Question

test the claim about the difference between two population means \\(\mu_1\\) and \\(\mu_2\\) at the samples are random and independent, and the populations are normally distribute

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\\)

\\(h_a: \mu_1 \leq \mu_2\\) \\(h_a: \mu_1 \geq \mu_2\\)
\\(\bigcirc\\) e. \\(h_0: \mu_1 \geq \mu_2\\) \\(\checkmark\\) f. \\(h_0: \mu_1 = \mu_2\\)
\\(h_a: \mu_1 < \mu_2\\) \\(h_a: \mu_1 \
eq \mu_2\\)

find the standardized test statistic \\(t\\).

\\(t = -1.62\\)
(round to two decimal places as needed.)

find the p - value.

\\(p = 0.118\\)
(round to three decimal places as needed.)

decide whether to reject or fail to reject the null hypothesis and interpret the decis
\\(h_0\\). there enough evidence at the 1% level of significanc

Explanation:

Step1: Recall the formula for pooled variance

For two independent samples with equal variances, the pooled variance \( s_p^2 \) is given by:

$$ s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2} $$

Substitute \( n_1 = 10,s_1 = 3.7,n_2 = 15,s_2 = 2.2 \):

$$ (n_1 - 1)s_1^2=(10 - 1)\times3.7^2=9\times13.69 = 123.21 $$

$$ (n_2 - 1)s_2^2=(15 - 1)\times2.2^2=14\times4.84 = 67.76 $$

$$ n_1 + n_2-2=10 + 15-2=23 $$

$$ s_p^2=\frac{123.21 + 67.76}{23}=\frac{190.97}{23}\approx8.303 $$

Step2: Calculate the standard error

The standard error \( SE \) for the difference in means is:

$$ SE=s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}} $$

First, \( s_p=\sqrt{8.303}\approx2.882 \)

$$ \frac{1}{n_1}+\frac{1}{n_2}=\frac{1}{10}+\frac{1}{15}=\frac{3 + 2}{30}=\frac{5}{30}=\frac{1}{6}\approx0.1667 $$

$$ SE = 2.882\times\sqrt{0.1667}\approx2.882\times0.4082\approx1.177 $$

Step3: Calculate the t - statistic

The formula for the t - statistic for the difference in means (\( \mu_1-\mu_2 \)) is:

$$ t=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{SE} $$

Under the null hypothesis \( H_0:\mu_1=\mu_2 \), so \( \mu_1-\mu_2 = 0 \), and \( \bar{x}_1 = 34.3,\bar{x}_2 = 36.2 \)

$$ \bar{x}_1-\bar{x}_2=34.3 - 36.2=-1.9 $$

$$ t=\frac{-1.9-0}{1.177}\approx - 1.615\approx - 1.62 $$

Step4: Determine the degrees of freedom and find the P - value

Degrees of freedom \( df=n_1 + n_2-2 = 23 \). The alternative hypothesis is two - tailed (\( H_a:\mu_1
eq\mu_2 \)). Using a t - distribution table or calculator, for \( t=-1.62 \) and \( df = 23 \), the two - tailed P - value.
Using a t - calculator, the P - value for \( t=-1.62 \) with \( df = 23 \) (two - tailed) is approximately \( 0.118 \)

Step5: Decision rule

Compare the P - value with \( \alpha = 0.01 \). Since \( P - value=0.118>0.01 \), we fail to reject the null hypothesis \( H_0 \). There is not enough evidence at the 1% level of significance to reject the claim that \( \mu_1=\mu_2 \)

Answer:

• The standardized test statistic \( t\approx - 1.62 \)
• The P - value \( P\approx0.118 \)
• Decision: Fail to reject \( H_0 \). There is not enough evidence at the 1% level of significance.