Question

A Food Marketing Institute found that 34% of households spend more than $125 a week on groceries. Assume the population proportion is 0.34 and a simple random sample of 124 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is less than 0.31

Answer 1

Answer:

0.2405 = 24.05% probability that the sample proportion of households spending more than $125 a week is less than 0.31.

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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 $s = \frac{\sigma}{\sqrt{n}}$.

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 $s = \sqrt{\frac{p(1-p)}{n}}$

Assume the population proportion is 0.34 and a simple random sample of 124 households is selected from the population.

This means that $p = 0.34, n = 124$

Mean and standard deviation:

$\mu = p = 0.34$

$s = \sqrt{\frac{0.34*0.66}{124}} = 0.0425$

What is the probability that the sample proportion of households spending more than $125 a week is less than 0.31?

This is the p-value of Z when X = 0.31, so:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.31 - 0.34}{0.0425}$

$Z = -0.705$

$Z = -0.705$ has a p-value of 0.2405.

0.2405 = 24.05% probability that the sample proportion of households spending more than $125 a week is less than 0.31.