Question

A newspaper poll on state budgetary issues interviewed 828 state residents. Of the residents surveyed, 470 of them felt that the state should balance the budget. Use the poll results to give a 95% confidence interval for p.

Answer 1

.5332 to .6020

Explanation:

p = proportion for favouring balancing budget = \frac{470}{828} =0.5676

std error of proportion = \sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.5676*0.4324}{828} } \\=0.0172

Confidence interval = 0.5676±1.96(0.0172)

= (0.5338, 0.6014)

Answer 2

Explanation:

Given that a newspaper poll on state budgetary issues interviewed 828 state residents.

Of the residents surveyed, 470 of them felt that the state should balance the budget.

p = proportion for favouring balancing budget =
`(470)/(828) =0.5676`

std error of proportion =
`\sqrt{(pq)/(n) } \\=\sqrt{(0.5676*0.4324)/(828) } \\=0.0172`

Confidence interval = 0.5676±1.96(0.0172)

= (0.5338, 0.6014)