Respuesta :
Answer:
a)
here margin of error E = 0.07 Â
for95% CI critical Z Â Â Â Â Â = 1.96 Â
estimated proportion=p= 0.5000 Â
required sample size n = Â Â Â Â p*(1-p)*(z/E)2= 196
b)
here margin of error E = 0.07 Â
for95% CI critical Z Â Â Â Â Â = 1.96 Â
estimated proportion=p= 0.8100 Â
required sample size n = Â Â Â Â p*(1-p)*(z/E)2= 121
c)
d. No, a sample of students at the nearest college is a convenience sample, not a simple random sample. so it is very possible that the results would not be representative of the population of adults.
Given Information:
Confidence level = 95%
Estimated error = 8%
Required Information:
(a) sample size = n = ?
(b) sample size = n = ?
(c) could he simply survey the adults at the nearest college?
Answer:
(a) sample size = n = 150
(b) sample size = n = 92
(c) d. No, a sample of students at the nearest college is a convenience sample, not a simple random sample. so it is very possible that the results would not be representative of the population of adults.
Step-by-step explanation:
(a) In this part we are required to assume that nothing is known about the percentage of adults who have heard about the brand.
Since we dont know about the percentage of adults who have heard about the brand, we will assume that it is 50%
The z value corresponding 95% confidence level is 1.96
n = z²*p*(1-p)/ο²
n = (1.96²*0.50(1-0.50))/0.08²
n = 150.06
n ≈ 150
So we need a bigger sample size in this case.
(a) In this part we are required to assume that 81% of adults have heard about the brand.
n = z²*p*(1-p)/ο²
n = (1.96²*0.81(1-0.81))/0.08²
n = 92.37
n ≈ 92
So we need a smaller sample size in this case.
(c) could he simply survey the adults at the nearest college?
No, because convenience samples are most likely to be biased therefore, we should avoid such methodology for sampling.
d. No, a sample of students at the nearest college is a convenience sample, not a simple random sample. so it is very possible that the results would not be representative of the population of adults