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

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jmonterrozar jmonterrozar · Answer 1 · 2020-05-12T04:00:43+02:00

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)

AFOKE88 AFOKE88 · Answer 2 · 2022-03-29T14:45:05+02:00

The 90% confidence interval for p is; (0.4593, 0.5073)

The proportion is;

p = 580/1200 = 0.4833

Formula for confidence interval is given by:

CI = p ± z√(p(1 - p)/ n)

Now, z-score at 90% confidence interval is; z = 1.645

Thus;

CI = 0.4833 ± 1.645√(0.4833(1 - 0.4833)/1200)

CI = 0.4833 ± 0.024

CI = (0.4593, 0.5073)

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