Question

In: Statistics and Probability

One of the student recruiters at UCW has surveyed 160 students regarding their interest to stay...

One of the student recruiters at UCW has surveyed 160 students regarding their interest to stay in Canada after graduation. 135 of the students stated that they would like to stay in Canada.

Construct the 95% confidence interval about the true proportion of all of the UCW students who would consider staying in Canada after graduation.

Calculate how many students he should survey to have the result with less than 0.05 error.

Expert Solution

Solution :

Given that,

a) Point estimate = sample proportion = = x / n = 135 / 160 = 0.844

1 - = 1 - 0.844 = 0.156

Z/2 = Z0.025 = 1.96

Margin of error = E = Z / 2 * (( * (1 - )) $\sqrt$ / n)

= 1.96 ( $\sqrt$ ((0.844 * 0.156) / 160)

= 0.056

A 95% confidence interval for population proportion p is ,

± E

= 0.844 ± 0.056

= ( 0.788, 0.900 )

b) margin of error = E = 0.05

sample size = n = (Z / 2 / E )2 * * (1 - )

= (1.96 / 0.05)2 * 0.844 * 0.156

= 202.32

sample size = n = 203

orchestra answered 2 years ago

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