Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Get detailed and accurate answers to your questions from a community of experts on our comprehensive Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
The formula for determining margin of error for a population proportion is expressed as
[tex]\text{Margin of error = z}_{\frac{\varphi}{2}}\sqrt[]{\frac{p^{\prime}q^{\prime}}{n}}[/tex]where
p' is the estimated proportion of success. It is also the point estimate
q' is the estimated proportion of failure
n is the number of people sampled
z alpha is the z score corresponding to a 95% confidence level
From the information given,
The success in this case is the proportion of Americans that keep a dog for protection. Thus,
p' = 32/100 = 0.32
q' = 1 - p' = 1 - 0.32 = 0.68
n = 1012
z alpha for a 95% confidence level is 1.96
Thus,
[tex]\text{margin of error = 1.96 x }\sqrt[]{\frac{0.32\times0.68}{1012}}\text{ }[/tex]Margin of error = 0.015
The confidence interval is written as
margin of error ± margin of error
Thus, the 95 confidence interval is
0.32 ± 0.015
The lower limit is 0.32 - 0.015 = 0.305 = 0.305 x 100 = 30.5%
The upper limit is 0.32 + 0.015 = 0.335 x 100 = 33.5%
The interpretation would be
We estimate with 95% confidence that between 30.5% and 33.5% of Americans keep a dog for protection.
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.