91Ó°ÊÓ

A high-school teacher in a lowincome urban school in Worcester, Massachusetts, used cash incentives to encourage student learning in his AP statistics class. \({ }^{28}\) In 2010,15 of the 61 students enrolled in his class scored a 5 on the AP statistics exam. Worldwide, the proportion of students who scored a 5 in 2010 was \(0.15\). Is this evidence that the proportion of students who would score a 5 on the AP statistics exam when taught by the teacher in Worcester using cash incentives is higher than the worldwide proportion of \(0.15\) ? (a) State hypotheses, find the \(P\)-value, and give your conclusions in the context of the problem. Do you have any reservations about using the \(z\) test for proportions for this data? (b) Does this study provide evidence that cash incentives cause an increase in the proportion of \(5 \mathrm{~s}\) on the AP statistics exam? Explain your answer.

Step by step solution

01

State the Hypotheses

We are testing whether the Cambridge teacher's approach results in a higher proportion of 5s compared to the worldwide rate of 0.15. Therefore, our null hypothesis is that the teacher's proportion, \( p \), of students scoring a 5 is equal to the worldwide proportion: \( H_0: p = 0.15 \). The alternative hypothesis is that the proportion is higher: \( H_a: p > 0.15 \).

02

Calculate the Test Statistic

We use the sample data to find the test statistic for the proportion. The formula for the \( z \)-test statistic in a test for proportions is: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \] where \( \hat{p} = \frac{15}{61} \approx 0.246 \) is the sample proportion, \( p_0 = 0.15 \) is the population proportion, and \( n = 61 \). Substitute these values into the formula: \[ z = \frac{0.246 - 0.15}{\sqrt{\frac{0.15 \times 0.85}{61}}} \].

03

Simplify the Expression

First, calculate the denominator: \( \sqrt{\frac{0.15 \times 0.85}{61}} \approx 0.0476 \). Then, find the numerator: \( 0.246 - 0.15 = 0.096 \). Plug these into the formula: \[ z = \frac{0.096}{0.0476} \approx 2.0168 \].

04

Determine the P-value

Using a standard normal distribution table or calculator, find the \( P \)-value corresponding to \( z = 2.0168 \). The area to the right of \( z = 2.0168 \) is approximately \( P \approx 0.0218 \).

05

Make a Conclusion

Since \( P \approx 0.0218 \) is less than the common significance level of 0.05, we reject the null hypothesis. There is sufficient evidence to conclude that the proportion of students who score a 5 is higher under the teacher's methods than the worldwide proportion of 0.15.

06

Evaluate Use of Z-test

The assumptions of using a \( z \)-test for proportions include a sufficiently large sample size, with both \( np_0 \) and \( n(1-p_0) \) being greater than 10. Here, \( np_0 = 9.15 \) is slightly below 10, which could raise reservations about the normal approximation.

07

Discuss Causality

Part (b) asks if cash incentives caused the increase. The answer is 'no' because causation cannot be inferred from this observational study. The observed result could be due to other factors not controlled in this study, like students' innate abilities or motivation.

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Key Concepts

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

Z-test for Proportions

The Z-test for proportions is a statistical method used to determine if there is a significant difference between an observed sample proportion and a hypothesized population proportion. It's particularly handy when we want to compare two proportions or check if an observed proportion is significantly different from a known probability.
To perform a Z-test for proportions, follow these steps:

• Set your null hypothesis (\(H_0\)): It often states that there is no difference between the sample proportion (\(\hat{p}\)) and the population proportion (\(p_0\)).
• Formulate an alternative hypothesis (\(H_a\)): This is what you suspect might be true (e.g., \( \hat{p} > p_0 \)).
• Calculate the test statistic using: $$ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} $$ where \( \hat{p} \) is the sample proportion, \( p_0 \) is the population proportion, and \( n \) is the sample size.
• Compare the calculated Z-score to a standard normal distribution to find the P-value.
• The P-value informs us of the probability that the observed data would occur if the null hypothesis were true.
• If the P-value is lower than the significance level (e.g., 0.05), reject \( H_0 \).

In our example, the teacher's students showed a higher success rate, and the test indicated this result was unlikely due to random chance alone, suggesting the methods might have a measurable positive impact.

AP Statistics

Advanced Placement (AP) Statistics is a course offered to high school students, typically taught at a college level. It covers major statistical concepts and methods, enabling students to explore data, understand statistical concepts, and apply them in real-life scenarios. The AP exam at the end of the course tests students on their ability to analyze and respond to statistical problems.
Subjects covered in AP Statistics include:

• Exploring Data: Examining distributions, relationships, and patterns.
• Sampling and Experimentation: How to plan and conduct a well-designed study.
• Anticipating Patterns: Understanding probability, random variables, and prediction models.
• Statistical Inference: Estimation techniques and testing hypotheses, including the use of Z-tests.

The course applies various real-world scenarios to the statistical methods students learn; for instance, in the teacher's class, the effectiveness of incentives in improving AP exam scores was analyzed. This practical approach helps grasp the utility of statistics in drawing conclusions from data, a vital skill in many disciplines.

Causality in Statistics

Causality in statistics refers to the relationship between two events where one directly affects the other. Understanding causality is crucial, but it requires careful methodological approaches since correlation does not imply causation.

Types of studies that help verify causality include:

• Randomized Controlled Trials (RCTs): Considered the gold standard, where subjects are randomly assigned to receive either the treatment or a placebo. This helps ensure that observed effects are due to the treatment.
• Observational Studies: In these, researchers merely observe subjects and measure variables without assigning interventions. While they can suggest associations, they are not definitive for establishing causality.

The teacher's study of cash incentives falls under observational studies. While the students' increased success rate might suggest incentives were beneficial, without controlling for other variables (like student motivation or innate ability), we cannot definitively conclude that incentives alone caused the improvement. Thus, any claim of causality should be approached with caution, ensuring that all potential confounding factors are considered.