A travel agent wants to estimate the proportion of vacationers who plan to travel outside United States in the next 12 months. A random sample of 130 vacationers revealed that 40 had plans for foreign travel in that time frame. Construct a 95% confidence interval estimate of the population proportion. Make a statement about this in context of the problem.
Solution :
Given that,
n = 130
x = 40
= x / n = 40/130 = 0.31
1 - = 1 - 0.31 = 0.69
At 95% confidence level the z is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
Z/2 = Z0.025 = 1.96
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.96 * (((0.31 * 0.69) / 130))
= 0.08
A 90% confidence interval for population proportion p is ,
- E < P < + E
0.31 - 0.08 < p < 0.31 + 0.08
0.23 < p < 0.39
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