Question

In a random sample of 100 voters in an area, 59 voters were in favour of a candidate A. The value of the test statistic for testing \(H_0: P = 0.5\) against \(H_0: P \neq 0.5\) is:

• A. 1.90
• B. 1.80
• C. 1.28
• D. 1.75

Hint:

Exam Tip:
For testing a proportion:

• Use the Z-test.
• \(Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}}\).
• For a two-tailed test, compare \(|Z|\) with the critical value.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We are testing a proportion using a large sample. The test statistic for a proportion is: \[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \]

Step 2: Key Formula or Approach:
Given: \(n = 100\), number in favour = 59, so \(\hat{p} = \frac{59}{100} = 0.59\).
Under \(H_0: p_0 = 0.5\).

Step 3: Detailed Explanation:
\[ Z = \frac{0.59 - 0.5}{\sqrt{\frac{0.5 \times 0.5}{100}}} = \frac{0.09}{\sqrt{\frac{0.25}{100}}} = \frac{0.09}{\sqrt{0.0025}} = \frac{0.09}{0.05} = 1.8 \] So, the test statistic is 1.80.

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