A statistics teacher wants to see if there is any difference in the abilities of students enrolled in statistics today and those enrolled five years ago.A sample of final examination scores from students enrolled today and from students enrolled five years ago was taken.You are given the following information. $\begin{array}{ll}
&\text { Today } & \text { Five Years Ago } \
\bar {x}&82 & 88 \
\sigma^2&112.5 & 54 \
n&45 & 36
\end{array}$
The standard error of $\bar { x } _ { 1 }$ - $\bar { x } _ { 2 }$ is
A) 12.9.
B) 9.3.
C) 4.
D) 2.

Question

A statistics teacher wants to see if there is any difference in the abilities of students enrolled in statistics today and those enrolled five years ago.A sample of final examination scores from students enrolled today and from students enrolled five years ago was taken.You are given the following information. $\begin{array}{ll}
&\text { Today } & \text { Five Years Ago } \
\bar {x}&82 & 88 \
\sigma^2&112.5 & 54 \
n&45 & 36
\end{array}$ 
The standard error of $\bar { x } _ { 1 }$ - $\bar { x } _ { 2 }$ is
A) 12.9.
B) 9.3.
C) 4.
D) 2.

Quizplus · Accepted Answer

The standard error of the difference between two means is calculated using the formula: $\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$. Substituting the given values: $\sqrt{\frac{112.5}{45} + \frac{54}{36}} = \sqrt{2.5 + 1.5} = \sqrt{4} = 2$.

A Statistics Teacher Wants to See If There Is Any Today Five Years Ago xˉ8288σ2112.554n4536\begin{array}{ll}&\text { Today } & \text { Five Years Ago } \\\bar {x}&82 & 88 \\\sigma^2&112.5 & 54 \\n&45 & 36\end{array}xˉσ2n​ Today 82112.545​ Five Years Ago 885436​

A Statistics Teacher Wants to See If There Is Any $\begin{array}{ll}&\text { Today } & \text { Five Years Ago } \\\bar {x}&82 & 88 \\\sigma^2&112.5 & 54 \\n&45 & 36\end{array}$