Question

A simple random sample of 25 vitamin tablets is obtained, and the potassium content of each tablet is measured. The sample has a standard deviation of 3.7 mg. Use a 0.05 significance level to test the claim that the potassium content of vitamin tablets has a standard deviation equal to 3.2 mg. Find the critical value(s) needed to test the claim.

A) 12.401, 39.364
B) 13.848, 36.415
C) 33.196
D) 36.415

Answer 1

To test the claim that the population standard deviation of potassium content in vitamin tablets is equal to 3.2 mg, we can use the chi-square test for variance.

### Step 1: Define Hypotheses

- Null hypothesis (\(H_0\)): \(\sigma = 3.2\) mg (the standard deviation of the potassium content is equal to 3.2 mg).
- Alternative hypothesis (\(H_a\)): \(\sigma \neq 3.2\) mg (the standard deviation is not equal to 3.2 mg).

### Step 2: Determine the Sample Statistics

Given:
- Sample size (\(n\)) = 25
- Sample standard deviation (\(s\)) = 3.7 mg
- Hypothesized population standard deviation (\(\sigma_0\)) = 3.2 mg

### Step 3: Calculate the Test Statistic

The test statistic for the chi-square test is calculated using the formula:

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

Plugging in the values:

- \(n - 1 = 25 - 1 = 24\)
- \(s^2 = (3.7)^2 = 13.69\)
- \(\sigma_0^2 = (3.2)^2 = 10.24\)

\[
\chi^2 = \frac{(24)(13.69)}{10.24} = \frac{328.56}{10.24} \approx 32.1
\]

### Step 4: Degrees of Freedom

The degrees of freedom \(df = n - 1 = 24\).

### Step 5: Determine Critical Values

At the 0.05 significance level for a two-tailed test, we look for the critical values in the chi-square distribution with 24 degrees of freedom.

Using a chi-square table or calculator:
- The critical value for \( \chi^2_{0.025, 24} \) (upper tail) is approximately **39.364**.
- The critical value for \( \chi^2_{0.975, 24} \) (lower tail) is approximately **12.401**.

### Conclusion

The critical values are:

- Lower tail: 12.401
- Upper tail: 39.364

Thus, the answer is:

**A) 12.401, 39.364**