Question

test the claim about the difference between two population means $\mu_1$ samples are random and independent, and the populations are normal
claim: $\mu_1 = \mu_2$; $\alpha = 0.01$. assume $\sigma_1^2 = \sigma_2^2$
sample statistics: $\bar{x}_1 = 34.3$, $s_1 = 3.7$, $n_1 = 10$ and
$\bar{x}_2 = 36.2$, $s_2 = 2.2$, $n_2 = 15$

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

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

Explanation:

Step1: Identify Hypotheses

The claim is \( \mu_1 = \mu_2 \), so the null hypothesis \( H_0 \) is \( \mu_1 = \mu_2 \), and the alternative hypothesis \( H_a \) is \( \mu_1
eq \mu_2 \) (two - tailed test). But for the test statistic, we use the pooled - variance t - test formula since \( \sigma_1^2=\sigma_2^2 \). The formula for the pooled variance \( s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2} \) and the test statistic \( t=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{s_p^2(\frac{1}{n_1}+\frac{1}{n_2})}} \), with \( \mu_1-\mu_2 = 0 \) (from \( H_0:\mu_1=\mu_2 \)).

Step2: Calculate Pooled Variance

First, calculate \( (n_1 - 1)s_1^2=(10 - 1)\times(3.7)^2=9\times13.69 = 123.21 \) and \( (n_2 - 1)s_2^2=(15 - 1)\times(2.2)^2=14\times4.84 = 67.76 \). Then \( n_1 + n_2-2=10 + 15-2 = 23 \). So \( s_p^2=\frac{123.21 + 67.76}{23}=\frac{190.97}{23}\approx8.303 \).

Step3: Calculate Test Statistic

\( \bar{x}_1-\bar{x}_2=34.3 - 36.2=- 1.9 \). \( \sqrt{s_p^2(\frac{1}{n_1}+\frac{1}{n_2})}=\sqrt{8.303\times(\frac{1}{10}+\frac{1}{15})}=\sqrt{8.303\times(\frac{3 + 2}{30})}=\sqrt{8.303\times\frac{5}{30}}=\sqrt{8.303\times\frac{1}{6}}\approx\sqrt{1.3838}\approx1.176 \). Then \( t=\frac{-1.9-0}{1.176}\approx - 1.615 \), and taking the absolute value (since t - distribution is symmetric) and rounding to two decimal places, \( t\approx - 1.62 \) (or 1.62, but the sign depends on the difference \( \bar{x}_1-\bar{x}_2 \)).

Answer:

The standardized test statistic \( t\approx\boxed{- 1.62} \) (or \( 1.62 \), depending on the direction, but with the given values, the negative sign is correct as \( \bar{x}_1<\bar{x}_2 \))