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Question: A survey of 156 people showed that 104 go for a brisk walk daily. Find the 99% confidence interval of the true proportion of people who go for a brisk walk.


A ) 57.3% < p < 76.7%
B ) 63.2% < p < 70.8%
C ) 66.6% < p < 67.4%
D ) 66% < p < 70%

Steps to derive

1 The problem is to find 99% confidence interval for the true proportion (p) of people, who go for a brisk walk.

2 Sample proportion, pˆ = xn = 104156 = 0.67, qˆ = 1 - pˆ = 1 - 0.67 = 0.33

3 For 99% confidence interval, zα/2 is 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

5 0.67 -2.58(0.67)(0.33)156 < p < 0.67 +2.58(0.67)(0.33)156

6 0.67 - 0.097 < p < 0.67 + 0.097

7 0.573 < p < 0.767

8 57.3% < p < 76.7%

9 So, one can be 99% confident that the true proportion of people who go for a brisk walk is between 57.3% and 76.7%.



Hence the right answer is Option A

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