Answer:
Z-score 1.96
Margin of error d= 0.04323
Step-by-step explanation:
Hello!
The study variable in this case is
X: Number of people that oppose the use of photo-cop for issuing traffic tickets in a sample of 500.
n= 500
The parameter of interest is the proportion of people that opposes it.
The estimated proportion is p'= 300/500= 0.6
To estimate the population proportion per Confidence interval you have to approximate the distribution of the sample proportion p' to normal:
p'≈N(p;p(1-p)1/n)
The mean of this distribution is p and the variance is p*(1-p)*1/n
To construct the Confidence interval, since the value of the population proportion is unknown, an estimated variance is used:
p'(1-p')*1/n ⇒ then the estimated standard error is √(p'(1-p')*1/n)
The formula for the confidence interval is:
p' ± [tex]Z_{1-\alpha /2}[/tex] * [tex]\sqrt{\frac{p'(1-p')}{n} }[/tex]
Where "[tex]Z_{1-\alpha /2}[/tex] * [tex]\sqrt{\frac{p'(1-p')}{n} }[/tex]" represents the margin of error of the interval.
Now for a confidence level of 0.95 the value of Z is [tex]Z_{0.975}= 1.965[/tex]
The estimated standard error is already calculated: [tex]\sqrt{\frac{p'(1-p')}{n} } = 0.022[/tex]
The margin of error (d) is then:
d= [tex]Z_{1-\alpha /2}[/tex] * [tex]\sqrt{\frac{p'(1-p')}{n} }[/tex]= 1.965 * 0.022
d= 0.04323
I hope it helps!
