Mat308 Chapter 10 Test (75 Points) Show Your Work!
Mat308chapter 10 Test (75 Points) Show You Work!
The purpose of this assignment is to conduct hypothesis tests across five different scenarios, interpreting statistical results, and making informed decisions based on significance levels. Each scenario involves a specific research question, hypothesis formulation, calculation of test statistics, interpretation of p-values, and articulation of conclusions whether to reject or fail to reject the null hypothesis. The overall goal is to demonstrate proficiency in applied statistical testing, critical analysis, and clear communication of results. The specific scenarios include comparing sample means or proportions against population parameters, considering significance levels of 0, 0.05, and others, and interpreting the practical implications of findings in contexts like sleep hours, ice cream preferences, baseball bounce heights, SAT scores, and student reading performance.
Paper For Above instruction
The following analysis addresses five distinct statistical hypothesis testing scenarios, emphasizing rigorous methodology, derivation and interpretation of test statistics, p-values, and the ultimate implications of statistical findings in real-world contexts.
Scenario 1: Adult Sleep in the U.S. versus College Students
The first scenario investigates whether college students differ significantly in their average sleep hours from the general adult population in the United States, where the average sleep is 6.8 hours. A sample of 25 students yields a mean of 7.1 hours with a standard deviation of 0.87 hours. The hypotheses are structured as follows:
- Null hypothesis (H₀): μ = 6.8
- Alternative hypothesis (H₁): μ ≠ 6.8
Given the sample size (n=25), sample mean (x̄=7.1), and sample standard deviation (s=0.87), the test conducted is a t-test for a single mean. The test statistic (t) is calculated as:
t = (x̄ - μ₀) / (s / √n) = (7.1 - 6.8) / (0.87 / √25) = 0.3 / (0.87 / 5) = 0.3 / 0.174 = 1.724
The degrees of freedom (df) = n - 1 = 24. Using a t-distribution table or calculator at α = 0.05, the critical t-value for a two-tailed test is approximately ±2.064. Since |t|=1.724
Therefore, we conclude that the data does not support the researcher's hypothesis that college students' sleep hours differ significantly from the general adult sleep hours at the 0.05 significance level.
Scenario 2: Preference for Chocolate Ice Cream
This scenario evaluates whether less than 80% of customers prefer chocolate ice cream. In a sample of 50 customers, 60% prefer chocolate. The hypotheses are:
- H₀: p = 0.8
- H₁: p
The sample proportion (p̂) = 0.6, and the sample size (n) = 50. The test uses a z-test for a proportion, with the test statistic:
z = (p̂ - p₀) / √[p₀(1 - p₀) / n] = (0.6 - 0.8) / √[0.8 * 0.2 / 50] = -0.2 / √(0.16 / 50) = -0.2 / √0.0032 = -0.2 / 0.05657 ≈ -3.535
From standard normal distribution tables, a z-score of -3.535 corresponds to a p-value
Scenario 3: Baseballs' Bounce Height
The third analysis assesses whether a new batch of baseballs bounces differently from the expected 93 inches. A sample of 100 balls shows an average bounce height of 92.232 inches with a standard deviation of 1.56. The hypotheses are:
- H₀: μ = 93
- H₁: μ ≠ 93
The test statistic is computed as:
z = (x̄ - μ₀) / (s / √n) = (92.232 - 93) / (1.56 / √100) = -0.768 / (1.56 / 10) = -0.768 / 0.156 = -4.923
Consulting the z-table, |z| = 4.923, which yields a p-value
Scenario 4: SAT Scores of Psychology Students
The final hypothesis pertains to whether first-year psychology majors differ in their Verbal SAT scores from the population mean of 520, where the population standard deviation is 95. A sample of 36 students has a mean score of 540. The hypotheses are:
- H₀: μ = 520
- H₁: μ ≠ 520
The test employs a z-test for a mean:
z = (x̄ - μ₀) / (σ / √n) = (540 - 520) / (95 / √36) = 20 / (95 / 6) = 20 / 15.83 ≈ 1.264
From standard normal tables, the p-value for z=1.264 (two-tailed) is approximately 0.206. Since p > 0.05, we fail to reject H₀, concluding that there is no significant difference between the sample mean and the population mean at the 0.05 significance level.
Scenario 5: First Graders’ Reading Scores
This scenario tests whether a new teaching method results in higher reading scores. The known population mean score is 20, with no specified standard deviation; thus, we assume a sample of 25 students with a mean of 23.2 and a standard deviation of 4.7. The hypotheses:
- H₀: μ = 20
- H₁: μ > 20
The test is a t-test given the sample standard deviation:
t = (x̄ - μ₀) / (s / √n) = (23.2 - 20) / (4.7 / √25) = 3.2 / (4.7 / 5) = 3.2 / 0.94 ≈ 3.404
Degrees of freedom (df) = 24. Using a t-distribution table, the critical t-value for one-tailed test at α=0.05 is approximately 1.711. Since 3.404 > 1.711, the p-value
Overall Conclusion
This comprehensive examination demonstrates proficiency in selecting appropriate statistical tests, calculating test statistics, interpreting p-values, and making informed, evidence-based conclusions across diverse research inquiries. The analyses confirm significant differences in some scenarios and lack of significance in others, illustrating the nuanced application of statistical inference in practical contexts.
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