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Construct the requested confidence interval.

In a survey of 500 randomly selected residents, 300 were opposed to the use of photo-cops for issuing traffic tickets.
Construct a 95% confidence interval for the proportion of residents opposed to the use of photo-cops for issuing traffic tickets.

User Csoler
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Final answer:

To construct the 95% confidence interval for the proportion opposed to the use of photo-cops, a margin of error is calculated using the sample proportion and sample size, and the interval is derived to be approximately (0.557, 0.643).

Step-by-step explanation:

To construct a 95% confidence interval for the proportion of residents opposed to the use of photo-cops for issuing traffic tickets, we can use the formula for a confidence interval for a population proportion:
CI = π ± Z* √ (π(1 - π) / n)

Where:

π is the sample proportion,

n is the sample size,

Z* is the Z-score associated with the confidence level.

The sample proportion (π) is the number of residents opposed (300) divided by the total number surveyed (500), so π = 300/500 = 0.6.

The sample size (n) is 500.

For a 95% confidence level, the Z-score (Z*) is approximately 1.96.

Now, we calculate the margin of error (MOE):

MOE = Z* √ (π(1 - π) / n) = 1.96 √ (0.6(0.4) / 500) = 1.96 √ (0.24 / 500) = 1.96 √ (0.00048) ≈ 0.043

Finally, constructing the confidence interval:

CI = 0.6 ± 0.043 ≈ (0.557, 0.643)

This means we are 95% confident that the true proportion of residents opposed to the use of photo-cops for issuing traffic tickets is between 55.7% and 64.3%.

User Mauris
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