A college admissions officer for an MBA program has determined that historically
applicants have undergraduate grade point averages that are normally distributed with
standard deviation 0.45. From a random sample of 25 applications from the current year,
the sample mean grade point average is 2.90.
Required: Find the confidence interval foe the population mean at the 5% significance level.
(10marks)
B. An educational study was designed to investigate the effectiveness of a reading program
of elementary age children. Each child was given a pre-test and post-test. Higher posttest scores would indicate reading improvement. From a very large population, a random
sample of scores for the pre-test and post-test are as follows:

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".

Population standard deviation: σ = 0.45

Sample size: n = 25

Sample Mean: x= 2.90

Х=2.90 represent the sample mean for the sample

μ population mean (variable of interest)

σ = 0.45 represent the population standard deviation

n=25 represent the sample size

We have the following distribution for the random variable:

X≈N(μ,σ =0.45)

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

X≈N( μ, [tex]\frac{σ}{\sqrt{n} }[/tex])

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha =0.05[/tex] and[tex]\frac{\alpha }{2} =0.025[/tex], 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

Z₁ =±1.96

Now we have everything in order to replace into formula:

[tex]2.90-1.96\frac{0.45}{\sqrt{25} }=2.724[/tex]

[tex]2.90+1.96\frac{0.45}{\sqrt{25} }=3.076[/tex]

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

At a significance level, the population mean's confidence interval is 2.724–30.76.

To learn more about confidence interval visit:

brainly.com/question/24131141

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