Question

Recycling trash, reducing waste, and reusing materials are eco-actions that will help the environment. According to a USA Today snapshot (Exercise 6.45), \(78 \%\) of respondents list recycling as the leading way to help our environment. \({ }^{11}\) Suppose that a random sample of \(n=100\) adults is selected and that the \(78 \%\) figure is correct. a. Does the distribution of \(\hat{p},\) the sample proportion of adults who list recycling as the leading way to help the environment have an approximate normal distribution? If so, what is its mean and standard deviation? b. What is the probability that the sample proportion \(\hat{p}\) is less than \(75 \% ?\) c. What is the probability that \(\hat{p}\) lies in the interval .7 to .75? d. What might you conclude about \(p\) if the sample proportion were less than \(.65 ?\)

Short answer

What is the probability of the sample proportion being less than 75% and being in the interval .7 to .75? What can we conclude if the sample proportion was below 65%? Answer: The sample proportion is approximately normally distributed with a mean of 0.78 and standard deviation of 0.0416. The probability of the sample proportion being less than 75% is 0.2358, and the probability of it being in the interval .7 to .75 is 0.2084. If the sample proportion was below 65%, it could suggest a non-representative sample or a significant change in the true proportion. Further investigation and more data would be required for a more robust conclusion.

Step-by-step solution

Verify if the distribution is approximately normal

Using the Central Limit Theorem, the sample proportion's distribution will be approximately normal if both \(np\geq 10\) and \(n(1-p)\geq 10\). Given: \(n=100\), \(p=0.78\) Check if these conditions are met: \(np = 100 \times 0.78 = 78 \geq 10\) \(n(1-p) = 100\times(1-0.78)= 100\times0.22 = 22 \geq 10\) Both conditions are met, so the distribution of the sample proportion is approximately normal.

Calculate the mean and standard deviation

For a sample proportion \(\hat{p}\), the mean and standard deviation can be calculated using the following formulas: Mean: \(\mu_{\hat{p}} = p\) Standard deviation: \(\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\) Given: \(p=0.78\), \(n=100\) Mean: \(\mu_{\hat{p}} = 0.78\) Standard deviation: \(\sigma_{\hat{p}} = \sqrt{\frac{0.78(1-0.78)}{100}} = \sqrt{\frac{0.78\times0.22}{100}} = 0.0416\)

Calculate the probability that the sample proportion is less than 75%

We need to find the probability that \(\hat{p} < 0.75\). To do this, we will use the z-score formula for a sample proportion: \(z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}}\) Given: \(\hat{p}=0.75\), \(\mu_{\hat{p}}=0.78\), \(\sigma_{\hat{p}}=0.0416\) Calculate the z-score: \(z = \frac{0.75 - 0.78}{0.0416} = -0.72\) Using a standard normal distribution table, we find that the probability is \(P(Z < -0.72) = 0.2358\).

Calculate the probability that the sample proportion lies in the interval .7 to .75

We need to find the probability that \(0.7 < \hat{p} < 0.75\). Firstly, we will calculate the z-scores for both bounds: For 0.7: \(z_1 = \frac{0.7 - 0.78}{0.0416} = -1.92\) For 0.75: \(z_2 = \frac{0.75 - 0.78}{0.0416} = -0.72\) Using a standard normal distribution table, we find the probabilities: \(P(Z<-1.92) = 0.0274\) \(P(Z<-0.72) = 0.2358\) Then, the probability of \(\hat{p}\) lying in the interval .7 to .75 is \(P(0.7 < \hat{p} < 0.75) = P(Z<-0.72) - P(Z<-1.92) = 0.2358 - 0.0274 = 0.2084\)

Conclusion regarding the true proportion if the sample proportion is less than 0.65

If the sample proportion \(\hat{p}\) were less than .65, it would be significantly lower than the given proportion of 0.78. This could suggest that either the sample was not a representative selection of the true population, or there is a significant change in the true proportion of adults who list recycling as the leading way to help the environment. Further investigation and more data would be required to draw a more robust conclusion.

Key concepts

Central Limit Theorem

The Central Limit Theorem (CLT) is a fundamental concept in statistics. It explains how the means of a large number of samples from a population will distribute.
Specifically, it states that as the sample size becomes large, the sampling distribution of the sample mean will approach a normal distribution.
When dealing with proportions, as in this exercise, if we take a large enough sample size, the distribution of the sample proportion will also approximate a normal distribution.
This is crucial for such problems because it allows us to use the properties of the normal distribution to make inferences about the sample.

• The CLT relies on the sample size being large enough. For our case, the conditions are met with np and n(1-p) both being greater than 10, confirming that the sample distribution can be treated as normal.
• This approximation enables the calculation of probabilities related to the sample mean or sample proportion, as detailed in the solution.

Normal Distribution

The normal distribution is a probability distribution that is symmetric about the mean. It is often referred to as the bell curve due to its shape.
In the context of the problem, once the Central Limit Theorem conditions are satisfied, we assume the sample proportion follows a normal distribution.
The key characteristics of a normal distribution include:

• Symmetry: The distribution is symmetric around the mean.
• Mean, Median, and Mode are equal and located at the center of the distribution.
• The spread of the distribution is defined by its standard deviation.

Using the properties of a normal distribution, we can make predictions about probabilities. For example, we can find how likely it is for our sample proportion to fall within a certain range by converting values into z-scores and using standard normal distribution tables.

Sample Proportion

In statistics, the sample proportion (\(\hat{p}\)is the fraction of items in a sample with a particular attribute. It is a measure to estimate the corresponding proportion in the population.
In the exercise, the sample proportion of interest is the percentage of adults who say recycling is crucial for environmental help.
Key concepts for sample proportion include:

• The formula for the sample proportion is \(\hat{p} = \frac{x}{n}\), where \(x\) is the number of successes, and \(n\) is the sample size.
• The mean of the sample proportion (\(\mu_{\hat{p}}\)) is equivalent to the population proportion \(p\).
• The standard deviation of the sample proportion (\(\sigma_{\hat{p}}\)) helps in understanding the variability around the mean and can be calculated as \(\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\).

Understanding the sample proportion is vital since it is used to infer the population proportion by considering the variability introduced by the sampling process.

Z-score

A Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It indicates how many standard deviations an element is from the mean.
In the exercise, the Z-score helps evaluate how extreme the observed sample proportions are in comparison to the expected distribution under the assumption of normality.
To compute a Z-score for a sample proportion we use the formula:

• \(z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}}\)

The Z-score allows us to use a standard normal distribution table to find probabilities:

• A positive Z-score indicates a value above the mean.
• A negative Z-score indicates a value below the mean.
• Z-scores enable comparison across different data sets by standardizing scores, making them dimensionless.

Using Z-scores, predictions or hypotheses about proportions within a population become manageable. They offer a way to quantify the rarity of particular sample outcomes.