Answer:
A)n= 703.96
B)n= 602.308
Step by step Explanation:
Given that you want to be 99% confident that the sample percentage is within 3.1 percentage points of the true population percentage.
Then z/2 = 1.645
And M = 3.1% = 0.031
A)Nothing is known therefore,
p = q = 0.50
E=0.031
For 90% confidence, z = 1.645
n = (zα/₂)²(p)(1-p)/M²
n= 1.645²× 0.5 × 0.5/0.031²
n= 703.96
Therefore, 703.96randomly selected air passengers must be surveyed to be 99%
B)we know that recent surveys surgest that about 38% of all air passengers prefer an aisle seat, thus p = 35% = 0.35
n = (zα/₂)²(p)(1-p)/M²
n= (1.645²× 0.31 × 0.69)/0.031²
n= 602.308
Hence, 602.308 randomly selected air passengers must be surveyed to be 90% confident that the sample percentage is within 3.1 percentage points of the true population percentage.
