Question

construct a 95% confidence interval for $\mu_1 - \mu_2$. two samples are randomly selected from sample statistics are given below. assume that $\sigma_1^2 = \sigma_2^2$.
$n_1 = 8, n_2 = 7, \bar{x}_1 = 4.1, \bar{x}_2 = 5.5, s_1 = 0.76, s_2 = 2.51$

\bigcirc a. $(2.112, 2.113)$
\bigcirc b. $(-3.406, 0.606)$
\bigcirc c. $(-1.132, 1.543)$
\bigcirc d. $(-1.679, 1.987)$

Explanation:

Step1: Calculate pooled variance

First, we need to calculate the pooled variance \( s_p^2 \). The formula for pooled variance when \( \sigma_1^2 = \sigma_2^2 \) is:

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

Substitute \( n_1 = 8 \), \( n_2 = 7 \), \( s_1 = 0.76 \), \( s_2 = 2.51 \):

$$ (n_1 - 1)s_1^2=(8 - 1)\times(0.76)^2=7\times0.5776 = 4.0432 $$

$$ (n_2 - 1)s_2^2=(7 - 1)\times(2.51)^2=6\times6.3001 = 37.8006 $$

$$ n_1 + n_2-2=8 + 7-2 = 13 $$

$$ s_p^2=\frac{4.0432 + 37.8006}{13}=\frac{41.8438}{13}\approx3.21875 $$

Step2: Calculate 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, find \( s_p=\sqrt{3.21875}\approx1.7941 \)
Then,

$$ \frac{1}{n_1}+\frac{1}{n_2}=\frac{1}{8}+\frac{1}{7}=\frac{7 + 8}{56}=\frac{15}{56}\approx0.2679 $$

$$ SE = 1.7941\times\sqrt{0.2679}\approx1.7941\times0.5176\approx0.928 $$

Step3: Calculate degrees of freedom and t - critical value

Degrees of freedom \( df=n_1 + n_2-2 = 13 \)
For a 95% confidence interval, the two - tailed \( t \) - critical value \( t_{\alpha/2} \) with \( df = 13 \) is approximately \( t_{0.025,13}=2.1604 \) (using t - distribution table)

Step4: Calculate the difference in sample means and margin of error

The difference in sample means \( \bar{x}_1-\bar{x}_2=4.1 - 5.5=-1.4 \)
The margin of error \( E=t_{\alpha/2}\times SE \)

$$ E = 2.1604\times0.928\approx2.006 $$

Step5: Calculate the confidence interval

The confidence interval for \( \mu_1-\mu_2 \) is:

$$ (\bar{x}_1-\bar{x}_2)-E<\mu_1 - \mu_2<(\bar{x}_1-\bar{x}_2)+E $$

$$ -1.4-2.006<\mu_1 - \mu_2<-1.4 + 2.006 $$

$$ -3.406<\mu_1 - \mu_2<0.606 $$

Answer:

B. \((-3.406,0.606)\)