Question

How does the confidence interval change with sample size?

Answer 1

Hint: We solve this problem by using the simple formula of confidence level for the given data.
The formula used to find the confidence interval of sample data is given as,
$CI=\bar{x}\pm Z\left( \dfrac{\sigma }{\sqrt{n}} \right)$
Where, $\bar{x}$ is the mean of the data, $Z$ is the confidence coefficient which is constant for respected confidence level (usually considered 1.96), $\sigma $ is the standard deviation and $n$ is the sample space.
By using the above formula we interpret the change of confidence level with the change of $'n'$

Complete step by step answer:
We are asked to find how the confidence interval changes with sample size.
We know that the formula used for the confidence interval is given as,
$CI=\bar{x}\pm Z\left( \dfrac{\sigma }{\sqrt{n}} \right)$
Where, $\bar{x}$ is the mean of the data, $Z$ is the confidence coefficient (usually considered 1.96), $\sigma $ is the standard deviation and $n$ is the sample space.
We know that the sample space and sample size represent the same quantity.
Here, we can see that if $'n'$ increases then, the confidence interval decreases because both $CI,n$ are inversely proportional as seen in the formula.
Now, let us check our result with help of an example.
Let us assume that some random data given as,
$\begin{align}
  & \bar{x}=20 \\
 & \sigma =6 \\
 & n=4 \\
\end{align}$
Now, let us find the value of confidence interval for this data then we get,
$\begin{align}
  & \Rightarrow CI=20\pm \left[ 1.96\left( \dfrac{6}{\sqrt{4}} \right) \right] \\
 & \Rightarrow CI=20\pm 5.88 \\
\end{align}$
Here, we can see that the range of confidence interval is given as $\left[ 20-5.88,20+5.88 \right]=\left[ 14.12,25.88 \right]$ such that the difference is $11.76$
Now, let us take other example having all the quantities same except sample size that is,
$\begin{align}
  & \bar{x}=20 \\
 & \sigma =6 \\
 & n=9 \\
\end{align}$
Now, let us find the value of confidence interval for this data then we get,
$\begin{align}
  & \Rightarrow CL=20\pm \left[ 1.96\left( \dfrac{6}{\sqrt{9}} \right) \right] \\
 & \Rightarrow CL=20\pm 3.92 \\
\end{align}$
Here, we can see that the range of confidence interval is given as $\left[ 20-3.92,20+3.92 \right]=\left[ 16.08,23.92 \right]$ such that the difference is $7.84$
Here, from the two examples we can clearly see that as the value of $n$ increases from 4 to 9 the confidence interval decreases from $11.76$ to $7.84$
Therefore, we can conclude that confidence interval and sample size are inversely proportional to each other.

Note: We need to note that we cannot decide anything from the formula directly. We can assume only from the formula but we need to cross check the result using the examples or experiments.
Here, we can see that the assumption that $CI,n$ are inversely proportional as seen in the formula. But, we cannot directly give an answer like that. We need to take at least two examples regarding the changes and conclude the data as we did in the solution.