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

1 answer

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