Question

Two samples are randomly selected from each population. The sample statistics are given below:

Use \(\alpha = 0.05\). \(n_1 = 50, \bar{x}_1 = 21, s_1 = 1.5\) and \(n_2 = 60, \bar{x}_2 = 19, s_2 = 1.9\).

What is the standardized test statistic to test the hypothesis that \(\mu_1 = \mu_2\)?

• A. 8.1
• B. 4.2
• C. 6.2
• D. 3.8

Hint:

For large samples use the two-sample $z$ statistic \(z=\dfrac{\bar x_1-\bar x_2}{\sqrt{s_1^2/n_1+s_2^2/n_2}}\).

Answer 1

The Correct Option is C

Step 1: Since both samples are large (\(n_1 = 50 > 30\) and \(n_2 = 60 > 30\)), the difference of sample means \(\bar{x}_1 - \bar{x}_2\) is tested using the two-sample \(z\) statistic:

\[ z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}} \]

Step 2: Compute each variance term.

\[ \frac{s_1^2}{n_1} = \frac{1.5^2}{50} = \frac{2.25}{50} = 0.045 \]\[ \frac{s_2^2}{n_2} = \frac{1.9^2}{60} = \frac{3.61}{60} = 0.0602 \]

Step 3: Add the two terms and take the square root.

\[ \sqrt{0.045 + 0.0602} = \sqrt{0.1052} = 0.3243 \]

Step 4: Compute the test statistic.

\[ z = \frac{21 - 19}{0.3243} = \frac{2}{0.3243} = 6.17 \approx 6.2 \]

The correct answer is (C) 6.2.