Final answer:
The confidence interval for the true proportion of families owning at least one gun, from a sample of 90 families where 40 owned guns, is found using the formula for confidence intervals with the given sample proportion, standard error, and z-score for 95% confidence.
Step-by-step explanation:
To find the 95% confidence interval of the true proportion of families who owned at least one gun based on the survey of 90 families, where 40 owned at least one gun, apply the following formula:
The sample proportion (p) can be calculated as the number of successes (families owning at least one gun) divided by the total number of trials (surveyed families), so p = 40/90.
We then use the standard error of the proportion, which is the square root of (p(1-p)/n), where n is the sample size. For a 95% confidence level, we use a z-score that corresponds to the level of confidence, which is typically 1.96 for 95% confidence.
The confidence interval formula becomes: CI = p ± z*(sqrt(p(1-p)/n)). Plugging in the numbers, we calculate the interval.
Interpretation of the confidence interval: If we were to take many samples and compute a confidence interval for each one using the same method, 95% of these intervals would contain the true proportion of families who own at least one gun.