An April 2009 survey of 2253 American adults conducted by the Pew
Research Center's Internet & American Life Project revealed that 1262 of
the respondents had at some point used wireless means for online access.
Calculate an interpret a 95% CI for the proportion of all American
adults who at the time of the survey had used wireless means for
online access.
Answer = (0.539, 0.581)
I know the answer but how did they get this can some one show me the steps?
Thanks.
Research Center's Internet & American Life Project revealed that 1262 of
the respondents had at some point used wireless means for online access.
Calculate an interpret a 95% CI for the proportion of all American
adults who at the time of the survey had used wireless means for
online access.
Answer = (0.539, 0.581)
I know the answer but how did they get this can some one show me the steps?
Thanks.
-
keepsmiling
CI = p̂ ± (z*)√[( p̂)(1 - p̂) / n]
p̂ = 1262/ 2253 = 0.56
z* = 1.96 from Standard Normal table with 5%/2 = 0.025 area in each tail
CI = 0.56 ± (1.96)√[( 0.56)(1 - 0.56) / 2253]
CI = (0.539, 0.581)
hope that helped
CI = p̂ ± (z*)√[( p̂)(1 - p̂) / n]
p̂ = 1262/ 2253 = 0.56
z* = 1.96 from Standard Normal table with 5%/2 = 0.025 area in each tail
CI = 0.56 ± (1.96)√[( 0.56)(1 - 0.56) / 2253]
CI = (0.539, 0.581)
hope that helped