AP Statistics 7.10 Skills Focus: Selecting, Implementing, and Communicating Inference Procedures Study Notes
AP Statistics 7.10 Skills Focus: Selecting, Implementing, and Communicating Inference Procedures Study Notes- New syllabus
AP Statistics 7.10 Skills Focus: Selecting, Implementing, and Communicating Inference Procedures Study Notes -As per latest AP Statistics Syllabus.
LEARNING OBJECTIVE
- Skills Focus: Selecting, Implementing, and Communicating Inference Procedures
Key Concepts:
- Skills Focus: Selecting, Implementing, and Communicating Inference Procedures
Skills Focus: Selecting, Implementing, and Communicating Inference Procedures
Skills Focus: Selecting, Implementing, and Communicating Inference Procedures
This topic emphasizes the ability to:
- Select an appropriate inference procedure based on the type of data and the research question (proportion, mean, difference of means, difference of proportions, regression slope, chi-square tests).
- Implement the procedure correctly: calculate test statistics, p-values, and/or confidence intervals while verifying assumptions and conditions.
- Communicate results in context: interpret confidence intervals or test results, justify claims about the population, and relate conclusions back to the research question.
Key Considerations for Selection:
- Type of variable: quantitative vs. categorical
- Number of samples/populations: one-sample vs. two-sample vs. paired
- Known or unknown population parameters: σ known → z-procedures; σ unknown → t-procedures
- Distribution assumptions: normality, sample size for CLT, independence, random sampling
General Steps for Implementing Inference:
- Identify the parameter of interest (mean, proportion, difference, slope).
- Choose the appropriate procedure (z-test, t-test, chi-square, regression t-test, etc.).
- Verify conditions and assumptions for the selected procedure.
- Compute the test statistic or confidence interval using formulas or software.
- Determine the p-value or interpret the confidence interval.
- Make a decision and communicate the result in context.
Example
A company wants to test whether a new training program increases employee productivity. A random sample of 20 employees trained under the new program has mean productivity score \(\bar{x} = 78\), standard deviation \(s = 10\). The historical mean is 75. Conduct a significance test at \(\alpha = 0.05\).
▶️ Answer / Explanation
Step 1: Identify parameter → population mean productivity \(\mu\).
Step 2: Select procedure → one-sample t-test (σ unknown, small sample, assume approximately normal).
Step 3: Verify conditions → random sample, independence (n ≤ 10% of population), approximately normal distribution.
Step 4: Compute test statistic:
\( t = \dfrac{\bar{x} – \mu_0}{s / \sqrt{n}} = \dfrac{78 – 75}{10 / \sqrt{20}} = \dfrac{3}{2.236} \approx 1.34 \)
Step 5: Degrees of freedom → df = 20 – 1 = 19.
Step 6: Determine p-value → one-sided t-test: p ≈ 0.096 (from t-table or calculator).
Step 7: Decision → p > α → fail to reject H₀.
Conclusion: There is insufficient evidence at the 5% significance level to conclude that the new training program increases productivity.
Example
A survey finds that 62 out of 100 randomly selected students prefer online lectures over in-person lectures. The university claims that 50% of students prefer online lectures. Test the claim at \(\alpha = 0.05\).
▶️ Answer / Explanation
Step 1: Parameter → population proportion \(p\).
Step 2: Procedure → one-sample z-test for proportion.
Step 3: Verify conditions → random sample, n p₀ = 100 × 0.5 = 50 ≥10, n(1-p₀) = 50 ≥10 → OK.
Step 4: Compute test statistic:
\( z = \dfrac{\hat{p} – p_0}{\sqrt{p_0(1-p_0)/n}} = \dfrac{0.62 – 0.5}{\sqrt{0.5 × 0.5 /100}} = \dfrac{0.12}{0.05} = 2.4 \)
Step 5: p-value → two-sided: p ≈ 0.0164
Step 6: Decision → p < 0.05 → reject H₀.
Conclusion: Evidence suggests that the proportion of students preferring online lectures differs from 50%.
