Question

the data in the accompanying bar graph are from a random sample of 818 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 required gun registration to reduce gun violence.
there is 95% confidence that between □% and □% of all adults in the country favor required gun registration.
(use ascending order. round to the nearest tenth as needed.)

Explanation:

Step1: Recall the formula for margin of error

For a sample of size \(n\), the margin of error \(E\) for a \(95\%\) confidence interval (common in such surveys when not specified otherwise) is given by \(E=\frac{1}{\sqrt{n}}\). Here, \(n = 818\).

Step2: Calculate the margin of error

Substitute \(n = 818\) into the formula: \(E=\frac{1}{\sqrt{818}}\).

$$
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Convert to percentage: \(E = 0.035\times100\%=3.5\%\) (rounded to the nearest tenth).

Step3: For part (b)

The sample percentage for required gun - registration is \(p = 69\%\). The confidence interval is \(p\pm E\).
The lower bound is \(69 - 3.5=65.5\%\) and the upper bound is \(69+3.5 = 72.5\%\)

Answer:

a. The margin of error is \(\pm3.5\%\)
b. There is \(95\%\) confidence that between \(65.5\%\) and \(72.5\%\) of all adults in the country favor required gun registration.