Answer:
a.
The 95% confidence interval for the proportion of all American adults who have a gun in their home is (0.4066, 0.4734). This means that we are 95% sure that the true population proportions of American adults who have a gun in their home is between these two values.
b.
The confidence level is the probability of the confidence interval containing the population proportion.
Explanation:
Question a:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.

In which
z is the zscore that has a pvalue of
.
Poll of 850 randomly selected American adults and finds that 44% of those surveyed have a gun in their home.
This means that

95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:

The upper limit of this interval is:

The 95% confidence interval for the proportion of all American adults who have a gun in their home is (0.4066, 0.4734). This means that we are 95% sure that the true population proportions of American adults who have a gun in their home is between these two values.
b. Explain what is meant by 95% confidence in this context.
The confidence level is the probability of the confidence interval containing the population proportion.