Question

A Food Marketing Institute found that 29% of households spend more than $125 a week on groceries. Assume the population proportion is 0.29 and a simple random sample of 207 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?
There is a ___ probability that the sample proportion of households spending more than $125 a week is less than 0.31. Round the answer to 4 decimal places.

Answer 1

Using the normal distribution, there is a 0.7357 = 73.57% probability that the sample proportion of households spending more than $125 a week is less than 0.31.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

• By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$, as long as $np \geq 10$ and $n(1 - p) \geq 10$.

The estimate and the sample size are:

p = 0.29, n = 207.

Hence the mean and the standard error are given as follows:

• $\mu = p = 0.29$.

• $s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.29(0.71)}{207}} = 0.0315$.

The probability that the sample proportion of households spending more than $125 a week is less than 0.31 is the p-value of Z when X = 0.31, hence:

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

By the Central Limit Theorem:

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

$Z = \frac{0.31 - 0.29}{0.0315}$

Z = 0.63

Z = 0.63 has a p-value of 0.7357.

0.7357 = 73.57% probability that the sample proportion of households spending more than $125 a week is less than 0.31.

More can be learned about the normal distribution at brainly.com/question/4079902

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