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Question: A survey of 120 families showed that at least 78 had a computer at home. Find the 99% confidence interval of the true proportion of families who own a home computer.


A ) 64.5% < p < 65.5%
B ) 5.38% < p < 7.62%
C ) 53.8% < p < 76.2%
D ) 60.6% < p < 69.4%

Steps to derive

1 The problem is to find the true proportion of families who own a home computer

2 Sample proportion, pˆ = 78120 = 0.65 and qˆ = 1 - pˆ = 1 - 0.65 = 0.35

3 For 99% confidence interval, α = 1 - 0.99 = 0.01 and z/2 = 2.58
 [Use the standard normal distribution table.]

4 The 99% confidence interval for the true proportion is:
pˆ - z/2 pˆqˆn < p < pˆ + z/2 pˆqˆn
 [Formula.]

5 0.65 - 2.58(0.65)(0.35)120 < p < 0.65 - 2.58(0.65)(0.35)120

6 0.65 - 0.112 < p < 0.65 + 0.112

7 0.538 < p < 0.762

8 53.8% < p < 76.2%

9 So, we are fairly 99% confident, that the families who own a home computer is between 53.8% and 76.2%.



Hence the right answer is Option C

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