Question

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A travel agent wants to estimate the proportion of vacationers who plan to travel outside the 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

Answer 1

Solution :

Given that,

n = 130

x = 40

Point estimate = sample proportion = = x / n = 40 / 130 = 0.308

1 - = 1 - 0.308 = 0.692

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 * (((0.308 * 0.692) / 130)

= 0.080

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.308 - 0.080 < p < 0.308 + 0.080

0.228 < p < 0.387

The 95% confidence interval for the population proportion p is : (0.228 , 0.387)