In a survey of 8451 US adults 314 said that they were taking

In a survey of 8451 US adults, 31.4% said that they were taking vitamin E as a suplement. The survey\'s margin of error is 1%. Construct a confidence interval and estimate its level of confidence. Explain

Solution

Proportion p = 0.314

Margin of error = 0.01

Confidence Interval :

=( 0.314 - 0.01 , 0.314 + 0.01)

=( 0.304 , 0.324) Answer

For level of significance:

Margin of error = z * sqrt(p(1-p)/N)

=> 0.01 = z * sqrt(0.314 (1-0.314)/8451)

=> z = 0.01/sqrt(0.314 (1-0.314)/8451)

=> z = 1.98

Therefore,

Significance Level = 95.23% Answer