We are given the following variables:
μ = the sample mean = 152 pounds
σ = the standard deviation = 26 pounds
x = the sample value we want to test = 180 pounds
n = the sample size = unknown
MOE = margin of error = 4% = 0.04
Confidence level = 96%
The first thing we can do is to find for the value of z using the formula:
z = (x – μ) / σ
z = (180 – 152) / 26
z = 1.0769 = 1.08
Since we are looking for the people who weigh more than 180 pounds, therefore this is a right tailed z test. The p value is:
p = 0.1401
Then we can use the formula below to solve for n:
n = z^2 * p * (1 – p) / (MOE)^2
n = 1.08^2 * 0.1401 * (1 – 0.1401) / (0.04)^2
n = 87.82 = 88
Therefore around 88 people must be surveyed.
We are given the following variables:
μ = the sample mean = 152 pounds
σ = the standard deviation = 26 pounds
x = the sample value we want to test = 180 pounds
n = the sample size = unknown
MOE = margin of error = 4% = 0.04
Confidence level = 96%
The first thing we can do is to find for the value of z using the formula:
z = (x – μ) / σ
z = (180 – 152) / 26
z = 1.0769 = 1.08
Since we are looking for the people who weigh more than 180 pounds, therefore this is a right tailed z test. The p value is:
p = 0.1401
Then we can use the formula below to solve for n:
n = z^2 * p * (1 – p) / (MOE)^2
n = 1.08^2 * 0.1401 * (1 – 0.1401) / (0.04)^2
n = 87.82 = 88
Therefore around 88 people must be surveyed.
#SPJ2
