Question 577037: Construct the 95% and the 99% confidence intervals for the population proportion p using the sample statistics below:
p=0.627
q=0.373
n=4351
The 95% confidence interval for the population proportion p is (____, ____)
The 99% confidence interval for the population proportion p is (_____,____)
Which interval is wider?
*Answer needed asap please. Thank you!*
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! 95% confidence interval
( p - z*sqrt( (p(1-p))/n ), p + z*sqrt( (p(1-p))/n ) )
( 0.627 - 1.95996*sqrt( (0.627(1-0.627))/4351 ), 0.627 + 1.95996*sqrt( (0.627(1-0.627))/4351 ) )
( 0.6126, 0.6414 )
So the 95% confidence interval is ( 0.6126, 0.6414 ), it's width is 0.6414 - 0.6126 = 0.0288
===================================================================
99% confidence interval
( p - z*sqrt( (p(1-p))/n ), p + z*sqrt( (p(1-p))/n ) )
( 0.627 - 2.5758*sqrt( (0.627(1-0.627))/4351 ), 0.627 + 2.5758*sqrt( (0.627(1-0.627))/4351 ) )
( 0.6081, 0.6459 )
So the 99% confidence interval is ( 0.6081, 0.6459 ), it's width is 0.6459 - 0.6081 = 0.0378
So the 99% confidence interval is wider
|
|
|
