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 =

Answer 1

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.