Question

test the claim about the difference between two population means $\mu_1$ and samples are random and independent, and the populations are normally d
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
$\bar{x}_2 = 36.2$, $s_2 = 2.2$, $n_2 = 15$

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

find the standardized test statistic $t$.
$t = -1.62$
(round to two decimal places as needed.)
find the p - value.
$p = \square$
(round to three decimal places as needed.)

Explanation:

Step1: Determine Degrees of Freedom

For pooled t - test, \(df=n_1 + n_2-2=10 + 15-2 = 23\)

Step2: Identify t - value and Test Type

The test statistic \(t=- 1.62\). Since the claim is \(\mu_1=\mu_2\) and we assume a two - tailed test (because the alternative hypothesis should be \(\mu_1
eq\mu_2\) for the claim \(\mu_1 = \mu_2\), but let's check the t - value and degrees of freedom). Wait, maybe the original hypothesis setup: The null hypothesis \(H_0:\mu_1=\mu_2\) and alternative \(H_a:\mu_1
eq\mu_2\) (two - tailed). But the t - value is \(t=-1.62\). We can use t - distribution table or calculator. Using t - distribution with \(df = 23\) and \(t=-1.62\), for two - tailed test, the p - value is \(2\times P(T < - 1.62)\) (since t - distribution is symmetric, \(P(T>1.62)=P(T < - 1.62)\)).
Looking up in t - table or using calculator: \(P(T < - 1.62)\) for \(df = 23\). The t - value of 1.62 with \(df = 23\): from t - table, the critical values for \(df = 23\), \(t_{0.10}=1.319\), \(t_{0.05}=1.714\). So \(P(T < - 1.62)\) is between \(0.05\) and \(0.10\). Using calculator (like TI - 84: tcdf(-1000, - 1.62,23)) gives \(P(T < - 1.62)\approx0.059\). Then for two - tailed test, \(p - value=2\times0.059 = 0.118\) (approx). Wait, maybe I made a mistake in test type. Wait, the original claim is \(\mu_1=\mu_2\), so \(H_0:\mu_1=\mu_2\), \(H_a:\mu_1
eq\mu_2\) (two - tailed). But let's recalculate:

Using a t - calculator: The formula for p - value when \(t=-1.62\) and \(df = 23\) for two - tailed test:

The p - value is calculated as \(2\times\int_{-\infty}^{-1.62}f(t)dt\), where \(f(t)\) is the probability density function of t - distribution with \(df = 23\).

Using an online calculator or statistical software: For \(t=-1.62\), \(df = 23\), two - tailed p - value is approximately \(0.118\)

Answer:

\(0.118\)