A survey conducted by General Motors of 38 drivers in America, 34 indicated that they would prefer a car with a sunroof over one without. When estimating the proportion of all Americans who would prefer a car with a sunroof over one without with 99% confidence, what is the margin of error

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Answer:

The margin of error is of 0.1282 = 12.82%.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].

The margin of error is of:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

A survey conducted by General Motors of 38 drivers in America, 34 indicated that they would prefer a car with a sunroof over one without.

This means that [tex]n = 38, \pi = \frac{34}{38} = 0.8947[/tex]

99% confidence level

So [tex]\alpha = 0.01[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].

What is the margin of error?

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]M = 2.575\sqrt{\frac{0.8947*0.1053}{38}} = 0.1282[/tex]

The margin of error is of 0.1282 = 12.82%.