The following explains the relationship between confidence interval, confidence level, and sample size.

There is a population following a normal distribution $N \left( m , \sigma ^ { 2 } \right)$. When a sample of size $n$ is randomly extracted from this population, the sample mean follows a normal distribution $\square$ (a).\\
Using the distribution of this sample mean, let the confidence interval for the population mean $m$ with confidence level $\alpha$ be $a \leqq m \leqq b$.\\
When the sample size is fixed at $n$ and the confidence level is set higher than $\alpha$, let the confidence interval be $c \leqq m \leqq d$. Then $d - c$ is $\square$ (b) than $b - a$.\\
On the other hand, when the confidence level is fixed at $\alpha$ and the sample size is $2 n$, let the confidence interval be $e \leqq m \leqq f$. Then $f - e$ is $\square$ (c) times $b - a$.

What are the correct values for (a), (b), and (c) in the above process? [3 points]

\begin{center}
\begin{tabular}{|l|l|l|}
\hline
 & (a) & (b) \\
\hline
(1) & $N \left( m , \sigma ^ { 2 } \right)$ & larger \\
\hline
(2) & $N \left( m , \sigma ^ { 2 } \right)$ & smaller \\
\hline
(3) & $N \left( m , \frac { \sigma ^ { 2 } } { n } \right)$ & smaller \\
\hline
(4) & $N \left( m , \frac { \sigma ^ { 2 } } { n } \right)$ & larger \\
\hline
(5) & $N \left( m , \frac { \sigma ^ { 2 } } { n } \right)$ & smaller \\
\hline
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{|l|l|}
\hline
 & (c) \\
\hline
(1) & $\frac{1}{2}$ \\
\hline
(2) & $\frac{1}{2}$ \\
\hline
(3) & $\frac{1}{\sqrt{2}}$ \\
\hline
(4) & $\sqrt{2}$ \\
\hline
(5) & $\frac{1}{\sqrt{2}}$ \\
\hline
\end{tabular}
\end{center}