Question

test the claim about the difference between two population means \\(mu_1\\) and \\(mu_2\\) a samples are random and independent, and the populations are normally distrib
claim: \\(mu_1 \leq \mu_2; \alpha = 0.10\\). assume \\(\sigma_1^2 \
eq \sigma_2^2\\)
sample statistics: \\(\bar{x}_1 = 2402, s_1 = 170, n_1 = 12\\) and
\\(\bar{x}_2 = 2296, s_2 = 51, n_2 = 10\\)

identify the null and alternative hypotheses. choose the correct answer below.
\\(\bigcirc\\) a. \\(h_0: \mu_1 < \mu_2\\)
\\(h_a: \mu_1 \geq \mu_2\\)
\\(\bigcirc\\) b. \\(h_0: \mu_1 \leq \mu\\)
\\(h_a: \mu_1 > \mu\\)
\\(\bigcirc\\) c. \\(h_0: \mu_1 > \mu_2\\)
\\(h_a: \mu_1 \leq \mu_2\\)
\\(\bigcirc\\) d. \\(h_0: \mu_1 \
eq \mu\\)
\\(h_a: \mu_1 = \mu\\)
\\(\bigcirc\\) e. \\(h_0: \mu_1 = \mu_2\\)
\\(h_a: \mu_1 \
eq \mu_2\\)
\\(\bigcirc\\) f. \\(h_0: \mu_1 \geq \mu\\)
\\(h_a: \mu_1 < \mu\\)

find the standardized test statistic \\(t\\).
\\(t = \square\\)
(round to two decimal places as needed )

Explanation:

Step1: Recall the formula for t - statistic (unequal variances)

The formula for the standardized test statistic \( t \) when testing the difference between two population means with unequal variances (\(\sigma_{1}^{2}
eq\sigma_{2}^{2}\)) is:

$$ t=\frac{(\bar{x}_{1}-\bar{x}_{2})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}} $$

From the claim \( \mu_{1}\leq\mu_{2} \), the null hypothesis \( H_{0}:\mu_{1}-\mu_{2}\leq0 \) and alternative hypothesis \( H_{a}:\mu_{1}-\mu_{2}>0 \). For the test statistic, we assume \( \mu_{1}-\mu_{2} = 0 \) (under the null hypothesis).

Step2: Substitute the given values

We are given \( \bar{x}_{1}=2402 \), \( s_{1}=170 \), \( n_{1}=12 \), \( \bar{x}_{2}=2296 \), \( s_{2}=51 \), \( n_{2}=10 \) and \( \mu_{1}-\mu_{2}=0 \).

First, calculate the numerator: \( (\bar{x}_{1}-\bar{x}_{2})-(\mu_{1}-\mu_{2})=(2402 - 2296)-0=106 \)

Next, calculate the denominator:

$$ \sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}=\sqrt{\frac{170^{2}}{12}+\frac{51^{2}}{10}} $$

Calculate \( \frac{170^{2}}{12}=\frac{28900}{12}\approx2408.333 \)
Calculate \( \frac{51^{2}}{10}=\frac{2601}{10} = 260.1 \)
Sum these two values: \( 2408.333+260.1 = 2668.433 \)
Take the square root: \( \sqrt{2668.433}\approx51.66 \)

Step3: Calculate the t - statistic

Now, divide the numerator by the denominator:

$$ t=\frac{106}{51.66}\approx2.05 $$

Answer:

\( 2.05 \)