Question

Suppose scores of a standardized test are normally distributed and have a known population standard deviation of 11 points and an unknown population mean. A random sample of 15 scores is taken and gives a sample mean of 101 points. Find the confidence interval for the population mean with a 98% confidence level.

Answer 1

Answer: 94.39, 107.61

Explanation:

Answer 2

`101-2.33(11)/(√(15))=94.382`

`101+2.33(11)/(√(15))=107.618`

So on this case the 98% confidence interval would be given by (94.382;107.618)

Explanation:

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=101` represent the sample mean

`\mu` population mean (variable of interest)

`\sigma=11` represent the population standard deviation

n=15 represent the sample size

Solution to the problem

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

`\bar X \pm z_(\alpha/2)(\sigma)/(√(n))` (1)

The point estimate of the population mean is
`\hat \mu = \bar X =11`

Since the Confidence is 0.98 or 98%, the value of
`\alpha=0.02` and
`\alpha/2 =0.01`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.01,0,1)".And we see that
`z_(\alpha/2)=2.33`

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

`101-2.33(11)/(√(15))=94.382`

`101+2.33(11)/(√(15))=107.618`

So on this case the 98% confidence interval would be given by (94.382;107.618)