Question

A random sample of size 20 from a normal population gives a mean 42 and variance 25. To test that the population standard deviation is 8, the value of test statistics is
1. 7.42
2. 15.62
3. 51.20
4. 55.60

• A. 7.42
• B. 15.62
• C. 51.20
• D. 55.60

Hint:

Always ensure you square the hypothesized standard deviation (\( \sigma_0 \)) to obtain the variance (\( \sigma_0^2 \)) in the denominator when calculating this statistic. Here, \( 8^2 = 64 \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question requires performing a Chi-Square test for a single population variance to evaluate the null hypothesis \( H_0: \sigma = \sigma_0 \).
Key Formula or Approach:
The test statistic used to evaluate a hypothesis about a single population variance is given by:
\[ \chi^2 = \frac{(n-1) s^2}{\sigma_0^2} \] where:
\( n = \) sample size
\( s^2 = \) sample variance
\( \sigma_0^2 = \) hypothesized population variance

Step 2: Detailed Explanation:
Let us identify and substitute the given parameters:
- Given Data:
Sample size, \( n = 20 \implies n-1 = 19 \) degrees of freedom
Sample variance, \( s^2 = 25 \)
Hypothesized population standard deviation, \( \sigma_0 = 8 \implies \sigma_0^2 = 8^2 = 64 \)
- Calculate the Chi-Square Test Statistic:
Substitute the values into the formula:
\[ \chi^2 = \frac{19 \times 25}{64} \] \[ \chi^2 = \frac{475}{64} \] Let us perform the division:
\[ \chi^2 = 7.421875 \] Rounding to two decimal places gives \( 7.42 \), which matches Option (A).

Step 3: Final Answer:
The value of the test statistic is \( 7.42 \).
Therefore, the correct choice is Option (A).