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:

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

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 $Z = 1.96$ .

72% of Americans said that they had at least one credit card

This means that $p = 0.72$

Give the 95% margin of error for this estimate.

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

$M = 1.96\sqrt{\frac{0.72*0.28)}{1030}} = 0.0274$

The 95% margin of error for this estimate is 0.0274 = 2.74 percentage points.

7 0

3 years ago

A Truck weighs 3 tons how many pounds does it weigh?

Monica [59]

Answer:

6000 pounds

Step-by-step explanation:

5 0

2 years ago