SOLUTION: In a random sample of 35 students at a local college, 19 regularly carry a laptop computer to class. Using this and a margin of error of 12.1%, construct and interpret an interval

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Question 868426: In a random sample of 35 students at a local college, 19 regularly carry a laptop computer to class. Using this and a margin of error of 12.1%, construct and interpret an interval estimate for the population of all students at the college who carry a laptop computer to class.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
The sample proportion is p-hat = x/n where x is the number of successes and n is the sample size
There are 19 successes (ie 19 people who have a laptop). So x = 19
There are 35 students in the sample. So n = 35.


p-hat = x/n
p-hat = 19/35
p-hat = 0.54285714285714

To find the interval estimate or the confidence interval, we just add and subtract the margin of error to/from the p-hat value to get the upper and lower bounds (U and L)

Upper Bound:

U = (p-hat) + (Margin Of Error)
U = (0.54285714285714) + (0.121)
U = 0.66385714285713

Lower Bound:

L = (p-hat) - (Margin Of Error)
L = (0.54285714285714) - (0.121)
L = 0.42185714285713

So the interval estimate or the confidence interval is from 0.42185714285713 to 0.66385714285713