Question

A college admissions officer for an MBA program has determined that historically,candidates have undergraduate grade point averages (GPA) that are normally distributed with standard deviation 0.45. A random sample of twenty five applicants from the current year is taken, yielding a sample mean GPA of 2.90. Find a 95% convidence interval for the population mean.

Answer 1

Answer:

$2.724< \mu$

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=2.90$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

$\sigma=0.45$ represent the population standard deviation

n=25 represent the sample size  

We have the following distribution for the random variable:

$X \sim N(\mu , \sigma=0.45)$

And by the central theorem we know that the distribution for the sample mean is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

2) Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=\pm 1.96$

Now we have everything in order to replace into formula (1):

$2.90-1.96\frac{0.45}{\sqrt{25}}=2.724$    

$2.90+1.96\frac{0.45}{\sqrt{25}}=3.076$

So on this case the 95% confidence interval would be given by (2.724;3.076)    

$2.724< \mu$