Question

bar graph
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 to reduce 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: Recall the formula for margin of error

The formula for margin of error \(E\) for a proportion in a simple random sample is \(E = \frac{1}{\sqrt{n}}\) when using a \(95\%\) confidence level (a common - sense approximation for large samples). Here, \(n = 890\).

Step2: Calculate the margin of error

Substitute \(n = 890\) into the formula \(E=\frac{1}{\sqrt{n}}\). So, \(E=\frac{1}{\sqrt{890}}\).
We know that \(\sqrt{890}\approx29.83\). Then \(E=\frac{1}{29.83}\approx0.0335\). Converting to a percentage, \(E = 3.35\%\).

Step3: For part B

The confidence interval for a proportion \(\hat{p}\) is \(\hat{p}-E\) to \(\hat{p} + E\). The proportion of adults who favor a ban on high - capacity (\(\hat{p}=66\%\)) and \(E = 3.35\%\)
The lower limit of the confidence interval is \(66 - 3.35=62.65\%\)
The upper limit of the confidence interval is \(66+3.35 = 69.35\%\)

Answer:

Part A: \(3.35\)
Part B: \(62.65\), \(69.35\)