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

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.