Question

The data below represent a random sample of 9 scores on a statistics quiz. (The maximum possible score on the quiz is 10.) Assume that the scores are normally distributed with a standard deviation of 2.3. Estimate the population mean with 95% confidence. 8,10,8,4,5,7,3,10,8 Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits. Confidence Interval

Answer 1

`7-1.96(2.3)/(√(9))=5.497`

`7+1.96(2.3)/(√(9))=8.503`

So on this case the 95% confidence interval would be given by (5.497;8.503)

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` represent the sample mean for the sample

`\mu` population mean (variable of interest)

`\sigma=2.3` represent the population standard deviation

n 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)

In order to calculate the mean and the sample deviation we can use the following formulas:

`\bar X= \sum_(i=1)^n (x_i)/(n)` (2)

`s=\sqrt{(\sum_(i=1)^n (x_i-\bar X))/(n-1)}` (3)

The mean calculated for this case is
`\bar X=7`

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)=1.96`

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

`7-1.96(2.3)/(√(9))=5.497`

`7+1.96(2.3)/(√(9))=8.503`

So on this case the 95% confidence interval would be given by (5.497;8.503)