A newspaper conducted a statewide survey concerning a proposal to raise taxes to prevent budget cuts to education. The newspaper took a rand

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

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1 week 2021-10-05T21:51:11+00:00 1 Answer 0

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    2021-10-05T21:52:48+00:00

    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)

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