Question

A newspaper conducted a statewide survey concerning a proposal to raise taxes to prevent budget cuts to education. The newspaper took a random sample (assume it is an SRS) of 1200 registered voters and found that 580 would vote to raise taxes. Let p represent the proportion of registered voters in the state that would vote to raise taxes. A 90% confidence interval for p is

Answer 1

Answer:

(0.4593, 0.5073)

Step-by-step explanation:

The first thing we must do is calculate p:

p = 580/1200 = 0.4833

now 90% confidence interval is given by:

p + - (z alpha / 2) * (p * (1-p) / n) ^ (1/2)

Now, at 90% confidence level the t is:

alpha = 1 - 90% = 1 - 0.90 = 0.10

alpha / 2 = 0.10 / 2 = 0.05

z (0.05) = 1.645

replacing these values we are left with:

p + - (1.645) * (0.4833 * (1- 0.4833) / 1200) ^ (1/2)

0.4833 + - 0.024

the interval would be:

0.4833 + 0.024 = 0.5073

0.4833 - 0.024 = 0.4593

(0.4593, 0.5073)