Assignment 111 CBU Planner Is Interested In Determining

Assignment 111 Cbu Planner Is Interested In Determining The Percentag

The CBU planner is interested in determining the percentage of spring semester students who will attend summer school. A pilot sample of 160 spring semester students revealed that 56 will return to summer school to take courses and participate in Christian activities on campus. The tasks are as follows: (a) construct confidence intervals at 88% and 95% confidence levels for the proportion of spring semester students who will attend summer school; (b) calculate the required sample size to achieve a margin of error of 3% or less with confidence levels of 88% and 95%.

Additionally, data from a grocery store’s sales indicate an average of $8,000 per day. After implementing advertising campaigns, a sample of 64 days of sales showed an average of $8,300 daily, with past data indicating a population standard deviation of $1,200. The study aims to test whether the advertising has increased sales, with hypotheses: H0: μ ≤ 8,000 versus H1: μ > 8,000. The specific tasks are: (a) compute the test statistic at α=0.05; (b) interpret the results using the critical value approach; (c) interpret the results using the P-value approach.

Paper For Above instruction

The statistical analysis presented addresses two core research questions: first, estimating the proportion of spring semester students planning to return to summer school; second, evaluating whether advertising efforts have effectively increased the daily sales of a grocery store. These analyses employ various inferential statistical techniques including confidence intervals, sample size determination, hypothesis testing, and both critical value and P-value methods. This comprehensive approach provides not only point estimates but also the precision and significance of the findings, critical for informed decision-making within educational planning and business strategy contexts.

Part 1: Estimation of the Proportion of Students Returning to Summer School

The pilot study sampling 160 students found that 56 would participate in summer school activities, suggesting an initial proportion estimate of p̂ = 56/160 = 0.35. To quantify the uncertainty around this estimate, constructing confidence intervals at 88% and 95% confidence levels provides insights into the range within which the true population proportion likely falls.

Using the formula for the confidence interval for a proportion:

p̂ ± Zα/2 * √[p̂(1 - p̂)/n], where Zα/2 is the critical Z-value corresponding to the confidence level, the calculations proceed as follows.

For the 88% confidence interval, Z0.06 ≈ 1.55; for the 95% confidence interval, Z0.025 ≈ 1.96.

At 88% confidence:

Margin of Error (MOE) = 1.55 √[0.35(0.65)/160] ≈ 1.55 √(0.2275/160) ≈ 1.55 * 0.0377 ≈ 0.0585.

Confidence interval: 0.35 ± 0.0585 ⇒ (0.2915, 0.4085).

At 95% confidence:

MOE = 1.96 √[0.35(0.65)/160] ≈ 1.96 0.0377 ≈ 0.0739.

Confidence interval: 0.35 ± 0.0739 ⇒ (0.2761, 0.4239).

These intervals suggest that with 88% confidence, between approximately 29.15% and 40.85% of spring students will return to summer school, whereas at 95% confidence, the range widens to approximately 27.61% to 42.39%.

Part 2: Sample Size Calculation for Desired Margin of Error

To determine the necessary sample size for a specified margin of error (E) of 3% (0.03), with known estimated proportion p̂=0.35, and confidence levels of 88% and 95%, the formula:

n = (Zα/2 / E)^2 * p̂(1 - p̂) is used.

For 88% confidence (Z0.06 ≈ 1.55):

n = (1.55 / 0.03)^2 0.35 0.65 ≈ (51.67)^2 0.2275 ≈ 2670.5 0.2275 ≈ 607.9.

> Rounded up, about 608 students are needed.

For 95% confidence (Z0.025 ≈ 1.96):

n = (1.96 / 0.03)^2 0.35 0.65 ≈ (65.33)^2 0.2275 ≈ 4267.7 0.2275 ≈ 971.3.

> Rounded up, approximately 972 students are needed.

These calculations indicate larger sample sizes are necessary to ensure the estimation precision at higher confidence levels.

Part 3: Hypothesis Testing on Grocery Store Sales

The second part examines whether the store’s advertising campaigns have resulted in increased daily sales. The sample mean sales of $8,300 over 64 days, with a known population standard deviation of $1,200, are compared against the historical mean of $8,000. The hypotheses are structured as:

H0: μ ≤ 8,000; H1: μ > 8,000.

The test statistic uses the Z-test for a population mean with known standard deviation:

Z = (X̄ - μ₀) / (σ / √n).

Calculating:

Z = (8300 - 8000) / (1200 / √64) = 300 / (1200 / 8) = 300 / 150 = 2.0.

At α=0.05, the critical Z-value for a one-tailed test is approximately 1.645.

Part 3a: Critical value approach

Since Z=2.0 > 1.645, we reject the null hypothesis, indicating that there is statistically significant evidence at the 5% level that the advertising campaigns have increased daily sales.

Part 3b: P-value approach

The P-value for Z=2.0 in a one-tailed test is approximately 0.0228. Because 0.0228

Conclusion

Both the critical value and P-value approaches lead to rejecting the null hypothesis, suggesting the advertising campaigns effectively increased the store's daily sales. Further analysis could involve confidence intervals for the mean increase or regression models to quantify the effect.

References

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