EDPSY 490 Win 2025 Homework 7: 2-group z/t-tests NHST/power (Ch. 7) (15 points) Due 4pm Feb 25 \( { }^{\text {th }} \) Name: \( \underline{\text { 2-Group Power (Chapter 7): Assume } \alpha=.05,2-t a i l e d . ~ P l e a s e ~ u s e ~ Z c r i t i c a l ~}= \pm 1.96 \) for this exercise (for power analysis); do not round the \( \% \) area under the curve. Please also show all your arithmetic steps in order to receive credit for your answers and attach additional paper as needed. Scenario: Imagine that you are writing a proposal to get some funding from a local school district board to study whether the game of "scatterball" has better child mental health outcomes than the traditional game of "dodgeball" during PE class (both games involve the children throwing a ball at each other, but in scatterball, no one is tagged "out" of the game). Specifically, you plan to randomly assign third-grade children in an after-school program to one of two groups: one group will play scatterball daily for a week, one group will play traditional dodgeball daily for a week. At the end of the week, you plan to measure each child on their mental health wellness using a mental health scale that has a population variance \( \sigma^{2}=160 \). You believe, based on prior research, that mental health for the first group (that played scatterball) will be +16 points better than the second group (that played traditional dodgeball). Your initial plan is to have \( \mathbf{n}=5 \) students in group 1 and \( \mathbf{n}=5 \) students in group 2 . 3. (a) Assuming your belief that there is a true difference between groups \( \mu 1-\mu 2=+16 \) points, sketch a picture of what these two distributions look like (the null and the alternative) using \( \alpha=.05 \) (2-tailed). This will be a picture of two sampling distributions that overlap. Use the steps discussed in class: (i) draw the null, (ii) label the critical regions, (iii) draw the alternative distribution, and (iv) find the area of the alternative that crosses into the critical region of the null, and shade that area. USE COLOR PENCILS. 2. (b) Examine your sketch above. i. What is your approximate Power (this is the shaded area)? Show the steps of your calculations. ii. What is your approximate Type II error rate? iii. What is your Type I error rate? EdPsy490 Win2025 HW7, p. 4

Below is one acceptable answer. (Your instructor expects that you also produce a colored sketch on paper. Here we describe what to sketch and then show the calculations.) ────────────────────────────── Step 1. Compute the Standard Error (SE) ────────────────────────────── You are comparing two independent groups with n = 5 each. The population variance is given as σ² = 160. Thus, the variance for each group’s mean is 160/5 = 32. Because the groups are independent, the variance of the difference in means is   Var(μ₁ – μ₂) = 32 + 32 = 64.   So, SE = √64 = 8. ────────────────────────────── Step 2. Find the Critical Values Under the Null ────────────────────────────── Under the null hypothesis (i.e., assuming no difference), the sampling distribution of the difference is centered at 0 with a standard error of 8. You are using a two‐tailed test with α = 0.05 so that the z-critical values are ±1.96. This means that the critical values on the scale of the difference are   Lower critical value: –1.96 × 8 = –15.68   Upper critical value: +1.96 × 8 = +15.68 You will reject H₀ if the observed difference is less than –15.68 or greater than +15.68. ────────────────────────────── Step 3. Sketching the Two Sampling Distributions ────────────────────────────── On your paper, make a horizontal axis of “Group Difference (μ₁ – μ₂).” Now draw two normal curves: (i) The Null Distribution:   • Center this curve at 0.   • Mark the two critical regions beyond –15.68 and +15.68 (shaded areas in both tails). (ii) The Alternative Sampling Distribution (Under the True Difference):   • This distribution is for the true difference μ₁ – μ₂ = +16 and is also Normal with SE = 8.   • Center this curve at 16.   • Because the two distributions overlap, part of the alternative curve will extend into the rejection region of the null (the area to the right of +15.68) and (a tiny bit, almost negligible) into the lower rejection region. Mark and shade the overlapping area that falls in the null’s critical region. (Use your color pencils.) ────────────────────────────── Step 4. Power and Errors Calculations ────────────────────────────── Under the alternative (true mean difference = 16, SE=8), we need to calculate the probability that the sample difference falls into one of the null’s rejection regions. (a) Find the probability that the observed difference is ≥ +15.68:   Convert to a z-score:     z = (15.68 – 16) / 8 = –0.04   Using the standard normal distribution,     P(Z ≥ –0.04) ≈ 0.5160     [since P(Z ≥ –0.04) = 1 – P(Z < –0.04) and note that P(Z < –0.04) ≈ 0.4840] (b) Find the probability that the observed difference is ≤ –15.68:   Compute its z-score:     z = (–15.68 – 16) / 8 = –31.68 / 8 = –3.96   Then,     P(Z ≤ –3.96) is extremely small, approximately 0.00004 (almost 0). Thus the total power (chance of correctly rejecting H₀ given μ₁ – μ₂ = 16) is approximately   Power = 0.5160 + ≈0 = 0.5160 (or about 51.6%). ────────────────────────────── Step 5. Answering the Questions ────────────────────────────── (b) i. What is your approximate Power?   Power ≈ 51.6%   Steps:    1. SE = 8.    2. Critical difference = 15.68.    3. Compute z for 15.68: (15.68 – 16)/8 = –0.04, yielding P ≈ 51.6%.    4. (The lower tail contributes ≈0.)   Thus, the “shaded area” on your sketch is about 51.6% of the alternative distribution.   ii. What is your approximate Type II error rate?   By definition, Type II error (β) = 1 – Power.     β ≈ 1 – 0.516 = 0.484 (or about 48.4%).   iii. What is your Type I error rate?   By design, α = 0.05 (or 5%). ────────────────────────────── Final Summary Answer ────────────────────────────── • Your sketch should show two overlapping normal curves:   – The Null curve: centered at 0, with tails shaded beyond –15.68 and +15.68.   – The Alternative curve: centered at 16 (with the same SE = 8), with the area to the right of +15.68 (and negligibly into the left tail) shaded. • Calculations:   Standard Error = 8.   Critical differences = ±15.68.   For X = 15.68 under the alternative, z = –0.04 → P ≈ 51.6%.   Therefore:    ∴ Power ≈ 51.6%     Type II error rate (β) ≈ 48.4%     Type I error rate (α) = 5%. This completes the answer.