Question

A sample of $50$ days showed that a fast food restaurant serves an average of $182$ customers during lunch (between $11$ am- $2$ pm). The standard deviation of the sample is $8$. Find the $95\%$ confidence interval for the mean?

Answer 1

Hint: The confidence interval is given by the formula \[\Rightarrow CI=\overline{x}\pm E\], where $\overline{x}$ is the mean, which is given to be equal to $182$, and $E$ is the margin of error given by $E={{z}_{\alpha /2}}\dfrac{\sigma }{\sqrt{n}}$. Here $\sigma $ is the standard deviation, given to be $8$, n is the size of the sample, which is equal to $50$, and $\alpha $ is the critical value given by $\alpha =1-CL$. CL is equal to the confidence level, which is given to be equal to $95\%$ or $0.95$.

Complete step by step solution:
According to the question, the mean or the average value of the customers is equal to $182$. Therefore, we can write the mean as
$\Rightarrow \overline{x}=182.......\left( i \right)$
The number of days is equal to $50$, which means the value of n is
$\Rightarrow n=50........\left( ii \right)$
The standard deviation of the sample is equal to $8$, which means that
$\Rightarrow \sigma =8.......\left( iii \right)$
Now, we know that the confidence interval is given by
\[\Rightarrow CI=\overline{x}\pm E.........\left( iv \right)\]
Where E is the margin of error given by
$\Rightarrow E={{z}_{\alpha /2}}\dfrac{\sigma }{\sqrt{n}}.......\left( v \right)$
And the significance level is given by
$\Rightarrow \alpha =1-CL$
Now, according to the question, the confidence level is equal to $95%$. Therefore, we substitute $CL=0.95$ in the above equation to get
$\begin{align}
  & \Rightarrow \alpha =1-0.95 \\
 & \Rightarrow \alpha =0.05 \\
\end{align}$
Now, substituting this in the equation (v) we get
\[\begin{align}
  & \Rightarrow E={{z}_{0.05/2}}\dfrac{\sigma }{\sqrt{n}} \\
 & \Rightarrow E={{z}_{0.025}}\dfrac{\sigma }{\sqrt{n}} \\
\end{align}\]
Putting the value of $n$ and $\sigma $ from equations (ii) and (iii) we get
\[\begin{align}
  & \Rightarrow E={{z}_{0.025}}\dfrac{8}{\sqrt{50}} \\
 & \Rightarrow E={{z}_{0.025}}\dfrac{8}{5\sqrt{2}} \\
 & \Rightarrow E={{z}_{0.025}}\dfrac{4}{5}\sqrt{2} \\
\end{align}\]
From the normal table, we get ${{z}_{0.025}}=1.96$. Putting this above, we get
$\Rightarrow E=1.96\times \dfrac{4}{5}\sqrt{2}$
On solving, we get
$\Rightarrow E=2.22$
Finally, substituting this in the equation (iv) we get
$\Rightarrow CI=\overline{x}\pm 2.22$
Substituting (i) we get
$\begin{align}
  & \Rightarrow CI=182\pm 2.22 \\
 & \Rightarrow CI=\left[ 179.78,184.22 \right] \\
\end{align}$

Hence, the confidence interval for the mean is $\left[ 179.78,184.22 \right]$

Note: For calculating the critical value ${{E}_{\alpha /2}}$, we need to have access to the normal distribution table and also we must be able to read the values from it. The solution involves the use of many formulae, each being related to each other. In case we forget anyone of these, we will not be able to solve the question.