Suppose that n units are randomly sampled and x number of the sampled units are found to have the characteristic of interest. A survey of n

Question

Suppose that n units are randomly sampled and x number of the sampled units are found to have the characteristic of interest. A survey of n = 540 pet owners revealed that x = 243 buy their pets holiday presents. For p = proportion of pet owners who revealed that they buy their pets holiday presents, provide a point estimate of p and determine its 95% error margin. Carry out all calculations exactly, round the final answers only. Point estimate =

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Eliza 1 week 2021-10-08T08:03:58+00:00 1 Answer 0

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    2021-10-08T08:05:04+00:00

    Answer:

    The point estimate is 0.45.

    The 95% error margin is 0.042 = 4.2 percentage points.

    Step-by-step explanation:

    In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

    \pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

    In which

    z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

    The margin of error is given by:

    M = z\sqrt{\frac{\pi(1-\pi)}{n}}

    Point estimate

    We have that n = 540, x = 243

    So the point estimate is:

    \pi = \frac{243}{540} = 0.45

    95% confidence level

    So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

    Error margin:

    M = z\sqrt{\frac{\pi(1-\pi)}{n}}

    M = 1.96\sqrt{\frac{0.45*0.55}{540}}

    M = 0.0420

    The 95% error margin is 0.042 = 4.2 percentage points.

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45:7+7-4:2-5:5*4+35:2 =? ( )