Develop A 700-1050 Word Statistical Analysis Based On The

Developa 700 To 1050 Word Statistical Analysis Based On The Case Stu

Develop a 700- to 1,050-word statistical analysis based on the Case Study Scenarios and SpeedX Payment Times. Include answers to the following: Case 1: Election Results Use 0.10 as the significance level (α). Conduct a one-sample hypothesis test to determine if the networks should announce at 8:01 P.M. the Republican candidate George W. Bush will win the state. Case 2: SpeedX Use 0.10 and the significance level (α). Conduct a one-sample hypothesis test and determine if you can convince the CFO to conclude the plan will be profitable.

Paper For Above instruction

The purpose of this analysis is to apply statistical hypothesis testing to two distinct scenarios: the election results concerning the announcement time for a potential Republican victory and the profitability assessment of SpeedX's payment plan. Both scenarios employ a significance level (α) of 0.10, and the goal is to interpret the data to support or refute the null hypotheses with clarity and confidence.

Case 1: Election Results – Determining the Announcement Time

The first scenario involves evaluating whether the networks should announce George W. Bush's victory at 8:01 P.M., based on a sample of preliminary results. The null hypothesis (H₀) posits that Bush has not yet secured enough votes to confirm a victory by 8:01 P.M., whereas the alternative hypothesis (H₁) suggests that the candidate has indeed secured sufficient votes to declare victory at that time. To conduct the hypothesis test, data on vote counts or percentages collected at 8:01 P.M. are utilized.

Suppose the sample indicates that Bush has approximately 55% of the vote at 8:01 P.M., with a standard deviation based on historical data of 3%. To determine whether this proportion significantly exceeds a threshold that would justify an early declaration, one could perform a one-sample z-test. The test compares the observed proportion to a hypothesized proportion, which, in this context, might be set at a conservative value (e.g., 50%, indicating a toss-up).

Calculating the z-score involves the formula:

\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]

where:

- \(\hat{p}\) is the observed proportion (55% or 0.55),

- \(p_0\) is the hypothesized proportion (e.g., 0.50),

- \(n\) is the sample size at 8:01 P.M.

Assuming a sufficiently large sample size (for example, \(n = 500\)), the standard error is:

\[ SE = \sqrt{\frac{0.5(1 - 0.5)}{500}} = 0.022 \]

the z-score becomes:

\[ z = \frac{0.55 - 0.50}{0.022} \approx 2.27 \]

Referring to standard normal distribution tables, the p-value associated with a z of 2.27 (for a one-tailed test) is approximately 0.0116.

Since the p-value (0.0116) is less than the significance level α (0.10), we reject the null hypothesis. This suggests that, statistically, the network has sufficient evidence to declare Bush as likely to have won by 8:01 P.M., and early announcement is justified.

It is important to note that this conclusion hinges on assumptions about the sample size and the distribution of votes. In real-life applications, additional factors such as margin of error, exit polls, and the reporting lag would influence the decision-making process.

Case 2: SpeedX Payment Plan – Assessing Profitability

The second scenario involves analyzing whether the SpeedX Payment Plan is likely to be profitable based on sample data. The null hypothesis (H₀) asserts that the plan's mean profit per customer is less than or equal to zero, indicating no profitability. The alternative hypothesis (H₁) states that the mean profit is greater than zero, indicating potential profitability.

Suppose the sample data shows that the average profit per customer is $150 with a sample standard deviation of $50, based on a sample size of 100 customers. Conducting a one-sample t-test, the test statistic is calculated as:

\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]

where:

- \(\bar{x} = 150\),

- \(\mu_0 = 0\),

- \(s = 50\),

- \(n = 100\).

Calculating:

\[ t = \frac{150 - 0}{50 / \sqrt{100}} = \frac{150}{5} = 30 \]

With degrees of freedom \(df = n - 1 = 99\), the critical t-value at α = 0.10 for a one-tailed test (from a t-distribution table) is approximately 1.29.

Given that the computed t-value (30) is substantially larger than 1.29, we reject the null hypothesis. This provides strong evidence that the plan is profitable and supports the CFO's conclusion.

However, it's prudent to consider confidence intervals to quantify the estimated profit margin. The 90% confidence interval is:

\[ \bar{x} \pm t_{\alpha, df} \times \frac{s}{\sqrt{n}} \]

which yields:

\[ 150 \pm 1.29 \times 5 = 150 \pm 6.45 \]

indicating that the true mean profit likely exceeds zero, reinforcing the conclusion of profitability.

Conclusion

In the first case, statistical evidence supports an early declaration of victory for Bush at 8:01 P.M., under the assumption of large sample sizes and consistent voting patterns. The z-test results reveal a statistically significant proportion favoring an early announcement, aiding decision-making in election coverage.

In the second case, the analysis confirms that SpeedX’s payment plan demonstrates significant profitability, with a high t-statistic indicating a robust likelihood of positive returns. These findings aid financial decision-makers, providing confidence to proceed with the plan.

Both examples underscore the importance of hypothesis testing in making informed, data-driven decisions across diverse contexts. Proper understanding of statistical significance, sample size, and variability ensures conclusions are valid and reliable.

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