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Now, let's delve into the heart of statistical mastery with a couple of master-level questions, expertly crafted to challenge your intellect and sharpen your analytical skills. Our resident statistician, Dr. Statistical Genius, will guide you through the solutions, providing insights that will illuminate your path to success.
Question 1: A manufacturing company is conducting a study to assess the relationship between production speed (in units per hour) and defect rate (percentage of defective units). The company collected data from 50 production runs and performed a regression analysis. The regression output is as follows:
- Regression Equation: Defect Rate = 4.25 - 0.07 * Production Speed
- Coefficient of Determination (R²): 0.81
- Standard Error of the Estimate: 0.02
Based on this information, calculate the predicted defect rate for a production speed of 200 units per hour.
Solution 1: To calculate the predicted defect rate, we'll use the regression equation provided:
Defect Rate = 4.25 - 0.07 * Production Speed
Substituting the production speed value of 200 units per hour into the equation:
Defect Rate = 4.25 - 0.07 * 200 = 4.25 - 14 = 0.25
Therefore, the predicted defect rate for a production speed of 200 units per hour is 0.25 (or 25%).
Question 2: A researcher is studying the effects of a new drug on blood pressure. In a clinical trial, 100 participants were randomly assigned to either the treatment group receiving the new drug or the control group receiving a placebo. The mean systolic blood pressure for the treatment group was found to be 122 mmHg with a standard deviation of 8 mmHg, while the mean systolic blood pressure for the control group was 128 mmHg with a standard deviation of 7 mmHg. Conduct a hypothesis test to determine if there is a significant difference in mean systolic blood pressure between the treatment and control groups, using a significance level of 0.05.
Solution 2: To conduct a hypothesis test for comparing means, we'll use the two-sample independent t-test. The null and alternative hypotheses are:
- Null Hypothesis (H0): There is no significant difference in mean systolic blood pressure between the treatment and control groups. (μ1 = μ2)
- Alternative Hypothesis (H1): There is a significant difference in mean systolic blood pressure between the treatment and control groups. (μ1 ≠ μ2)
We'll use a significance level of α = 0.05. Given the sample means, standard deviations, and sample sizes for both groups, we'll calculate the t-statistic and compare it to the critical t-value from the t-distribution.
t = (mean1 - mean2) / √((s1²/n1) + (s2²/n2))
t = (122 - 128) / √((8²/100) + (7²/100)) = -6 / √((64/100) + (49/100)) = -6 / √(0.64 + 0.49) = -6 / √(1.13) ≈ -6 / 1.06 ≈ -5.66
Degrees of freedom (df) = n1 + n2 - 2 = 100 + 100 - 2 = 198
Using statistical software or a t-table, the critical t-value for a two-tailed test with α = 0.05 and df = 198 is approximately ±1.96.
Since the absolute value of the calculated t-statistic (|t| = 5.66) exceeds the critical t-value (1.96), we reject the null hypothesis. Therefore, there is sufficient evidence to conclude that there is a significant difference in mean systolic blood pressure between the treatment and control groups.
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