Question

A study on students drinking habits asks a random sample of 40 "greek" (ones who belong to a fraternity or sorority) UF students how many alcoholic beverages they have consumed in the past week. The sample reveals an average of 6.56 alcoholic drinks, with a standard deviation of 6.25. Construct a 90% confidence interval for the true average number of alcoholic drinks all UF "greek" students have in a one week period. A.(-3.72, 16.84) B.(4.01, 9.11) C.(4.62, 8.50) D.(4.90, 8.23)

Answer 1

Answer:

$df=n-1=40-1=39$

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(0.05,39)".And we see that $t_{\alpha/2}=1.68$

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

$6.56-1.68\frac{6.25}{\sqrt{40}}=4.90$    

$6.56+1.68\frac{6.25}{\sqrt{40}}=8.23$    

So on this case the 90% confidence interval would be given by (4.90;8.23)    

Step-by-step 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".

$\bar X=6.56$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s=6.25 represent the sample standard deviation

n=40 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:

$df=n-1=40-1=39$

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(0.05,39)".And we see that $t_{\alpha/2}=1.68$

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

$6.56-1.68\frac{6.25}{\sqrt{40}}=4.90$    

$6.56+1.68\frac{6.25}{\sqrt{40}}=8.23$    

So on this case the 90% confidence interval would be given by (4.90;8.23)