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A statistician calculates that 7% of Americans are vegetarians. If the statistician is correct, what is the probability that the proportion of vegetarians in a sample of 403 Americans would differ from the population proportion by more than 3%

User Romeozor
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1 Answer

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Answer:

0.0182 = 1.82% probability that the proportion of vegetarians in a sample of 403 Americans would differ from the population proportion by more than 3%.

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
\mu and standard deviation
\sigma, the z-score of a measure X is given by:


Z = (X - \mu)/(\sigma)

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean
\mu and standard deviation
\sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
\mu and standard deviation
s = (\sigma)/(√(n)).

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
\mu = p and standard deviation
s = \sqrt{(p(1-p))/(n)}

A statistician calculates that 7% of Americans are vegetarians.

This means that
p = 0.07

Sample of 403 Americans

This means that
n = 403

Mean and standard deviation:


\mu = p = 0.07


s = \sqrt{(p(1-p))/(n)} = \sqrt{(0.07*0.93)/(403)} = 0.0127

What is the probability that the proportion of vegetarians in a sample of 403 Americans would differ from the population proportion by more than 3%?

Proportion below 7 - 3 = 4% or above 7 + 3 = 10%. Since the normal distribution is symmetric, these probabilities are equal, which means that we find one of them, and multiply by 2.

Probability the proportion is below 4%

p-value of Z when X = 0.04.


Z = (X - \mu)/(\sigma)

By the Central Limit Theorem


Z = (X - \mu)/(s)


Z = (0.04 - 0.07)/(0.0127)


Z = -2.36


Z = -2.36 has a p-value of 0.0091

2*0.0091 = 0.0182

0.0182 = 1.82% probability that the proportion of vegetarians in a sample of 403 Americans would differ from the population proportion by more than 3%.

User Cagdas Kanar
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