Example
Two teaching methods are compared. Sample 1: n₁ = 16, x̄₁ = 82, s₁ = 8. Sample 2: n₂ = 14, x̄₂ = 77, s₂ = 7. Assume independent samples. Test whether Method 1 leads to higher scores at α = 0.05.
▶️ Answer / Explanation
Step 1: Parameter → difference of population means \(\mu_1 – \mu_2\).
Step 2: Procedure → two-sample t-test for difference of means (σ unknown).
Step 3: Verify conditions → independent, random samples, approximately normal or n ≥30? n₁+n₂ < 30 → check approximately normal → assume OK.
Step 4: Compute test statistic:
\( SE = \sqrt{\dfrac{8^2}{16} + \dfrac{7^2}{14}} = \sqrt{4 + 3.5} = \sqrt{7.5} \approx 2.738 \)
\( t = \dfrac{82 – 77}{2.738} \approx 1.83 \)
Step 5: df ≈ smaller of n₁-1, n₂-1 → df ≈ 13
Step 6: p-value → one-sided ≈ 0.045
Step 7: Decision → p < 0.05 → reject H₀.
Conclusion: Method 1 likely leads to higher mean scores than Method 2.
Example
10 students take a pre-test and post-test after a review session. Differences (post – pre) have x̄_d = 5, s_d = 3. Test at α = 0.05 whether the session improves scores.
▶️ Answer / Explanation
Step 1: Parameter → mean difference μ_d.
Step 2: Procedure → one-sample t-test for matched pairs.
Step 3: Verify conditions → random sample, differences independent, approximately normal (n = 10 < 30) → assume roughly normal.
Step 4: Compute test statistic:
\( t = \dfrac{x̄_d – 0}{s_d / \sqrt{n}} = \dfrac{5 – 0}{3 / \sqrt{10}} = \dfrac{5}{0.9487} \approx 5.27 \)
Step 5: df = n-1 = 9
Step 6: p-value → very small (p < 0.001)
Step 7: Decision → reject H₀.
Conclusion: The review session significantly improves student scores.
Example
A researcher investigates whether hours studied (x) predicts exam score (y) for 12 students. Sample regression line: \( \hat{y} = 50 + 4x \), standard error of slope \( SE_b = 1.2 \). Test whether slope differs from 0 at α = 0.05.
▶️ Answer / Explanation
Step 1: Parameter → population slope β.
Step 2: Procedure → one-sample t-test for regression slope.
Step 3: Verify conditions → random sample, independence, linearity, residuals approximately normal.
Step 4: Compute test statistic:
\( t = \dfrac{b – \beta_0}{SE_b} = \dfrac{4 – 0}{1.2} = 3.33 \)
Step 5: Degrees of freedom → df = n – 2 = 12 – 2 = 10
Step 6: p-value → two-sided ≈ 0.007
Step 7: Decision → p < 0.05 → reject H₀.
Conclusion: Hours studied is significantly associated with exam score; slope differs from 0.
Example
A clinical trial tests two treatments for a condition. Treatment A: 45 of 80 patients improved. Treatment B: 30 of 70 improved. Test whether the proportions differ at α = 0.05.
▶️ Answer / Explanation
Step 1: Parameter → difference of population proportions \( p_A – p_B \).
Step 2: Procedure → two-sample z-test for proportions.
Step 3: Verify conditions → independent random samples, np and n(1-p) ≥10 for both groups → OK.
Step 4: Compute pooled proportion:
\( \hat{p} = \dfrac{45 + 30}{80 + 70} = \dfrac{75}{150} = 0.5 \)
\( SE = \sqrt{\hat{p}(1-\hat{p})\left(\dfrac{1}{n_A} + \dfrac{1}{n_B}\right)} = \sqrt{0.5 \cdot 0.5 \left(\dfrac{1}{80} + \dfrac{1}{70}\right)} \approx 0.0845 \)
\( z = \dfrac{0.5625 – 0.429}{0.0845} \approx 1.59 \)
Step 5: p-value → two-sided ≈ 0.11
Step 6: Decision → p > 0.05 → fail to reject H₀.
Conclusion: There is insufficient evidence to conclude that the improvement rates differ between treatments A and B.