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Question: A sample of 300 tickets purchased during vacation at a travelling center included 45 children's tickets. Find the 95% confidence interval for the true proportion of children, who were traveling during the vacation.


A ) 11% < p <19%
B ) 13% < p <17%
C ) 12.9% < p <17.1%
D ) 15% < p < 85%

Steps to derive

1 The problem is to find 95% confidence interval for the true proportion (p) of children, who were traveling during the vacation.

2 Sample proportion, pˆ = 45300 = 0.15 and qˆ = 1 - 0.15 = 0.85

3 For a 95% confidence interval, α = 1 - 0.95 = 0.05 and z∝/2 = 1.96
 [Use the standard normal distribution table.]

4 The 95% confidence interval for the true proportion is:
pˆ - z∝/2 pˆqˆn < p < pˆ + z∝/2 pˆqˆn

5 0.15 - 1.96(0.15)(0.85)300 < p < 0.15 +1.96(0.15)(0.85)300

6 0.15 - 0.040 < p < 0.15 + 0.040

7 0.110 < p < 0.190

8 11% < p < 19%

9 So, we are fairly 95% confident that the true percentage of children traveling during the vacation is between 11% and 19%.



Hence the right answer is Option A

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