In a survey of 2035 workers, 73% reported working out 3 or more days a week. What is the margin of error? What is the interval that is likely to contain the exact percent of all people who work out 3 or more days a week?

Margin of error is the critical value (t score or z score) times the standard error (standard deviation of the sample).

ME = CV × SE

Since n > 30, we can use the z score as the critical value.

Assuming 95% confidence, z = 1.960.

The standard error for a proportion is:

s = √(p (1 − p) / n)

Given p = 0.73 and n = 2035:

s = √(0.73 (1 − 0.73) / 2035)

s = 0.0098

Therefore, the margin of error is:

ME = 1.960 × 0.0098

ME = 0.019

The margin of error is 1.9%.  The interval that likely contains the true percent of all people who work out 3 or more days a week is:

73% ± 1.9%

(71.1%, 74.9%)

Step-by-step explanation:

i dont know

step-by-step explanation:

umm. i dont know