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: \\( \overline { x } _ { 1 } = 15, n _ { 1 } = 31, \overline { x } _ { 2 } = 17, 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.)

Explanation:

Step1: Write the formula for the standardized test statistic

The formula for the \( z - \) test statistic for two - sample means (when population variances \(\sigma_1^{2}\) and \(\sigma_2^{2}\) 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 \(\bar{x}_1 = 15\), \(\bar{x}_2=17\), \(\sigma_1 = 3.5\), \(\sigma_2 = 1.7\), \(n_1 = 31\), \(n_2=26\).
First, calculate the numerator: \((\bar{x}_1 - \bar{x}_2)-(\mu_1-\mu_2)=(15 - 17)-0=- 2\).
Then, calculate the denominator:

$$
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$$

Now, find \(z\): \(z=\frac{-2}{0.7129}\approx - 2.81\)

Answer:

\(z=-2.81\)