An admissions officer has determined that the population of applicants to the MBA program has undergraduate GPA' that are approximately normally distributed with standard deviation 0.45. A random sample of 25 applicants for next fall has a sample mean GPA of 3.30. Find the 95% confidence interval for the mean GPA among applicants to this MBA.

Solution

Given that sample size $n = 25$, sample mean $\overline{X}= 3.3$ and population standard deviation $\sigma = 0.45$.

Step 1 Specify the confidence level $(1-\alpha)$

Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.

Step 2 Given information

Given that sample size $n =25$, sample mean $\overline{X}=3.3$ and population standard deviation is $\sigma = 0.45$.

Step 3 Specify the formula

$100(1-\alpha)$% confidence interval for the population mean $\mu$ is

$$ \begin{aligned} \overline{X} - E \leq \mu \leq \overline{X} + E \end{aligned} $$

where $E = Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$, and $Z_{\alpha/2}$ is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

Step 4 Determine the critical value

The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.

Thus $Z_{\alpha/2} = Z_{0.025} = 1.96$.

Z-critical0.05
Z-critical0.05

Step 5 Compute the margin of error

The margin of error for mean is

$$ \begin{aligned} E & = Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}\\ & = 1.96 \frac{0.45}{\sqrt{25}} \\ & = 0.176. \end{aligned} $$

Step 6 Determine the confidence interval

$95$ % confidence interval estimate for population mean is

$$ \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 3.3 - 0.176 & \leq \mu \leq 3.3 + 0.176\\ 3.124 & \leq \mu \leq 3.476. \end{aligned} $$

Thus, $95$% confidence interval for the mean GPA among applicants to this MBA is $(3.124,3.476)$.

Further Reading