Construct a 95% confidence interval for $\mu _ { 1 } - \mu _ { 2 }$ Assume the two samples are random and independent. The sample statistics are given below.
Population statistics: $\sigma _ { 1 } = 1.5 \text { and } \sigma _ { 2 } = 1.9$ Sample statistics: $\overline { \mathrm { x } } 1 = 25 , \mathrm { n } _ { 1 } = 50 \text { and } \overline { \mathrm { x } } 2 = 23 , \mathrm { n } _ { 2 } = 60$
A)(1.364, 2.636)
B)(1.572, 2.987)
C)(1.723, 3.012)
D)(1.919, 3.142)

Question

Quizplus · Accepted Answer

The confidence interval for the difference between two means, $\mu_1 - \mu_2$, when the population standard deviations ($\sigma_1$ and $\sigma_2$) are known, is calculated using the formula: $$ \left( \overline{x}_1 - \overline{x}_2 \right) \pm Z \cdot \sqrt{ \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} } $$ Given: $\overline{x}_1 = 25$, $\overline{x}_2 = 23$, $\sigma_1 = 1.5$, $\sigma_2 = 1.9$, $n_1 = 50$, $n_2 = 60$, and for a 95% confidence interval, $Z = 1.96$. Plugging in the values: $$ \text{CI} = (25 - 23) \pm 1.96 \cdot \sqrt{ \frac{1.5^2}{50} + \frac{1.9^2}{60} } $$ $$ \text{CI} = 2 \pm 1.96 \cdot \sqrt{ \frac{2.25}{50} + \frac{3.61}{60} } $$ $$ \text{CI} = 2 \pm 1.96 \cdot \sqrt{ 0.045 + 0.06017 } $$ $$ \text{CI} = 2 \pm 1.96 \cdot \sqrt{ 0.10517 } $$ $$ \text{CI} = 2 \pm 1.96 \cdot 0.3243 $$ $$ \text{CI} = 2 \pm 0.636 $$ $$ \text{CI} = (1.364, 2.636) $$

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

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