Question

In 2017, a website reported that only 10% of surplus food is being recovered in the food-service and restaurant sector, leaving approximately 1.5 billion meals per year uneaten. Assume this is the true population proportion and that you plan to take a sample survey of 575 companies in the food service and restaurant sector to further investigate their behavior.1. What is the probability that your survey will provide a sample proportion within ±0.03 of the population proportion? (Round your answer to four decimal places.)?
2. What is the probability that your survey will provide a sample proportion within ±0.015 of the population proportion? (Round your answer to four decimal places.)?

Answer 1

Answer:

1. 0.9836 = 98.36% probability that your survey will provide a sample proportion within ±0.03 of the population proportion

2. 0.7698 = 76.98% probability that your survey will provide a sample proportion within ±0.015 of the population proportion.

Step-by-step explanation:

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes 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 $s = \sqrt{\frac{p(1-p)}{n}}$

In this question, we have that:

$p = 0.1, n = 575$

So

$\mu = 0.1, s = \sqrt{\frac{0.1*0.9}{575}} = 0.0125$

1. What is the probability that your survey will provide a sample proportion within ±0.03 of the population proportion?

This is the pvalue of Z when X = 0.1 + 0.03 = 0.13 subtracted by the pvalue of Z when X = 0.1 - 0.03 = 0.07. So

X = 0.13

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

By the Central Limit Theorem

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

$Z = \frac{0.13 - 0.1}{0.0125}$

$Z = 2.40$

$Z = 2.40$ has a pvalue of 0.9918

X = 0.07

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

$Z = \frac{0.07 - 0.1}{0.0125}$

$Z = -2.40$

$Z = -2.40$ has a pvalue of 0.0082

0.9918 - 0.0082 = 0.9836

0.9836 = 98.36% probability that your survey will provide a sample proportion within ±0.03 of the population proportion.

2. What is the probability that your survey will provide a sample proportion within ±0.015 of the population proportion?

This is the pvalue of Z when X = 0.1 + 0.015 = 0.115 subtracted by the pvalue of Z when X = 0.1 - 0.015 = 0.085. So

X = 0.115

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

$Z = \frac{0.115 - 0.1}{0.0125}$

$Z = 1.2$

$Z = 1.2$ has a pvalue of 0.8849

X = 0.085

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

$Z = \frac{0.085 - 0.1}{0.0125}$

$Z = -1.2$

$Z = -1.2$ has a pvalue of 0.1151

0.8849 - 0.1151 = 0.7698

0.7698 = 76.98% probability that your survey will provide a sample proportion within ±0.015 of the population proportion.