Question

suppose you want to estimate the proportion of cars that are sport utility vehicles (suvs) being driven in regina at rush hour by standing on the corner of victoria avenue and albert street and counting suvs. you believe the figure is no higher than 0.40. if you want the error of the confidence interval to be no greater than 0.03, how many cars should you randomly sample? use a 90% level of confidence.

Answer 1

To be 90% confident that the estimate of the proportion of SUVs in Regina is within a 3% margin of error, you should randomly sample at least 723 cars.

Step-by-step explanation:

To estimate the proportion of cars that are SUVs in Regina at rush hour with a 90% confidence level and a margin of error no greater than 0.03, we need to calculate the required sample size. We use the formula for sample size in estimating a population proportion:

n = (Z^2 * p * (1-p)) / E^2

Where:

Z is the Z-value (for a 90% confidence level, Z is approximately 1.645)

p is the estimated proportion of SUVs (0.40 as per the assumption)

E is the desired margin of error (0.03)

Plugging the values in the equation:

n = (1.645^2 * 0.40 * (1-0.40)) / 0.03^2

n ≈ (2.706 * 0.40 * 0.60) / 0.0009

n ≈ 722.4

Since we can't survey a fraction of a car, we'll round up to the nearest whole number:

n = 723

Therefore, you need to randomly sample 723 cars to be 90% confident that the proportion of SUVs is estimated within a 3% margin of error.