Question

g In a survey of 500 residents, 300 were opposed to the use of the photo-cop for issuing traffic tickets.The standard error of the estimate is found to be 0.022. Find the margin of error that corresponds toa 95% confidence interval. The z-score is 1.96.A)0.035B)0.043C)0.60D)0.40E)none of these

Answer 1

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' ± $Z_{1-\alpha /2}$ * $\sqrt{\frac{p'(1-p')}{n} }$

Where "$Z_{1-\alpha /2}$ * $\sqrt{\frac{p'(1-p')}{n} }$" represents the margin of error of the interval.

Now for a confidence level of 0.95 the value of Z is $Z_{0.975}= 1.965$

The estimated standard error is already calculated: $\sqrt{\frac{p'(1-p')}{n} } = 0.022$

The margin of error (d) is then:

d= $Z_{1-\alpha /2}$ * $\sqrt{\frac{p'(1-p')}{n} }$= 1.965 * 0.022

d= 0.04323

I hope it helps!