Question

A polling company conducts an annual poll of adults about political opinions. The survey asked a random sample of 406 adults whether they think things in the country are going in the right direction or in the wrong direction. 45% said that things were going in the wrong direction. Complete parts a and b below. Calculate the margin of error for the proportion of all adults who think things are going in the wrong direction for 90% confidence. Would the margin of error be larger or smaller for 85% confidence? Explain.

Answer 1

Answer:

The margin of error or the proportion of all adults who think things are going in the wrong direction for 90% confidence is 0.0406, which is 4.06 percentage points.

At the 85% confidence level, the margin of error would be smaller due to the lower value of Z.

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:

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

90% confidence level

So $\alpha = 0.10$, z is the value of Z that has a pvalue of $1 - \frac{0.10}{2} = 0.95$, so $Z = 1.645$.

In this problem, we have that:

$n = 406, p = 0.45$

So

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

$M = 1.645\sqrt{\frac{0.45*0.55}{406}}$

$M = 0.0406$

The margin of error or the proportion of all adults who think things are going in the wrong direction for 90% confidence is 0.0406, which is 4.06 percentage points.

85% confidence level

So $\alpha = 0.15$, z is the value of Z that has a pvalue of $1 - \frac{0.15}{2} = 0.925$, so $Z = 1.44$.

With the lower confidence level, z is lower, which means that the margin of error would be smaller.