Question

estimated that the proportion of students who prefer 2 day a week classes to 1 day a week classes was somewhere in the interval (0.451, 0.872) . If a 95% confidence interval was used, find the point estimate of the proportion and the margin of error

Answer 1

Complete Question

A certain university estimated that the proportion of students who prefer 2 day a week classes to 1 day a week classes was somewhere in the interval (0.451, 0.872) . If a 95% confidence interval was used, find the point estimate of the proportion and the margin of error

Answer:

The point estimate is
`\^ p =0.6615`

The margin of error is
`E = 0.2105`

Explanation:

From the question we are that

The estimate of the proportion of students who prefer 2 day a week classes to 1 day a week classes is (0.451, 0.872)

Generally the point estimate is mathematically represented as

`\^ p = ( 0.872 + 0.451 )/(2)`

=>
`\^ p =0.6615`

Generally the margin of error is mathematically represented as

`E = (0.872 - 0.451)/(2)`

=>
`E = 0.2105`