In a sleep study, the average number of hours of sleep 600 people reported getting the previous night was 5.75 with a standard deviation of 1.5 hours. What is the margin of error for any sample of sleep time with 99% confidence ( z=2.58)?
A) 42.13
B) 0.16
C) 10.99
D) 0.61
The margin of error can be calculated using the formula:
Margin of Error = z * (standard deviation / √n)
Given that z = 2.58, standard deviation = 1.5 hours, and n = 600, we can substitute these values into the formula to calculate the margin of error:
Margin of Error = 2.58 * (1.5 / √600)
Margin of Error ≈ 2.58 * (1.5 / 24.49)
Margin of Error ≈ 2.58 * 0.06126
Margin of Error ≈ 0.1579876
Rounded to the nearest hundredth, the margin of error is approximately 0.16.
Therefore, the answer is B) 0.16.