Question

For a random sample of 350 students, the mean cost for textbooks during the first semester of college was found to be $374.75, and the population standard deviation was $37.51. Assuming that the population is normally distributed, find the margin of error of a 95% confidence interval for the population mean The margin of error for a 95% confidence interval is (Round to two decimal places as needed.)

Answer 1

The margin of error for a 95% confidence interval with a known population standard deviation of $37.51, a sample size of 350, and using the z-score is approximately $3.93.

Step-by-step explanation:

To find the margin of error for a 95% confidence interval when the population standard deviation is known, and the population is assumed to be normally distributed, we use the z-score corresponding to 95% confidence and the formula for the margin of error with a known standard deviation:
Margin of Error = z * (σ/√n).
Where z is the z-score for a 95% confidence level (which is approximately 1.96), σ is the population standard deviation, and n is the sample size.

Using the given information: n=350, σ=$37.51, and z=1.96, we can calculate the margin of error as follows:
Margin of Error = 1.96 * ($37.51/√350) ≈ 1.96 * 2.0068 ≈ $3.93 (rounded to two decimal places).