UNIT 4: Written Assignment

HS 4241 - 01 Psychopathology and Mental Health - UNIT 4 Written Assignment Instructor: Pankaj Mehrotra Date: 25/02/2025 QUESTION 1 One-tailed probabilities for the given z-scores: ●​ z = -1.79 → Probability ≈ 0.0367 (to the left of -1.79) ●​ z = -0.03 → Probability ≈ 0.4888 ●​ z = 2.2 → Probability ≈ 0.9861 (to the right: 1 - 0.9861 = 0.0139 for one-tailed probability) ●​ z = 7.00 → Probability ≈ 0.0000 (extremely small, essentially 0 for practical purposes) ●​ z = 0.004 → Probability ≈ 0.5016 ●​ z = 0.534 → Probability ≈ 0.7035 QUESTION 2 We use the formula for a confidence interval (CI) for the population mean when the population standard deviation (σ\sigma) is known: Where: ●​ X = 25.5 (sample mean weight) ●​ σ = 3.1 (population standard deviation) ●​ n = 40 (sample size) ●​ Z* is the critical value from the z-table for the desired confidence level. 1. 99% Confidence Interval: ●​ Z value for 99% confidence level = 2.576 CI = 25.5 ± 2.576 × (3.1/40) Lower bound: 25.5 − 1.26 = 24.24 Upper bound: 25.5 + 1.26 = 26.76 99% CI: (24.24 kg, 26.76 kg) 2. 95% Confidence Interval: ●​ Z value for 95% confidence level = 1.96 CI= 25.5 ± 1.96 × (3.1/40) Using the SE = 0.49: CI = 25.5 ± 0.96 Lower bound: 25.5 − 0.96 = 24.54 Upper bound: 25.5 + 0.96 = 26.46 95% CI: (24.54 kg, 26.46 kg) QUESTION 3 We use the z-score formula to calculate probabilities from a normal distribution: Where: ●​ X = observed value ●​ μ = 48 cm (mean) ●​ σ = 5.2 cm (standard deviation) 1. Probability that a randomly selected baby has a length of at least 70 cm: Step 1: Find the z-score for 70 cm: z = (70−48) / 5.2 =≈ 4.23 Step 2: Find the probability:​ Looking up z = 4.23 in the z-table, the probability to the left of this value is approximately 0.99999. Step 3: Since we want the probability of a baby being at least 70 cm (the area to the right), we subtract this from 1: P (X≥70) = 1 − 0.99999 = 0.00001 or about 0.001% — extremely rare! 2. Probability that a randomly selected baby has a length between 40 cm and 60 cm: Step 1: Find z-scores: ●​ For 40 cm: z = (40 − 48) / 5.2 =≈−1.54 ●​ For 60 cm: z = (60−48) / 5.2 =≈2.31 Step 2: Get the probabilities from the z-table: ●​ P (X≤40) ≈ 0.0618 ●​ P (X≤60) ≈ 0.9893 Step 3: The probability between 40 cm and 60 cm is: P (40 ≤ X ≤ 60) = P (X≤60) − P (X≤40) = 0.9893 − 0.0618 = 0.9275 or about 92.75%. Likelihood of a baby being at least 70 cm: 0.001% (very rare) Likelihood of a baby being between 40 cm and 60 cm: 92.75% (very likely) QUESTION 4 Let’s calculate the standard error of the difference between two means using this formula: Where: ●​ σ1 = 25 mm Hg (standard deviation of the treatment group) ●​ n1 = 150 (sample size of the treatment group) ●​ σ2 = 23 mm Hg (standard deviation of the control group) ●​ n2 = 95 (sample size of the control group) Step 1: Square the standard deviations σ1^2 = 25^2 = 625 σ2^2 = 23^2 = 529 Step 2: Divide by the sample sizes σ1^2/n1 = 625/150 ≈ 4.17 σ2^2/n2 = 529/95 ≈ 5.57 Step 3: Sum the two values 4.17 + 5.57 = 9.74 Step 4: Take the square root SEdiff = sqrt(9.74) ≈ 3.12 The standard error of the difference of the two means is approximately: 3.12 mm Hg REFERENCES: ●​ Illowsky, B. and Dean, S. (2017). Introductory Statistics. OpenStax Rice University. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution 4.0 International License (CC BY 4.0). Retrieved from https://my.uopeople.edu/pluginfile.php/1954716/mod_page/content/3/TEXT%20intro ductory-statistics.compressed.pdf ●​ Gravetter, F. J., & Wallnau, L. B. (2020). Statistics for the Behavioral Sciences (11th ed.). Cengage Learning. ●​ Illowsky, B., & Dean, S. (2017). Introductory Statistics. OpenStax. https://openstax.org/books/introductory-statistics ●​ McClave, J. T., Benson, P. G., & Sincich, T. (2018). Statistics for Business and Economics (13th ed.). Pearson.