Question

in a study of speed​ dating, male subjects were asked to rate the attractiveness of their female​ dates, and a sample of the results is listed below ​(1equals=not ​attractive; 10equals=extremely ​attractive). Construct a confidence interval using a 9090​% confidence level. What do the results tell about the mean attractiveness ratings of the population of all adult​ females?

Answer 1

`7-1.796(2.216)/(√(12))=5.851`

`7+1.796(2.216)/(√(12))=8.149`

So on this case the 90% confidence interval would be given by (5.851;8.149)

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)

s represent the sample 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 t_(\alpha/2)(s)/(√(n))` (1)

Data: 7 8 2 10 6 5 7 8 8 9 5 9

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`

The sample deviation calculated
`s=2.216`

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=12-1=11`

Since the Confidence is 0.90 or 90%, the value of
`\alpha=0.1` and
`\alpha/2 =0.05`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,11)".And we see that
`t_(\alpha/2)=1.796`

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

`7-1.796(2.216)/(√(12))=5.851`

`7+1.796(2.216)/(√(12))=8.149`

So on this case the 90% confidence interval would be given by (5.851;8.149)