Engineering Probability Homework 71 Sony Manufactures
Engineering Statsprobability Hw 71 Sony Manufactures
Compare proportions on time from two Sony manufacturing plants using hypothesis testing, including steps without and with p-value as a rejection rule. Test variability in camshaft hardness depth after heat treatment, including steps with and without p-value. Analyze differences in website visits from two search engines, including steps with and without confidence intervals. Evaluate whether two machines produce different breaking strengths of artificial leather, including steps with and without confidence interval. Determine if forged transmission pans meet weight requirements, including steps without and with p-value. Assess if maintenance costs of motors differ from a specified amount, including steps with and without p-value and calculate Type II error for a specific mean. Test if variance of PVC pipes conforms to a given value, including steps without and with confidence interval. Judge acceptability of artificial leather based on breaking strength, including steps with and without p-value. Calculate optimal guarantee period for compressors based on mean life and acceptable failure rate, including normal distribution plots. Test if IEEE claim about engineers continuing graduate studies is supported, using hypothesis testing with p-value placebo. For all tests, specify hypotheses, perform relevant statistical analysis, visualize data distribution, and interpret results according to the specified rejection rule.
Paper For Above instruction
1. Comparison of Proportions from Two Sony Plants
The hypothesis test aims to determine whether the proportion of on-time orders from the Tai Pi plant exceeds that from the Seoul plant. First, define the hypotheses: H₀ (null) states that the proportions are equal (p₁ = p₂), and H₁ (alternative) asserts that the proportion from Tai Pi is greater (p₁ > p₂). The sample data shows 798 of 1,000 orders completed on time at Tai Pi and 631 of 1,000 at Seoul. Calculating sample proportions: p̂₁ = 0.798, p̂₂ = 0.631. The pooled proportion: p̂ = (798 + 631) / 2000 = 0.7145. The standard error (SE) for the difference: SE = √[p̂(1 - p̂) (1/n₁ + 1/n₂)] = √[0.7145 0.2855 (1/1000 + 1/1000)] ≈ 0.0202. The test statistic (z): (p̂₁ - p̂₂) / SE = (0.798 - 0.631) / 0.0202 ≈ 8.37. This z-value indicates a significant difference supporting the claim that Tai Pi's proportion is higher. Since the problem specifies no p-value determination in part A, we rely on the z-score exceeding typical critical values for conclusion. The result suggests strong evidence against H₀, favoring H₁.
In part B, the same hypotheses are tested, but the decision rule is based solely on the p-value. With z = 8.37, the p-value approaches zero, far below any standard significance level, leading to rejection of H₀. This confirms that the proportion of on-time orders from Tai Pi is statistically greater than from Seoul, with high confidence.
2. Variance Reduction in Camshaft Hardness Depth via Heat Treatment
The hypothesis centers on whether the new heating process reduces the variance of camshaft hardness depth. Null hypothesis H₀ states that variance remains unchanged (σ² = 0.47²), and alternative H₁ suggests reduction (σ² S² / σ₀², where σ₀² = 0.47². Calculating: χ² = (30 - 1) 0.297489² / 0.47² ≈ 17.4. The critical chi-square value for df=29 at significance level α = 0.01 (one-tailed) is approximately 13.1. Since the computed χ² exceeds this value, we do not reject H₀, indicating insufficient evidence to claim the variance has decreased. In part B, with p-value approach, the p-value corresponding to χ²=17.4 at df=29 exceeds 0.01, so we again fail to reject H₀, consistent with part A.
3. Difference in Web Traffic from Two Search Engines
To evaluate if there is a significant difference in average daily visits, we formulate H₀: μ₁ = μ₂ against H₁: μ₁ ≠ μ₂. Since the data are paired daily visits, a paired t-test is suitable. The differences per day are computed and plotted as a distribution histogram: differences reveal variability but trend analysis shows whether the mean difference is statistically significant. The test statistic t is computed from the mean difference and its standard deviation, considering sample size n=7. After calculating the mean of differences, standard deviation, and t-value, compare it with critical t at df=6 and α=0.05. The resulting t-value is compared against the t-distribution plot, leading to a p-value which, if below 0.05, indicates significant difference. Alternatively, the confidence interval for the mean difference, if it does not include zero, supports the conclusion of a statistically significant difference, fulfilling the requirement without p-value as a rejection rule.
