Question

A random sample of vacationers were asked whether they were traveling to another country for their upcoming trip. The resulting confidence interval for the proportion of vacationers traveling abroad is (0.14,0.16). What is the margin of error?

Answer 1

The Margin of error = 0.01

Explanation:

Explanation:-

step(i):-

Given confidence interval for the proportion of vacationers traveling abroad

(0.14,0.16)

The 95% of confidence interval for Population proportion with margin of error is determined by

( p⁻ - M.E , p⁻ + M.E)

step(ii):-

The margin of error is determined by

`M.E = Z_(\alpha ) \sqrt{(p(1-p))/(n) }`

Given Confidence interval is ( 0.14 , 0.16 )

Now

(( p⁻ - M.E , p⁻ + M.E) = (0.14,0.16)

Equating

p⁻ - M.E = 0.14 ...(i)

p⁻ + M.E = 0.16 ...(ii)

Solving (i) and (ii) equations , we get

p⁻ - M.E = 0.14

p⁻ + M.E = 0.16

- - -

- 2 M.E = -0.02

M.E = 0.01

The margin of error = 0.01

Conclusion:-

The margin of error = 0.01