Question

Suppose you have the following X (a sample of visits to Chik Fil A in a week): (6, 5, 4, 1, 1). What is the mean of X? What is the Standard deviation of X? What is the Standard Error of X? Calculate the 95% confidence interval for X. Calculate the 99% confidence interval for X. Test the hypothesis that X=0. What is the null hypothesis? What is the t-statistic?

Answer 1

I have used excel to find the mean , standard deviation and standard error of the x .

First enter the given data set into column A of excelsheet.

To find mean use function =AVERAGE( data set column range )

To find standard deviation use function =STDEV( data set column range )

Standard error =   n is the count of the data set.

Therefore mean of x ; ( ) = 3.4

Standard deviation of x ; (s) = 2.3022

Standard error of x = 1.0296

Confidence interval for Mean :

it is given by

Lower bound =   - E and upper bound = + E

E is margin of error = t* standard error of x

t is the critical value follows t distribution with degrees of freedom (d.f ) = n - 1

We can find t using t distribution table for given confidence level and d.f

For 95% confidence level ,

We have n = 5 , so d.f = 5 - 1 = 4 ,so look for d.f = 4 and across confidence level c = 0.95 on the table.

So t = 2.776

For 99% confidence level ,

d.f = 4 and confidence level c = 0.99

t = 4.604

Therefore for 95% confidence interval  

E = 2.776* 1.0296 = 2.8582

Lower bound = 3.4 - 2.8582 = 0.5418

Upper bound = 3.4 + 2.8582 = 6.2582

So 95% confidence interval for mean is ( 0.5418 , 6.2582 )

For 99% confidence interval ;

E = 4.604 * 1.0296 = 4.7403

Lower bound = 3.4 - 4.7403 = -1.3403

Upper bound = 3.4 + 4.7403 = 8.1403

So 95% confidence interval for mean is ( -1.3403,8.1403)

Hypothesis test for mean :

Null hypothesis H0 : = 0  

t test statistic = under H0 ; = 0

We have = 3.4 and standard error = 1.0296

t =   = 3.30

Therefore t = 3.30