Question

In a study of the accuracy of fast food​ drive-through orders, Restaurant A had 286 accurate orders and 70 that were not accurate. a. Construct a 90​% confidence interval estimate of the percentage of orders that are not accurate. b. Compare the results from part​ (a) to this 90​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B: 0.176less thanpless than0.248. What do you​ conclude?

Answer 1

a.
TRADITIONAL METHOD
given that,
possibile chances (x)=70
sample size(n)=356
success rate ( p )= x/n = 0.197
I.
sample proportion = 0.197
standard error = Sqrt ( (0.197*0.803) /356) )
= 0.021
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
margin of error = 1.645 * 0.021
= 0.035
III.
CI = [ p ± margin of error ]
confidence interval = [0.197 ± 0.035]
= [ 0.162 , 0.231]
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DIRECT METHOD
given that,
possibile chances (x)=70
sample size(n)=356
success rate ( p )= x/n = 0.197
CI = confidence interval
confidence interval = [ 0.197 ± 1.645 * Sqrt ( (0.197*0.803) /356) ) ]
= [0.197 - 1.645 * Sqrt ( (0.197*0.803) /356) , 0.197 + 1.645 * Sqrt ( (0.197*0.803) /356) ]
= [0.162 , 0.231]
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interpretations:
1. We are 90% sure that the interval [ 0.162 , 0.231] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
Answer:
90​% confidence interval estimate of the percentage of orders that are not accurate = 0.162,0.231
= 16.2%,23.1%

b.
Compare the results from part​ (a) to this 90​% confidence interval for the percentage of orders that are not accurate
at Restaurant​ B: 0.176less thanpless than0.248 = (0.176,0.248)
At restaurent A confidence interval = [0.162 , 0.231]
we conclude that both restaurents A and B different percentage of orders that are not accurate.
restaurent A is lee than restaurent B.