Question

A survey of 1027 random voters in Oregon revealed that 673 favor the approval of an issue before the state legislature. Construct a 95% confidence interval for the true population proportion of all voters in the state who favor approval. (Round all answers to two decimal places unless otherwise specified.)

Does this satisfy the conditions for constructing a confidence interval?

Select an answer Yes No There's conditions?

Identify ˆpp^ and zα2zα2.

ˆp=p^=

zα2=zα2=

Calculate the margin of error. (Round to three decimal places.)

E=E=

State the 95% confidence interval for the population proportion.(Round to three decimal places.)

<p<<p<

Answer 1

Solution :

n = 1027

x = 673

= x / n = 673 / 1027 = 0.655

1 - = 1 - 0.655 = 0.345

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.960

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

= 1.960* (((0.655* 0.345) / 1027 )

= 0.029

A 95 % confidence interval for population proportion p is ,

- E < P < + E

0.655- 0.029 < p < 0.655 + 0.029

0.626 < p < 0.684