Question

the data in the accompanying bar graph are from a random sample of 881 adults. the graph shows four proposals to reduce gun violence in a country and the percentage of surveyed adults who favored each of these proposals. complete parts (a) and (b) below. a. find the margin of error for this survey. the margin of error is ±□%. (do not round until the final answer. then round to the nearest tenth as needed.) b. write a statement about the percentage of adults in the countrys population who favor to restrict ammunition purchases to reduce gun violence. there is 95% confidence that between □% and □% of all adults in the country favor to restrict ammunition purchases. (use ascending order. round to the nearest tenth as needed.)

Explanation:

Step1: Calculate the margin of error

The formula for the margin of error \(E\) for a proportion in a simple random sample is \(E = \frac{1}{\sqrt{n}}\), where \(n\) is the sample size. Here, \(n = 881\).
So, \(E=\frac{1}{\sqrt{881}}\).

Step2: Simplify the expression

\(\sqrt{881}\approx29.68\), then \(E=\frac{1}{29.68}\approx0.0337\).
To convert this to a percentage, we multiply by \(100\). So \(E = 3.37\%\approx3.4\%\).

Step3: For part (b)

The sample proportion \(p = 54\%\) (from the bar - graph for "Restrict amount or type of ammunition purchases"). The confidence interval is given by \((p - E,p + E)\).
Substitute \(p = 54\) and \(E = 3.4\) into the formula.
\(p - E=54 - 3.4 = 50.6\) and \(p + E=54+3.4 = 57.4\)

Answer:

a. The margin of error is \(\pm3.4\%\)
b. There is \(95\%\) confidence that between \(50.6\%\) and \(57.4\%\) of all adults in the country favor to restrict ammunition purchases.