Question

use the information below to answer the questions:
the data in the accompanying bar graph are from a random sample of 890 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.
part a
find the margin of error for this survey. round to the hundredth.
part b
complete the statement about the percentage of adults in the countrys population who favor a ban on high - capacity gun violence. (round to the hundredth)
there is a 95% confidence that between % and % of all adults in the country favor a ban on high - capacity.

Explanation:

Step1: Calculate the margin of error

The formula for margin of error \(E\) for a proportion in a sample is \(E = z\sqrt{\frac{p(1 - p)}{n}}\). For a 95% confidence interval, \(z = 1.96\). Let's assume \(p\) (the proportion in the sample) is \(0.68\) (since the sample proportion for high - capacity ban is \(68\%\) or \(0.68\)) and \(n=890\).

First, calculate \(\frac{p(1 - p)}{n}=\frac{0.68\times(1 - 0.68)}{890}=\frac{0.68\times0.32}{890}=\frac{0.2176}{890}\approx0.0002445\)

Then, \(\sqrt{\frac{p(1 - p)}{n}}\approx\sqrt{0.0002445}\approx0.0156\)

And \(E = 1.96\times0.0156\approx0.0306\) or \(3.06\%\)

Step2: Calculate the confidence interval

The confidence interval for a proportion is \(\hat{p}-E

The lower limit is \(0.68-0.0306 = 0.6494\) (or \(64.94\%\))

The upper limit is \(0.68 + 0.0306=0.7106\) (or \(71.06\%\))

Answer:

Part A: The margin of error is approximately \(3.06\%\)
Part B: There is a 95% confidence that between \(64.94\%\) and \(71.06\%\) of all adults in the country favor a ban on high - capacity.