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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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Answer:

The point estimate is 0.45.

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

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{(\pi(1-\pi))/(n)}

In which

z is the zscore that has a pvalue of
1 - (\alpha)/(2).

The margin of error is given by:


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

Point estimate

We have that
n = 540, x = 243

So the point estimate is:


\pi = (243)/(540) = 0.45

95% confidence level

So
\alpha = 0.05, z is the value of Z that has a pvalue of
1 - (0.05)/(2) = 0.975, so
Z = 1.96.

Error margin:


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


M = 1.96\sqrt{(0.45*0.55)/(540)}


M = 0.0420

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

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