4. Breaking Strength of Artificial Leather from Two Machines
Hypotheses: H₀: μ₁ = μ₂ (no difference in breaking strength), H₁: μ₁ ≠ μ₂. Since variances are unequal, a two-sample t-test with unequal variances (Welch’s test) applies. Sample means are 8.73 and 8.68, standard deviations 0.591608 and 0.632456, sample sizes 15 and 17, respectively. The test statistic is calculated considering unequal variances, and the degrees of freedom are approximated via the Welch-Satterthwaite equation. The t-value and degrees of freedom are used to find the p-value, which indicates whether the difference is statistically significant. Alternatively, a confidence interval for the difference in means, if it includes zero, suggests no significant difference. The distribution plot, illustrating the t-distribution with calculated degrees of freedom, aids in visual interpretation.
5. Testing if Forged Transmission Pans Meet Weight Requirement
The hypotheses: H₀: μ ≤ 20 (mean weight is less than or equal to 20 LBs) versus H₁: μ > 20. The sample mean is 21.4, standard deviation 2.1, sample size 10, significance level α=0.025. The test uses a t-distribution with n-1 = 9 degrees of freedom. The test statistic t = (x̄ - μ₀) / (s/√n) = (21.4 - 20) / (2.1 / √10) ≈ 2.154. Comparing this to the critical t-value at α=0.025 with df=9, approximately 2.262, we see t
6. Maintenance Cost of 3 HP Motors
Null hypothesis: H₀: μ = 21. The sample size is 25; mean cost is $20, standard deviation $3. Using a t-test for the mean, t = (20 - 21) / (3/√25) = - (1) / (0.6) ≈ -1.67. At α=0.05, degrees of freedom = 24, the critical t-value is approximately ±2.064. Since |−1.67|
7. Variance Test for PVC Pipes
Null hypothesis H₀: σ² = 0.17. Sample variance S² = 0.328². Test statistic χ² = (n−1) S² / σ₀². Calculate χ² and compare with critical values at α=0.01. Use the chi-square distribution plot to determine support or rejection of H₀. Without p-value, the test relies on the critical chi-square values. The plot shows whether the sample variance falls into the rejection region.
8. Artificial Leather Breaking Strength
Null hypothesis H₀: μ = 100 psi. Sample mean = 100.6, standard deviation = 2, sample size = 9. The t-test statistic t = (100.6 - 100) / (2/√9) = 0.9. Degrees of freedom = 8. Critical t at α=0.05 is approximately 2.306. Since 0.9
9. Guarantee Period for Gas Compressors
Given mean life μ=10 years, σ=1 year, acceptable failure rate of 3%. Using the normal distribution, find z-value corresponding to the 97th percentile: z = Φ^(-1)(0.97) ≈ 1.88. The guarantee period T = μ + zσ = 10 + 1.881 ≈ 11.88 years. This ensures only 3% fail before this period. Plot the normal distribution to visualize probability curves, confirming that the chosen period captures 97%, matching the acceptable failure rate.
10. Claim on Engineers Continuing Graduate Studies
Sample proportion p̂ = 267/484 ≈ 0.55. Null hypothesis H₀: p = 0.5; alternative H₁: p ≠ 0.5. Compute the test statistic: z = (p̂ - 0.5) / √[0.5*0.5 / 484] ≈ 2.17. At α=0.05, rejection region is |z| > 1.96. Since 2.17 > 1.96, reject H₀, supporting the claim that more than half of the engineers plan to continue grad studies, consistent with the IEEE statement. The normal plot shows the z-score within the rejection boundaries, confirming the hypothesis.
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