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Sagot :
The margin of error that corresponds to a 95 % confidence interval for the proportion of residents that are opposed to the use of photo - cops for issuing traffic tickets is 0.04.
Given:
in a survey of 500 residents, 300 were opposed to the use of photo-cops for issuing traffic tickets.
Sample size n = 500
Number of success X = 300
Sample proportion p = X/n
= 300/500
= 30/50
= 3/5
= 0.6
z score at 95% = 1.96
Margin of error = z * [tex]\sqrt{p(1-p)/n}[/tex]
= 1.96 * [tex]\sqrt{0.6(1-0.6)/500}[/tex]
= 0.04
Therefore the margin of error that corresponds to a 95 % confidence interval is 0.04.
Learn more about the confidence interval here:
https://brainly.com/question/24131141
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