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

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 , and a confidence level of , we have the following confidence interval of proportions. In which

z is the zscore that has a pvalue of .

The margin of error is given by: Point estimate

We have that So the point estimate is: 95% confidence level

So , z is the value of Z that has a pvalue of , so .

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