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Recycling Do most teens recycle? To find out, an AP Statistics class asked an SRS of 100 students at their school whether they regularly recycle. How many students in the sample would need to say Yes to provide convincing evidence that more than half of the students at the school recycle? The Fathom dotplot below shows the results of taking 200 SRSs of 100 students from a population in which the true proportion who recycle is 0.50. (a) Suppose 55 students in the class’s sample say Yes. Explain why this result does not give convincing evidence that more than half of the school’s students recycle. (b) Suppose 63 students in the class’s sample say Yes. Explain why this result gives strong evidence that a majority of the school’s students recycle.

Step by step solution

01

Understand the Question

We need to determine if the number of students saying 'Yes' can be considered evidence that more than half of the school's students recycle. This involves comparing sample results to a null hypothesis of a 50% population recycling rate.

02

Define Null and Alternative Hypotheses

Let's define the hypotheses. The null hypothesis ( H_0 ) is that the true proportion of students who recycle is 0.50. The alternative hypothesis ( H_a ) is that the true proportion is greater than 0.50.

03

Calculate the Expected Variability

Using the true proportion ( p = 0.50 ), we calculate the standard deviation of the sampling distribution under the null hypothesis. The formula is:\[ \sigma = \sqrt{ \frac{p(1-p)}{n} } = \sqrt{ \frac{0.5 \times 0.5}{100} } = 0.05 \]This tells us how much variation we could expect due to random sampling.

04

Analyze 55 'Yes' Responses

If 55 out of 100 students say 'Yes', the sample proportion is 0.55. We calculate the z-score to see how many standard deviations this is from 0.50:\[ z = \frac{0.55 - 0.50}{0.05} = 1.0 \]A z-score of 1.0 shows that 55 is not far enough from 50 to be significant at common confidence levels.

05

Analyze 63 'Yes' Responses

If 63 students say 'Yes', the sample proportion is 0.63. We again calculate the z-score:\[ z = \frac{0.63 - 0.50}{0.05} = 2.6 \]A z-score of 2.6 typically corresponds to a p-value well below 0.05, suggesting that 63 'Yes' responses is significant evidence that more than half of the students recycle.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sampling Distribution

The concept of a sampling distribution is essential in statistics as it helps us understand the variability of sample statistics. Imagine you take several samples from the same population and calculate a particular statistic, like the sample mean or proportion, for each sample.
These individual statistics form what's known as a sampling distribution. It's like seeing the bigger picture by analyzing many samples.

• A sampling distribution shows how a sample statistic, such as a sample proportion, would be distributed if you repeated the sampling many times.
• In our case, the concern is the sample proportion of students who say 'Yes' to recycling.
• The understanding of its distribution helps in determining how likely or unlikely a particular sample result is, given the assumptions about the population (in this scenario, that 50% recycle).

The standard deviation of this distribution helps quantify the expected variability. It's a powerful way to determine how a sample result compares to what we'd expect under our hypothesis.

Null Hypothesis

The null hypothesis, often denoted as \( H_0 \), is a fundamental concept in statistical hypothesis testing. It's the starting assumption that there is no effect or difference, and in hypothesis testing, it's what statisticians aim to either reject or fail to reject based on sample data.

• In our recycling problem, the null hypothesis claims that the true proportion of students who recycle is 0.50, or 50%.
• This hypothesis sets the baseline for what we expect if there's no underlying change or effect in the population.

If our sample evidence shows a significant deviation from what's projected by the null hypothesis, it might suggest that this hypothesis is likely not accurate. The validity of our conclusion depends heavily on how "convincing" this deviation is in terms of statistical significance.

Alternative Hypothesis

Closely linked to the null hypothesis, the alternative hypothesis is what you might accept if there's enough evidence to reject \( H_0 \). It's represented by \( H_a \).

• In the problem of teen recycling, the alternative hypothesis suggests that the true recycling proportion is greater than 0.50.
• This hypothesis captures the scenario that more than half of the students recycle, representing a meaningful deviation from what \( H_0 \) claims.

To support the alternative hypothesis, the sample must provide sufficiently strong evidence. This evidence comes in the form of a sample result that significantly differs from what's expected under the null hypothesis, typically guided by z-scores and p-values.

Standard Deviation

Standard deviation is a statistical measure that quantifies the extent of variability in a set of data. It's a crucial concept when dealing with sampling distributions.

• For the sampling distribution of a sample proportion, the standard deviation can be calculated using the formula \( \sigma = \sqrt{ \frac{p(1-p)}{n} } \) where \( p \) is the assumed population proportion (in our case, 0.50), and \( n \) is the sample size.
• In this activity, the calculated standard deviation is 0.05, revealing how much we might expect the sample proportions to vary if \( H_0 \) is true.

In analyzing both 55 and 63 'Yes' responses from the sampled students, the standard deviation serves as a lens to see how ordinary or extraordinary the sample's outcome is.

Z-score

A z-score is a tool used in statistics to determine how far away a particular data point is from the mean, measured in terms of standard deviation.

• For instance, in the recycling study, it helps us assess how far our observed sample proportion (like 0.55 or 0.63) is from the assumed population proportion (0.50).
• Calculating the z-score involves the formula \( z = \frac{\text{sample proportion} - \text{population proportion}}{\sigma} \), aiding us in understanding the rarity of our sample finding.
• A z-score of 1, as with 55 affirmatives, indicates that this result is not that exceptional. But a z-score of 2.6, like with 63 affirmatives, signals a result unlikely to happen by mere chance, leading to a strong consideration of the alternative hypothesis.

The z-score guides us to make informed decisions regarding the hypothesis, using statistical significance as a yardstick.