Question

test the claim about the difference between two population means $\mu_1$ and $\mu_2$ at the level of significance $\alpha$. assume the samples are random and independent, and the populations are normally distributed.
claim: $\mu_1 = \mu_2$; $\alpha = 0.01$
population parameters: $\sigma_1 = 3.5$, $\sigma_2 = 1.7$
sample statistics: $\bar{x}_1 = 15$, $n_1 = 29$, $\bar{x}_2 = 17$, $n_2 = 30$

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 z - test statistic for two population means

The formula for the standardized test statistic \( z \) when testing the difference between two population means \( \mu_1 \) and \( \mu_2 \) (when population standard deviations \( \sigma_1 \) and \( \sigma_2 \) are known) 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 claim is \( \mu_1=\mu_2 \), then \( \mu_1 - \mu_2 = 0 \).

Step2: Substitute the given values into the formula

We are given:
\( \bar{x}_1 = 15 \), \( n_1=29 \), \( \sigma_1 = 3.5 \)
\( \bar{x}_2=17 \), \( n_2 = 30 \), \( \sigma_2=1.7 \)
\( \mu_1-\mu_2 = 0 \) (from the claim \( \mu_1=\mu_2 \))

First, calculate the numerator: \( (\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)=(15 - 17)-0=- 2 \)

Then, calculate the denominator:

$$ \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}=\sqrt{\frac{(3.5)^2}{29}+\frac{(1.7)^2}{30}} $$

Calculate \( \frac{(3.5)^2}{29}=\frac{12.25}{29}\approx0.4224 \)
Calculate \( \frac{(1.7)^2}{30}=\frac{2.89}{30}\approx0.0963 \)
Sum these two values: \( 0.4224 + 0.0963=0.5187 \)
Take the square root: \( \sqrt{0.5187}\approx0.7202 \)

Now, calculate the z - statistic:

$$ z=\frac{-2}{0.7202}\approx - 2.78 $$

Answer:

\( z\approx - 2.78 \)