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A sample of 469 teenage births at a large city hospital is observed for the number of births resulting in twins. The 95% confidence interval for the proportion of all births among teenage girls resulting in twins was estimated to be (0.006, 0.026). What is the margin of error for this interval?

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wizard123
The confidence interval can be written as:
[tex]x \pm m[/tex]
where m is margin of error, x is mid-point
[tex]x = \frac{.006+.026}{2} = 0.016[/tex]
[tex]m = .026 - .016 = 0.01[/tex]

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