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: $\mu_1 = \mu_2$; $\alpha = 0.10$. assume $\sigma_1^2 = \sigma_2^2$

sample statistics: $\bar{x}_1 = 35.8$, $s_1 = 3.6$, $n_1 = 12$ and $\bar{x}_2 = 37.8$, $s_2 = 2.4$, $n_2 = 16$

identify the null and alternative hypotheses. choose the correct answer below.

a. $h_0: \mu_1 = \mu_2$
$h_a: \mu_1 \
eq \mu_2$
b. $h_0: \mu_1 \geq \mu_2$
$h_a: \mu_1 < \mu_2$
c. $h_0: \mu_1 > \mu_2$
$h_a: \mu_1 \leq \mu_2$
d. $h_0: \mu_1 < \mu_2$
$h_a: \mu_1 \geq \mu_2$
e. $h_0: \mu_1 \leq \mu_2$
$h_a: \mu_1 > \mu_2$
f. $h_0: \mu_1 \
eq \mu_2$
$h_a: \mu_1 = \mu_2$

find the standardized test statistic $t$.
$t = \square$

Explanation:

Step1: Calculate the pooled variance \(s_p^2\)

The formula for pooled variance is \(s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}\)
Substitute \(n_1 = 12\), \(s_1 = 3.6\), \(n_2 = 16\), \(s_2 = 2.4\)
\(s_p^2=\frac{(12 - 1)\times3.6^2+(16 - 1)\times2.4^2}{12 + 16-2}\)
\(=\frac{11\times12.96+15\times5.76}{26}\)
\(=\frac{142.56 + 86.4}{26}=\frac{228.96}{26}\approx8.806\)

Step2: Calculate the standardized test statistic \(t\)

The formula for \(t\) - statistic for two - sample \(t\) - test (equal variances) is \(t=\frac{\bar{x}_1-\bar{x}_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\)
Substitute \(\bar{x}_1 = 35.8\), \(\bar{x}_2 = 37.8\), \(s_p=\sqrt{8.806}\approx2.968\), \(n_1 = 12\), \(n_2 = 16\)
\(t=\frac{35.8 - 37.8}{2.968\sqrt{\frac{1}{12}+\frac{1}{16}}}\)
First, calculate \(\sqrt{\frac{1}{12}+\frac{1}{16}}=\sqrt{\frac{4 + 3}{48}}=\sqrt{\frac{7}{48}}\approx0.383\)
Then \(t=\frac{- 2}{2.968\times0.383}\approx\frac{-2}{1.137}\approx - 1.76\)

Answer:

The standardized test statistic \(t\approx - 1.76\)