Independent random samples of are taken from each of two normally distributed populations.The relevant sample statistics are n₁ = 10,n₂ = 8, $\bar { x } _ { 1 }$
= 50, $\bar { x } _ { 2 }$
= 42,s₁= 5,and s₂ = 7.If we assume the population variances are equal,give a 95% confidence for $\mu _ { 1 } - \mu _ { 2 }$
)

Question

Independent random samples of are taken from each of two normally distributed populations.The relevant sample statistics are n₁ = 10,n₂ = 8,

\bar { x } _ { 1 }

= 50,

\bar { x } _ { 2 }

= 42,s₁= 5,and s₂ = 7.If we assume the population variances are equal,give a 95% confidence for

\mu _ { 1 } - \mu _ { 2 }

)

Quizplus · Accepted Answer

The correct 95% confidence interval for $\mu_1 - \mu_2$ can be calculated using the formula for the confidence interval of the difference between two means, assuming equal variances. The formula involves the sample means ($\bar{x}_1$ and $\bar{x}_2$), the pooled standard deviation, and the critical value from the t-distribution for the desired confidence level. Given the sample sizes (n1=10, n2=8), sample means ($\bar{x}_1=50$, $\bar{x}_2=42$), and sample standard deviations (s1=5, s2=7), and assuming equal variances, the calculation leads to the interval [2.01, 13.99], which matches choice A.

Independent Random Samples of Are Taken from Each of Two xˉ1\bar { x } _ { 1 }xˉ1​

Independent Random Samples of Are Taken from Each of Two $\bar { x } _ { 1 }$