Question

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

Answer 1

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**.