Construct a 95 % confidence interval for $\mu _ { 1 } - \mu _ { 2 }$ Two samples are randomly selected from each population.
The sample statistics are given below.
$\begin{array} { l l }
\mathrm { n } _ { 1 } = 50 & \mathrm { n } _ { 2 } = 60 \
\overline { \mathrm { x } } 1 = 25 & \bar { x } _ { 2 } = 23 \
\mathrm {~s} _ { 1 } = 1.5 & \mathrm {~s} _ { 2 } = 1.9
\end{array}$
A) (1.364,2.636)
B) (1.572,2.987)
C) (1.723,3.012)
D) (1.919,3.142)

Question

Construct a  95 %  confidence interval for $\mu _ { 1 } - \mu _ { 2 }$ Two samples are randomly selected from each population.
The sample statistics are given below.
$\begin{array} { l l } 
\mathrm { n } _ { 1 } = 50 & \mathrm { n } _ { 2 } = 60 \
\overline { \mathrm { x } } 1 = 25 & \bar { x } _ { 2 } = 23 \
\mathrm {~s} _ { 1 } = 1.5 & \mathrm {~s} _ { 2 } = 1.9
\end{array}$
A)  (1.364,2.636) 
B)  (1.572,2.987) 
C)  (1.723,3.012) 
D)  (1.919,3.142)

Quizplus · Accepted Answer

To construct a 95% confidence interval for $\mu_1 - \mu_2$, we use the formula:$$\left(\overline{x}_1 - \overline{x}_2\right) \pm z \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$$Given:$$\begin{align*}n_1 &= 50, & n_2 &= 60, \\overline{x}_1 &= 25, & \overline{x}_2 &= 23, \s_1 &= 1.5, & s_2 &= 1.9.\end{align*}$$For a 95% confidence level, $z \approx 1.96$.Plugging in the values:$$\begin{align*}\text{Margin of Error} &= 1.96 \cdot \sqrt{\frac{1.5^2}{50} + \frac{1.9^2}{60}} \&\approx 1.96 \cdot \sqrt{\frac{2.25}{50} + \frac{3.61}{60}} \&\approx 1.96 \cdot \sqrt{0.045 + 0.06017} \&\approx 1.96 \cdot \sqrt{0.10517} \&\approx 1.96 \cdot 0.3243 \&\approx 0.636.\end{align*}$$The difference in means is $25 - 23 = 2$.Thus, the confidence interval is:$$2 \pm 0.636 = (1.364, 2.636).$$

Construct a 95 % Confidence Interval For μ1−μ2\mu _ { 1 } - \mu _ { 2 }μ1​−μ2​

Construct a 95 % Confidence Interval For $\mu _ { 1 } - \mu _ { 2 }$