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Develop a 92% confidence interval estimate for the proportion of all voters opposed to a campaign to eliminate non-returnable beverage containers. Additionally, address statistical questions related to confidence intervals for mean income, luggage weight, and sample size estimates based on given data, aligning with the data set provided in the ROI Excel spreadsheet. This assignment involves calculating confidence intervals for proportions and means, interpreting the results, and applying statistical formulas, including t-distribution and Z-distribution, to real-world data.
Paper For Above instruction
The purpose of this research paper is to provide detailed examples of constructing confidence intervals for proportions and means, applying these techniques to both hypothetical and real datasets. The analyses include calculating confidence intervals for voter opposition proportions, average income of airline pilots, weight of airline carry-on luggage, and the sample size determination for a soft drink content study. Throughout, the emphasis is on clarity, interpretation, and application of statistical methods relevant to social science research and business analytics.
Confidence Interval for Voter Opposition to Beverage Containers
The first question involves estimating the proportion of voters opposed to a policy using sample data. With a sample size of 600 voters and 210 opposed, the sample proportion (p̂) is calculated as:
p̂ = 210 / 600 = 0.35
To create a 92% confidence interval for this proportion, the z-value corresponding to 92% confidence level is essential. Consulting standard normal distribution tables, the z-value (critical value) for 92% confidence (α = 0.08, two-tailed) is approximately 1.75. The formula for the confidence interval is:
CI = p̂ ± z * √[p̂(1 - p̂) / n]
Calculating the standard error:
SE = √[0.35 (1 - 0.35) / 600] = √[0.35 0.65 / 600] = √[0.2275 / 600] ≈ √0.000379 ≈ 0.0195
Thus, the confidence interval is:
0.35 ± 1.75 * 0.0195 = 0.35 ± 0.0341
Range:
Lower bound: 0.3159
Upper bound: 0.3841
Interpretation: Approximately 31.6% to 38.4% of all voters are opposed to the beverage container policy with 92% confidence.
Confidence Interval for Mean Yearly Income of Airline Pilots
The second question concerns estimating the average income. The sample mean income of 87 pilots is $99,400, with a standard deviation of $12,000. For a 95% confidence interval, the t-distribution must be used due to the small sample size and unknown population standard deviation.
1. Calculating the degrees of freedom:
df = 87 - 1 = 86
From t-distribution tables, t-value for 95% confidence and df=86 is approximately 1.99.
2. The standard error (SE):
SE = SD / √n = 12,000 / √87 ≈ 12,000 / 9.33 ≈ 1286.4
The confidence interval:
CI = 99,400 ± 1.99 * 1,286.4 ≈ 99,400 ± 2,557.1
Lower bound: 96,842.9
Upper bound: 101,957.1
Interpretation: We are 95% confident that the true mean yearly income of all airline pilots lies between approximately $96,843 and $101,957.
Confidence Intervals for Carry-On Luggage Weight
The third question examines the average weight of carry-on luggage, with a known population standard deviation of 7.5 pounds. A sample of 25 bags has an average weight of 18 pounds.
- For a 97% confidence interval, the z-value is approximately 2.17.
The standard error:
SE = 7.5 / √25 = 7.5 / 5 = 1.5
The confidence interval:
CI = 18 ± 2.17 * 1.5 = 18 ± 3.255
Range:
Lower: 14.745 pounds
Upper: 21.255 pounds
This indicates with 97% confidence that the average luggage weight is between roughly 14.75 and 21.26 pounds.
Similarly, for a 95% confidence interval, the z-value is approximately 1.96:
CI = 18 ± 1.96 * 1.5 ≈ 18 ± 2.94, giving a range from 15.06 to 20.94 pounds.
Sample Size Determination for Soft Drink Content
Finally, the last question involves estimating the sample size necessary to estimate the mean content of a soft drink within a specific margin of error at 95% confidence. The reported average is 12 ounces, with a known population standard deviation of 1.28 ounces. The confidence interval bounds are between 11.7 to 12.3 ounces, implying a margin of error (E):
E = (Upper bound - Mean) = 0.3 ounces
or equivalently, the half-width of the confidence interval:
E = 12.3 - 12 = 0.3
Using the formula for the sample size:
n = (Z * σ / E)²
where Z is the z-value at 95% confidence (approximately 1.96):
n = (1.96 * 1.28 / 0.3)² ≈ (2.5088 / 0.3)² ≈ (8.36)² ≈ 69.89
Approximately 70 samples are needed to estimate the mean content within ±0.3 ounces with 95% confidence.
Conclusion
This paper illustrates critical statistical techniques for hypothesis testing and estimation, providing realistic applications in social sciences and business analytics. Proper interpretation of confidence intervals gives decision-makers insight into population parameters, enhancing data-driven strategies. These methods are foundational in research involving surveys, sampling, and estimation, emphasizing the importance of accurate calculation, understanding of distribution types, and contextual interpretation of statistical results.
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