In a study of the accuracy of fast food drive-through orders, restaurant a had 277 accurate orders and 67 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: 0.174less thanpless than0.244. what do you conclude?

Step-by-step explanation:

Given that in a study of the accuracy of fast food drive-through orders, Restaurant A had 277 accurate orders and 67 that were not accurate.

Sample proportion for non accurate =

Std error =

Margin of error = 1.645 * std error

Confidence Interval:

±4.23

Range for the true population proportion:

19.96% to 28.42%

b) 17.40 to 24.40 is given. Mid value is 20.9% which must have been the sample proportion.

24.19% observed proportion lies in both the intervals.

10 down (bang1)

5 up

5down (2)

2.5 up

2.5 down (3)

1.25 up

1.25 down (4)

now add them up : )

hoped i lol

exactly you go it

step-by-step explanation: