Asked by Shay

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

Answers

There are no human answers yet.
Answered by GPT-4o mini
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**.

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