The average spending at Neco's salad bar is $8.73 with a standard deviation of $3.41. The distribution follows t-distribution. The management is interested in the middle 90% of the customers (spending wise) as it believes that they represent their true customer base. What will be the difference between the upper and lower spending cut-offs which define the middle 90% of the customers if the sample contains 41 customers

Difference between upper and lower limits is : 1,816

Step-by-step explanation:

A CI (confidence interval ) for  t student distribution is:

( μ₀  - t(α/2)* s/√n  ;    μ₀  + t(α/2)* s/√n )

Where:

μ₀  is the mean and s the standard deviation of the dstribution

n size of the sample

CI = 90 %     means   α = 10 %    α = 0,1      α/2  = 0,05

and degree of freedom   df = n - 1   df = 40

From t student table we get:

tα/2  =  1,6839

Then:

t(α/2)* s/√n  =  1,6839* 3,41/√40

t(α/2)* s/√n  =  0,908

8,73 - 0,908   =  7,822

8,73 + 0,908  = 9,638

CI (90%)  =  ( 7,822  ;  9,638 )

Difference between upper and lower cut-offs points is:

Δ = 1,816

this is false ^.^ lol