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

Answer:

$7.72-2.433\frac{2.5}{\sqrt{20}}=6.360$    

$7.72+2.433\frac{2.5}{\sqrt{20}}=9.080$    

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

Step-by-step 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:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (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):

$7.72-2.433\frac{2.5}{\sqrt{20}}=6.360$    

$7.72+2.433\frac{2.5}{\sqrt{20}}=9.080$    

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