate \( \sigma_0^2 \):

\[ \sigma_0^2 = (0.25)^2 = 0.0625 \]

Step 2: Plug the values into the Chi-Squared formula

Now we can calculate the test statistic \( \chi^2 \):

\[ \chi^2 = \frac{(26 - 1) \cdot 0.0324}{0.0625} = \frac{25 \cdot 0.0324}{0.0625} \]

Calculating this step by step:

  1. \( 25 \cdot 0.0324 = 0.81 \)
  2. \( \chi^2 = \frac{0.81}{0.0625} \)

Continuing with the calculation:

\[ \chi^2 = 12.96 \]

Conclusion

The calculated test statistic is \( \chi^2 = 12.96 \).

Thus, the correct answer is C) 12.960.