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 } = 3.6, \sigma _ { 2 } = 1.4 $
sample statistics: $ \overline { x } _ { 1 } = 17, n _ { 1 } = 30, \overline { x } _ { 2 } = 15, n _ { 2 } = 26 $
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.)

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Accepted Answer

# Explanation:
## Step1: Formula for $ z - $ test statistic
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 $H_0:\mu_1=\mu_2$ (so $\mu_1 - \mu_2 = 0$), $\bar{x}_1 = 17$, $\bar{x}_2=15$, $\sigma_1 = 3.6$, $\sigma_2 = 1.4$, $n_1 = 30$, and $n_2=26$.
## Step2: Substitute values into the formula
Substitute the values into the formula:
$$
\begin{align*}
z&=\frac{(17 - 15)-0}{\sqrt{\frac{3.6^{2}}{30}+\frac{1.4^{2}}{26}}}\
&=\frac{2}{\sqrt{\frac{12.96}{30}+\frac{1.96}{26}}}\
&=\frac{2}{\sqrt{0.432+0.0754}}\
&=\frac{2}{\sqrt{0.5074}}\
&=\frac{2}{0.7123}\
&\approx2.81
\end{align*}
$$

# Answer:
$z = 2.81$

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Explanation:

Step1: Formula for \( z - \) test statistic

Step2: Substitute values into the formula

Answer: