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]
| (a) | (b) |
| (1) | $N \left( m , \sigma ^ { 2 } \right)$ | larger |
| (2) | $N \left( m , \sigma ^ { 2 } \right)$ | smaller |
| (3) | $N \left( m , \frac { \sigma ^ { 2 } } { n } \right)$ | smaller |
| (4) | $N \left( m , \frac { \sigma ^ { 2 } } { n } \right)$ | larger |
| (5) | $N \left( m , \frac { \sigma ^ { 2 } } { n } \right)$ | smaller |
| (c) |
| (1) | $\frac{1}{2}$ |
| (2) | $\frac{1}{2}$ |
| (3) | $\frac{1}{\sqrt{2}}$ |
| (4) | $\sqrt{2}$ |
| (5) | $\frac{1}{\sqrt{2}}$ |