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

`2.724< \mu <3.076`

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, (\sigma)/(√(n)))`

2) Confidence interval

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

`\bar X \pm z_(\alpha/2)(\sigma)/(√(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(0.45)/(√(25))=2.724`

`2.90+1.96(0.45)/(√(25))=3.076`

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

`2.724< \mu <3.076`