It is believed that 11% of all Americans are left-handed. In a random sample of 100 students from a particular college with 10123 students, 14 were left-handed. Find a 98% confidence interval for the percentage of all students at this particular college who are left-handed.

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.

32

step-by-step explanation:

answer: the answer is simply nothing.

step-by-step explanation: because there is no ending to this question

gimme brainliest > : 3