Q:

A catalog company promises to deliver online orders within 3 days of the order being placed. Follow-up calls were made to randomly selected customers to see whether their orders arrived on time (within 3 days). A 95% confidence interval for percentage of on-time arrival is 88% +/- 6%. What does this mean? Are these conclusions correct? Explain. a) 95% of all random samples of customers will show that 88% of orders arrive on time. b) 95% of all random samples of customers will show that 82% to 94% of orders arrive on time. c) we are 95% sure that between 82% and 94% of the orders placed by the sampled customers arrived on time. d) on 95% of the days, between 82% and 94% of the orders will arrive on time.

Accepted Solution

A:
Answer:c) 95% of all random samples of customers will show that 82% to 94% oders arrive on timeStep-by-step explanation:From problem statement   we take 88 as a mean of a normal distribution, in that case we know, about relation between mean and standard deviation μ  ± σ        ⇒ [ μ  -  σ ;   μ  +  σ ]   ⇒  [  88  - 6  ;  88  + 6 ]  ⇒ [  82  ; 94 ]The above mentioned interval get 95.7 % of all values for a normal distribution, so we must be sure that a 95 % of all random samples of customers will show that 82% to 94% of orders arrive on time