Question

If n=590 and (p-hat) =0.24, find the margin of error at a 90% confidence level.

As in the reading, in your calculations:
--Use z = 1.645 for a 90% confidence interval
--Use z = 2 for a 95% confidence interval
--Use z = 2.576 for a 99% confidence interval.

Answer 1

The correct answer is option 1. The margin of error at a 90% confidence level with a sample size of 590 and a sample proportion of 0.24 is calculated using the standard Z distribution and the formula ME = Z * sqrt((p-hat * (1 - p-hat)) / n), where Z is 1.645.

Step-by-step explanation:

The question involves calculating the margin of error for a sample proportion at a specific confidence level, which requires using the standard normal (Z) distribution. Given that n=590 and π-hat=0.24 (the sample proportion), the task is to find the margin of error at a 90% confidence level. The formula for the margin of error (ME) in this context is ME = Z * sqrt((π-hat * (1 - π-hat)) / n), where Z is the critical value from the Z-distribution corresponding to the desired confidence level.

Since a 90% confidence interval is requested, the critical Z value is 1.645. Plugging the values into the formula we get: ME = 1.645 * sqrt((0.24 * (1 - 0.24)) / 590). By performing the calculations, we can find the margin of error for the given sample size and proportion at the 90% confidence level.

Note that increasing the sample size or changing the confidence level would affect the margin of error. A larger sample size would tend to decrease it, whereas a higher confidence level would require a larger critical Z value (thus increasing the margin of error).