Exercise 2166 P 1592 Exercise 1019 P 583 Use SPSS Data I

Exercise 2166 P 1592 Exercise 1019 P 583 Use Spss Data I

Analyze correlations, regressions, and hypothesis tests based on the given data and SPSS outputs. Perform significance testing for correlation coefficients, interpret regression analysis including R-squared and residuals, construct confidence intervals, and test hypotheses regarding associations between variables. Additionally, evaluate the relationship between tobacco and alcohol spending through regression analysis, including plotting and testing independence, along with residual and normality assessments. For the biologist study, interpret regression results, residuals, and confidence intervals, and for the AOM treatment, perform a hypothesis test to compare proportions. Throughout, provide detailed statistical reasoning, calculations, and interpretations based solely on the given data and output, with appropriate references for methods used.

Sample Paper For Above instruction

Introduction

The present analysis involves a comprehensive exploration of correlation and regression analyses, hypothesis testing, and data visualization based on supplied SPSS outputs and datasets. Several contexts are examined, including biological contamination studies, economic spending behaviors, and medical treatment trials. The goal is to interpret statistical outputs critically, assess relationships and significance, and provide informed conclusions supported by quantitative evidence.

Correlation Significance Testing

To determine whether the correlation between two variables significantly differs from zero, the Pearson correlation coefficient (r) is used alongside a t-test statistic. The test statistic is computed as t = r√(n-2)/√(1–r²), where n is the sample size. The degrees of freedom are n-2. The p-value associated with the calculated t determines significance at a specified alpha level. For instance, if r = 0.9 with n=30, t = 0.9×√(28)/√(1–0.81)= 0.9×5.92/0.436 ≈ 12.23, with df=28; a p-value would be very small (

Regression Analysis of Heavy Metal Contaminants and Animal Weight

The given SPSS output indicates a strong positive correlation (R=0.921) between the contaminant level and weight of bats. The R-squared value of 0.848 suggests that approximately 84.8% of the variance in contaminant level is explained by the linear regression model, highlighting a substantial explanatory power.

The regression equation derived from the unstandardized coefficients is:

Contaminant level (ppb) = -71.54 + 2.22399 × weight (grams)

This equation indicates that each additional gram in weight is associated with an average increase of approximately 2.224 ppb in contaminant level.

To predict contaminant levels for an animal weighing 150 grams:

Predicted level = -71.54 + 2.22399 × 150 ≈ -71.54 + 333.6 = 262.06 ppb.

For the specific animal weighing 145 grams with a contaminant level of 222 ppb, the residual is calculated as:

Residual = Actual - Predicted = 222 - (-71.54 + 2.22399 × 145) ≈ 222 - ( -71.54 + 322.46) ≈ 222 - 250.92 = -28.92 ppb.

The least squares regression line is therefore:

Contaminant level = -71.54 + 2.22399 × weight.

The slope coefficient (approximately 2.224) indicates that the contaminant level increases on average by about 2.224 ppb for each additional gram of weight.

Constructing a 99% confidence interval for the slope involves using the standard error (0.740), degrees of freedom (n-2), and the critical t-value at 99% confidence. Using t(28, 0.5%) ≈ 2.763, the interval is:

Slope ± t × Std. Error = 2.22399 ± 2.763 × 0.740 ≈ 2.22399 ± 2.046.

Thus, the 99% confidence interval for the slope is approximately (0.178, 4.271), indicating a statistically significant positive association.

Hypothesis Testing of the Correlation between Contaminant Level and Weight

The null hypothesis (H₀): ρ = 0 (no correlation).

Alternative hypothesis (H₁): ρ > 0 (positive correlation).

The test statistic t = 12.23 with df=28, and the corresponding p-value is extremely small (

Analysis of Antibiotic Treatment trial for AOM

In testing whether the 'wait-and-see' prescription reduces antibiotic use compared to the standard prescription, the null hypothesis posits no difference in proportions: H₀: p₁ - p₂ = 0, where p₁ and p₂ are the proportions in the WASP and SP groups, respectively. The sample data show 82/138 (59.4%) in the WASP group and 17/145 (11.7%) in the SP group used antibiotics.

The pooled proportion (p̂) is:

p̂ = (82 + 17) / (138 + 145) = 99 / 283 ≈ 0.35.

The test statistic (z) is calculated as:

z = (p̂₁ - p̂₂) / √[p̂(1 – p̂) (1/n₁ + 1/n₂)] ≈ (0.594 - 0.117) / √[0.35 × 0.65 × (1/138 + 1/145)] ≈ 0.477 / 0.070 ≈ 6.82.

The critical value at α=0.01 (two-tailed) is approximately 2.33. Since 6.82 > 2.33, we reject H₀ and conclude that the WASP significantly reduces antibiotic use.

Spending and Regression Analysis

The least-squares regression equation of tobacco spending (predictor) predicting alcohol spending (response) is determined via the slope and intercept coefficients obtained from SPSS output. Assuming B₀ and B₁ are provided, the equation takes the form:

Alcohol spending = B₀ + B₁ × tobacco spending.

The specific numerical coefficients were not provided in the data, but the general formula applies.

Plotting the scatterplot with the regression line reveals the trend, which, if positive, indicates that higher tobacco spending correlates with higher alcohol spending, suggesting a potentially positive association.

The R-squared value indicates the percentage of variation explained by the model. For example, if R-squared = 0.45, then 45% of the variation in alcohol spending is explained by tobacco spending.

Constructing a confidence interval for the slope involves the standard error of the slope and the critical t-value, providing a range where the true slope likely falls with 99% confidence.

Conducting a test of independence or correlation coefficient activity involves H₀: ρ = 0 against H₁: ρ ≠ 0. Calculated p-value determines whether the variables are statistically associated.

Normality and Residual Analysis

The normal probability plot assesses whether residuals distribute approximately along a straight line, indicating normality; deviations suggest non-normality.

The residual plot examines whether residuals are randomly dispersed around zero. If patterns or outliers exist, assumptions underlying linear regression might be violated. Outliers are identified as points with large residuals far from the fitted line.

Conclusion

The analyses performed provide evidence of significant correlations, meaningful regression models, and the effects of treatment choices. Such statistical evaluations are essential in drawing reliable conclusions for scientific, medical, and economic decision-making, emphasizing the critical role of rigorous statistical methods in research.

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