Question

It is given that the standard deviation of a large population is 10. To test \(H_0: \mu = 10\) against \(H_1: \mu \neq 10\) at \(1\%\) level of significance, a sample of size 100 is taken. \(H_0\) will be rejected if:

• A. \(\bar{x} > 2.58\)
• B. \(\bar{x} > 12.58\)
• C. \(\bar{x} > 11.96\)
• D. \(\bar{x} > 7.42\)

Hint:

Exam Tip:
For a two-tailed test:

• At \(1\%\) significance, \(Z_{\alpha/2} = 2.58\).
• At \(5\%\) significance, \(Z_{\alpha/2} = 1.96\).
• At \(10\%\) significance, \(Z_{\alpha/2} = 1.645\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We are testing a population mean \(\mu\) with a known population standard deviation \(\sigma\).
This is a two-tailed test at the \(1\%\) level of significance.

Step 2: Key Formula or Approach:
The test statistic is: \[ Z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} \] For a two-tailed test at \(\alpha = 0.01\), the critical values are \(Z = \pm 2.58\).
We reject \(H_0\) if \(|Z| > 2.58\).

Step 3: Detailed Explanation:
Given: \(\sigma = 10\), \(n = 100\), \(\mu_0 = 10\).
The standard error is: \[ \frac{\sigma}{\sqrt{n}} = \frac{10}{\sqrt{100}} = 1 \] The rejection region is: \[ |\bar{x} - 10| > 2.58 \times 1 = 2.58 \] So, we reject \(H_0\) if \(\bar{x} < 10 - 2.58 = 7.42\) or \(\bar{x} > 10 + 2.58 = 12.58\).
Thus, \(H_0\) is rejected if \(\bar{x} > 12.58\) or \(\bar{x} < 7.42\).
The option that matches is \(\bar{x} > 12.58\).

Step 4: Final Answer:
Therefore, option (B) is correct.