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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.

Step-by-step explanation:

Question a:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Poll of 850 randomly selected American adults and finds that 44% of those surveyed have a gun in their home.

This means that $n = 850, \pi = 0.44$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.44 - 1.96\sqrt{\frac{0.44*0.56}{850}} = 0.4066$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.44 + 1.96\sqrt{\frac{0.44*0.56}{850}} = 0.4734$

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.

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3 years ago