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.

Question

06-12-2019
Mathematics

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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.

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jtellezd jtellezd · Answer 1 · 2019-12-07T19:24:14+01:00

Answer:

c) 95% of all random samples of customers will show that 82% to 94% oders arrive on time

Step-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