in a certain population, body weights are normally

Ariana Montgomery

Ariana Montgomery

Answered question

2022-07-19

in a certain population, body weights are normally distributed with a mean of 152 pounds 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

Answer & Explanation

Andre BalkonE

Andre BalkonE

Skilled2023-05-29Added 110 answers

Given that the population body weights are normally distributed with a mean (μ) of 152 pounds and a standard deviation (σ) of 26 pounds, we want to estimate the percentage of people weighing more than 180 pounds.
To calculate the required sample size, we need to consider the desired confidence level and the maximum error margin. In this case, we want a 96% confidence level with a maximum error of 4 percentage points.
First, we need to determine the z-score corresponding to the desired confidence level. The z-score can be found using a standard normal distribution table or a calculator. For a 96% confidence level, the z-score is approximately 1.75.
Next, we calculate the standard error, which is the standard deviation of the sample mean. In this case, we use the population standard deviation (σ) divided by the square root of the sample size (n):
Standard Error=σn
Since we don't have an estimate of the percentage of people weighing more than 180 pounds, we can assume a conservative estimate of 50% to obtain a maximum sample size. This assumption ensures that the sample size is sufficient regardless of the actual percentage.
Now, we can set up an equation to determine the sample size:
1.75×Standard Error=0.04×0.5
Simplifying the equation:
1.75×σn=0.04×0.5
We can substitute the given values:
1.75×26n=0.04×0.5
Solving for the square root of n:
n=1.75×260.04×0.5
Simplifying the right side of the equation:
n=227.5
Squaring both sides to isolate n:
n=227.52
Calculating the value:
n51806.25
Rounding up to the nearest whole number, we get:
n51807
Therefore, to estimate the percentage of people who weigh more than 180 pounds with a 96% confidence level and a maximum error margin of 4 percentage points, we need to survey at least 51,807 people.

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