Question

Among 20 olden hamster litters recorded, there was a sample mean of baby hamsters, with a sample standard deviation of. Create a 97.5 confidence interval for the mean number of baby hamsters per liter

Answer 1

`7.72-2.433(2.5)/(√(20))=6.360`

`7.72+2.433(2.5)/(√(20))=9.080`

So on this case the 97.5% confidence interval would be given by (6.360;9.080)

Explanation:

For this case we assume the deviation given as s =2.5 and the mean
`\bar = 7.72`

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=7.72` represent the sample mean

`\mu` population mean (variable of interest)

s=2.5 represent the sample standard deviation

n=20 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)

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

`df=n-1=20-1=19`

Since the Confidence is 0.975 or 97.5%, the value of
`\alpha=0.025` 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: "=-T.INV(0.0125,19)".And we see that
`t_(\alpha/2)=2.433`

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

`7.72-2.433(2.5)/(√(20))=6.360`

`7.72+2.433(2.5)/(√(20))=9.080`

So on this case the 97.5% confidence interval would be given by (6.360;9.080)