A sample of n = 25 n=25 diners at a local restaurant had a mean lunch bill of $16 with a standard deviation of σ = $ 4 σ=$4 . We obtain a 95% confidence interval as ( 14.43 , 17.57 ) (14.43,17.57) . Which choice correctly interprets this interval?

( 14.43 , 17.57 ) correctly interprets this interval.

Step-by-step explanation:

mean = x`= $ 16

Standard deviation= S= $4

sample size= n= 25

Using the formula

x`± zₐ/₂  S/ √n

= 16 ± (1.96) 4/ √25

= 16 ± (1.96) 4/5

= 16 ± (1.96)0.8

= 16 ± (1.568)

= 16+ 1.568         ,       16-1.568

= 17.568             , 14.432

From 14.43 to 17.57

( 14.43 , 17.57 ) correctly interprets this interval.

If two samples are taken  and two standard deviations are considered then the answer would be completely different from 13.78 to 18.22

step-by-step explanation:

step-by-step explanation: