88 people must be surveyed according to the confidence interval.
According to the statement
we have given that the mean of 152 pounds and a standard deviation of 26 pounds. And we have to find the percentage who weigh more than 180 pounds
So, For this purpose, we know that the
The given variables are which are given in the statement are:
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%
Then
Firstly we find the value of the Z is :
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.
Now, 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.
So, 88 people must be surveyed according to the confidence interval.
Learn more about Confidence interval here
brainly.com/question/17097944
Disclaimer: This question was incomplete. Please find the full content below.
Question:
In a certain population, body weights are normally distributed with a mean of 152 pounds 30) and a standard deviation of 26 pounds. how many people must be surveyed if we want to estimate the percentage who weigh more than 180 pounds? assume that we want 96% confidence that the error is no more than 4 percentage points.
#SPJ4
