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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.
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