Case Study – Election Results And SpeedX QNT/561 Version

Case Study – Election Results and SppedX QNT/561 Version

Suppose in the exit poll from the state of Florida during the 2000 year elections, the pollsters recorded only the votes of the two candidates who had any chance of winning: Democrat Al Gore and Republican George W. Bush. In a sample of 765 voters, the number of votes cast for Al Gore was 358 and the number of votes cast for George W. Bush was 407.

The network predicts the candidate as a winner if he wins more than 50% of the votes. The polls close at 8:00 P.M. Based on the sample results, conduct a one-sample hypothesis test to determine if the networks should announce at 8:01 P.M. that the Republican candidate George W. Bush will win the state. Use 0.10 as the significance level (α).

Paper For Above instruction

The 2000 U.S. presidential election in Florida was a highly contentious and closely-watched event, with exit polls playing a crucial role in early predictions. In this case, the pollsters recorded data on a sample of 765 voters, with 358 supporting Democrat Al Gore and 407 supporting Republican George W. Bush. This data serves as a basis for conducting a hypothesis test to evaluate whether the observed results provide sufficient statistical evidence to suggest that Bush is leading the election at the time the polls close, and whether the networks should declare him as the projected winner.

Formulating the Hypotheses

The primary objective is to determine if George W. Bush’s support proportion exceeds 50%. The null hypothesis (H0) states that Bush does not have more than half of the votes: H0: p ≤ 0.50, where p is the true proportion of votes for Bush. The alternative hypothesis (Ha) posits that Bush’s support proportion is indeed greater than 50%: Ha: p > 0.50. This is a right-tailed test, suitable for testing whether a significant majority favors Bush.

Sample Data and Test Statistic Calculation

From the data, the sample proportion of votes for Bush is calculated as:

p̂ = 407 / 765 ≈ 0.5327

Under the null hypothesis, the assumed proportion is p0 = 0.50. The standard error (SE) of the sample proportion is computed as:

SE = √[p0(1 - p0) / n] = √[0.50 * 0.50 / 765] ≈ 0.0181

The z-score for the observed proportion is then:

z = (p̂ - p0) / SE = (0.5327 - 0.50) / 0.0181 ≈ 1.793

Decision Rule and Significance Level

Using a significance level (α) of 0.10, the critical z-value for a one-tailed test is approximately 1.28. Since the calculated z-value of 1.793 exceeds 1.28, the evidence suggests that we can reject the null hypothesis at the 10% significance level.

Conclusion and Implications

Given the statistical analysis, there is sufficient evidence to support the claim that Bush’s support in Florida exceeds 50% at the poll close time. Therefore, the networks could reasonably announce Bush as the projected winner at 8:01 P.M., based on the sample data and the chosen significance level. However, it is important to recognize the limitations inherent within exit polls, including sampling biases and the margin of error, which should be considered before making final declarations.

Analysis of SpeedX – Testing the Effect of Self-Addressed Envelopes

Turning to the SpeedX case study, the company aims to determine whether including stamped self-addressed envelopes with invoices can statistically significantly decrease the average time taken for payments, which currently averages 24 days with a standard deviation of 6 days. The claim is that this intervention will reduce the mean payment period by at least 2 days, from 24 to 22 days, which the CFO believes will lead to an improved cash flow.

Formulating Hypotheses

The null hypothesis (H0) states that the mean payment time does not decrease with the new procedure: H0: μ ≥ 24 days. The alternative hypothesis (Ha) is that the mean decreases, specifically that μ

Sample Data and Test Statistic Calculation

A sample of 220 customers is selected, and the recorded days until payment are analyzed. Suppose the sample mean (x̄) is found to be 22 days, and the population standard deviation remains at 6 days. The test statistic (z) is calculated as:

z = (x̄ - μ0) / (σ / √n) = (22 - 24) / (6 / √220) ≈ -2 / (6 / 14.83) ≈ -2 / 0.405 ≈ -4.938

Decision Rule and Significance Level

With α = 0.10, the critical z-value for a left-tailed test is approximately -1.28. The computed z-value of -4.938 is far below -1.28, indicating a strong rejection of the null hypothesis.

Conclusion and Business Implication

The statistical evidence strongly supports the claim that providing stamped self-addressed envelopes reduces the average days for payment by at least 2 days. The company can be confident in this conclusion, and the initiative likely results in improved cash flow, justifying the cost of envelopes and stamps. It highlights how inferential statistics are essential tools for making data-driven business decisions, especially when evaluating process improvements.

Summary

Both case studies demonstrate the power of hypothesis testing in evaluating claims—whether predicting election outcomes or assessing process changes. Properly formulated hypotheses, calculated test statistics, and comparison against critical values enable organizations to make informed decisions based on statistical evidence. These applications reinforce the importance of statistical literacy and rigorous analysis in real-world scenarios.

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