Answer:
Step-by-step explanation:
We can calculate this confidence interval using the population proportion calculation. To do this we must find p' and q'
Where p' = 14/100= 0.14 (no of left handed sample promotion)
q' = 1-p' = 1-0.14= 0.86
Since the requested confidence level is CL = 0.98, then α = 1 – CL = 1 – 0.98 = 0.02/2= 0.01, z (0.01) = 2.326
Using p' - z alpha √(p'q'/n) for the lower interval - 0.14-2.326√(0.14*0.86/100)
= -2.186√0.00325
= -2.186*0.057
= 12.46%
Using p' + z alpha √(p'q'/n)
0.14+2.326√(0.14*0.86/100)
= 0.466*0.057
= 26.5%
Thus we estimate with 98% confidence that between 12% and 27% of all Americans are left handed.
