Question

test the claim about the difference between two population means \\( \mu _ { 1 } \\) and \\( \mu _ { 2 } \\) at the level of significance \\( \alpha \\). assur and independent, and the populations are normally distributed. claim: \\( \mu _ { 1 } = \mu _ { 2 } ; \alpha = 0.01 \\) population parameters: \\( \sigma _ { 1 } = 3.3, \sigma _ { 2 } = 1.5 \\) sample statistics: \\( \overline { x } _ { 1 } = 16, n _ { 1 } = 28, \overline { x } _ { 2 } = 18, n _ { 2 } = 30 \\) determine the alternative hypothesis. \\( h _ { a } : \mu _ { 1 } \
eq \mu _ { 2 } \\) determine the standardized test statistic. \\( z = \square \\) (round to two decimal places as needed.)

Explanation:

Step1: Recall the formula for the standardized test statistic \(z\)

When testing the difference between two population means \(\mu_1\) and \(\mu_2\) with known population standard - deviations \(\sigma_1\) and \(\sigma_2\), the formula for the standardized test statistic \(z\) is \(z=\frac{(\bar{x}_1 - \bar{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}\). Since the null hypothesis \(H_0:\mu_1=\mu_2\) (i.e., \(\mu_1 - \mu_2 = 0\)), the formula simplifies to \(z=\frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}\).

Step2: Substitute the given values into the formula

We are given that \(\bar{x}_1 = 16\), \(\bar{x}_2 = 18\), \(\sigma_1 = 3.3\), \(\sigma_2 = 1.5\), \(n_1 = 28\), and \(n_2 = 30\).
First, calculate the denominator:
\(\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}=\sqrt{\frac{3.3^{2}}{28}+\frac{1.5^{2}}{30}}\)
\(=\sqrt{\frac{10.89}{28}+\frac{2.25}{30}}\)
\(=\sqrt{0.389 + 0.075}\)
\(=\sqrt{0.464}\approx0.681\)
Then, calculate the numerator: \(\bar{x}_1-\bar{x}_2=16 - 18=-2\)
Now, find \(z\): \(z=\frac{-2}{0.681}\approx - 2.94\)

Answer:

\(-2.94\)