A student polls his school to see if students in the school district are for or

Question

A student polls his school to see if students in the school district are for or against the new legislation regarding school uniforms. She surveys 600 students and finds that 480 are against the new legislation.

Interpret the confidence interval:

We estimate with 10% confidence that between 77.3% and 82.7% of ALL students are against the legislation.

We cannot estimate with 90% confidence that between 77.3% and 82.7% of ALL students are against the legislation.

We estimate with 90% confidence that between 77.3% and 82.7% of ALL students are against the legislation.

We estimate with 90% confidence that between 77.3% and 82.7% of ALL students are for the legislation

We estimate with 10% confidence that between 77.3% and 82.7% of ALL students are against the legislation.

We cannot estimate with 90% confidence that between 77.3% and 82.7% of ALL students are against the legislation.

We estimate with 90% confidence that between 77.3% and 82.7% of ALL students are against the legislation.

We estimate with 90% confidence that between 77.3% and 82.7% of ALL students are for the legislation

Explanation / Answer

Here P = 480/600 = 0.80

E(p) = P =0.80

Var(p) = [P(1 - P) / n] = [0.8(1 - 0.8) / 600] = 1 / 3750

The 90% confidence interval will be given by [E(p) - Z(0.10/2) . S.E.(p) , E(p) + Z(0.10/2) . S.E.(p) ]

[0.80 - 1.6449 . 0.0163 , 0.80 + 1.6449 . 0.0163]

[ 0.773, 0.827 ]

This means that we can estimate with 90% confidence that between 77.3% and 82.7% of all students are against the legislation.

Hence the 3rd option is the correct option.

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