c. the obtained value (z or t), e. your decision to retain or reject the null and why f. the two possibilities that exist, given your decision 1. Imagine you are interested in students' feelings about the course text. You randomly select 5 students to complete your class with the course text and then you ask them to rate their feelings about the text using a scale of \( 1-5 \), where \( 1= \) dislike very much and \( 5= \) like very much. The true mean of the rating scale is 3 and the true variance is 2.5 . Your sample mean turns about to be an average rating of 2.1 . Do students differ from chance in their feelings about the course text? Imagine you are interested in curricula that helps prepare people for a math achievement test. You randomly select 10 people to receive an intensive 2 -month math-test-preparation class. At the end of their participation, you measure the people on their math achievement on a norm- referenced test that has true mean of 100 and a true variance of 225 . You find that your sample mean turned about to be an average score of 113 . Do people who participate in your intensive

Below are worked‐out answers for each scenario. ────────────────────────────── Scenario 1: Students’ Feelings about the Course Text The situation is as follows: • Population (chance) mean μ₀ = 3 • Population (true) variance σ² = 2.5 (so σ = √2.5) • Sample size n = 5 • Observed sample mean M = 2.1 Since the population variance is known, we use a z‑test. Step 1. Compute the standard error:   SE = σ/√n = √2.5/√5 = √(2.5/5) = √0.5 ≈ 0.707 Step 2. Compute the test statistic (c: the obtained value):   z = (M – μ₀) / SE = (2.1 – 3) / 0.707 = (–0.9) / 0.707 ≈ –1.27 Step 3. Decision (e: retain or reject the null, and why): At a common significance level (α = 0.05, two‐tailed), the critical z‑values are approximately ±1.96. Since |–1.27| = 1.27 is less than 1.96, the z‑statistic is not in the rejection region. • Therefore, we do not reject (i.e. we “fail to reject”) the null hypothesis. • This decision is based on the fact that the test statistic does not provide strong enough evidence (p-value > 0.05) to rule out the possibility that the sample came from a population whose true mean rating is 3. Step 4. Two possibilities (f: given your decision): 1. In reality, students’ feelings might truly be centered at a rating of 3 (no difference from chance). 2. Alternatively, the sample is very small (n = 5) and may not reflect a true departure from chance even if one exists—that is, the study might be underpowered to detect a real difference. ────────────────────────────── Scenario 2: Math Achievement and the Intensive Curriculum The situation is as follows: • Population (chance) mean μ₀ = 100 • Population (true) variance σ² = 225 (so σ = √225 = 15) • Sample size n = 10 • Observed sample mean M = 113 Again, with the known variance, use a z‑test. Step 1. Compute the standard error:   SE = σ/√n = 15/√10 ≈ 15/3.162 = 4.743 Step 2. Compute the test statistic (c: the obtained value):   z = (M – μ₀) / SE = (113 – 100) / 4.743 = 13 / 4.743 ≈ 2.74 Step 3. Decision (e: retain or reject the null, and why): At α = 0.05 for a two‑tailed test, the critical values are ±1.96. Since 2.74 > 1.96, the test statistic falls in the rejection region. • Therefore, we reject the null hypothesis and conclude that the average performance of participants significantly differs from chance (it is higher than 100). • This decision is based on the small p‑value associated with z ≈ 2.74 (p ≈ 0.006). Step 4. Two possibilities (f: given your decision): 1. One possibility is that the intensive math test–preparation curriculum is effective and leads to higher-than‐expected test scores. 2. An alternative possibility is that some other factor (such as self-selection of particularly motivated individuals, preexisting higher math ability, or another confounding variable) contributed to the higher observed average score. Further investigation (e.g., with a control group) would be needed to determine the cause conclusively. ────────────────────────────── Summary of Answers For Scenario 1:  c. z ≈ –1.27  e. We fail to reject the null hypothesis because –1.27 does not exceed the critical value of ±1.96 (p > 0.05), indicating insufficient evidence to claim that students’ ratings differ from the chance value of 3.  f. Two possibilities are: (1) students’ true feelings are centered around 3; or (2) the sample is too small, so a real difference (if it exists) could have gone undetected. For Scenario 2:  c. z ≈ 2.74  e. We reject the null hypothesis because 2.74 is greater than the critical value of 1.96 (p < 0.05), showing that the participants’ average test score of 113 is significantly different from (and higher than) the chance value of 100.  f. Two possibilities are: (1) the intensive preparation curriculum truly improves test scores; or (2) other factors (such as selection bias or confounding variables) may explain the higher scores, warranting further research. These conclusions illustrate the importance of adequate sample size and the need for controlled studies when drawing conclusions from statistical tests.