A survey was given to a random sample of 195 residents of a town to determine whether they support a new plan to raise taxes in order to inc

Question

A survey was given to a random sample of 195 residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. Of those surveyed, 117 respondents said they were in favor of the plan. Determine a 95% confidence interval for the proportion of people who favor the tax plan.

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Dulcie 5 days 2021-07-18T22:52:35+00:00 1 Answers 0 views 0

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    2021-07-18T22:54:30+00:00

    Answer:

    (0.5415 ; 0.6785)

    Step-by-step explanation:

    Sample size, n = 195

    Number of samples who werw in favor, x = 117

    Phat = 117 / 195 = 0.609 = 0.61

    The confidence interval :

    Phat ± Margin of error

    Margin of Error = Zcritical * √(phat(1-phat)/n)

    Zcritical at 95% = 1.96

    1 – phat = 1 – 0.61 = 0.39

    Margin of Error = 1.96 * √((0.61*0.39)/195) = 0.0685

    Confidence interval :

    Lower boundary : 0.61 – 0.0685 = 0.5415

    Upper boundary : 0.61 + 0.0685 = 0.6785

    (0.5415 ; 0.6785)

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