The two samples whose statistics are given in the table are thought to come from populations with equal variances. What is the pooled estimate of the population standard deviation?

$\begin{array}{|c|c|c|}
\hline \boldsymbol{n} & \text {Mean} & \text {SD} \
\hline 12 & 45 & 6 \
\hline 16 & 41 & 8 \
\hline
\end{array}$

A) 10
B) 7.22
C) 7.14
D) 2.65
E) 7

Question

The two samples whose statistics are given in the table are thought to come from populations with equal variances. What is the pooled estimate of the population standard deviation?

$\begin{array}{|c|c|c|}
\hline \boldsymbol{n}  & \text {Mean} & \text {SD} \
\hline 12 & 45 & 6 \
\hline 16 & 41 & 8 \
\hline
\end{array}$

A) 10
B) 7.22
C) 7.14
D) 2.65
E) 7

Quizplus · Accepted Answer

The pooled estimate of the population standard deviation is calculated using the formula: $\sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}}$, where $n_1$ and $n_2$ are the sample sizes, and $s_1$ and $s_2$ are the sample standard deviations. Plugging in the given values: $\sqrt{\frac{(12-1)6^2 + (16-1)8^2}{12+16-2}} = \sqrt{\frac{11*36 + 15*64}{26}} = \sqrt{\frac{396 + 960}{26}} = \sqrt{\frac{1356}{26}} = \sqrt{52.15} \approx 7.22$.

The Two Samples Whose Statistics Are Given in the Table nMeanSD1245616418\begin{array}{|c|c|c|}\hline \boldsymbol{n} & \text {Mean} & \text {SD} \\\hline 12 & 45 & 6 \\\hline 16 & 41 & 8 \\\hline\end{array}n1216​Mean4541​SD68​​

The Two Samples Whose Statistics Are Given in the Table $\begin{array}{|c|c|c|}\hline \boldsymbol{n} & \text {Mean} & \text {SD} \\\hline 12 & 45 & 6 \\\hline 16 & 41 & 8 \\\hline\end{array}$