Solved (Free): An admissions officer has determined that the population of applicants to the MBA program has undergraduate GPA that are approximately

ByDr. Raju Chaudhari

Mar 31, 2021

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

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 $(3.124,3.476)$.