Question 1 In A Poll Of 600 Voters In A Campaign To Eliminat

Question 1in A Poll Of 600 Voters In A Campaign To Eliminate Non Retur

Question 1 in a poll of 600 voters in a campaign to eliminate non-returnable beverage containers, 210 of the voters were opposed. Develop a 92% confidence interval estimate for the proportion of all the voters who opposed the container control bill. Question 2 A random sample of 87 airline pilots had an average yearly income of $99,400 with a standard deviation of $12,000. 1. If we want to determine a 95% confidence interval for the average yearly income, what is the value of t? 2. Develop a 95% confidence interval for the average yearly income of all pilots. Question 3 In order to determine the average weight of carry-on luggage by passengers in airplanes, a sample of 25 pieces of carry-on luggage was collected and weighed. The average weight was 18 pounds. Assume that we know the standard deviation of the population to be 7.5 pounds. 1. Determine a 97% confidence interval estimate for the mean weight of the carry-on luggage. 2. Determine a 95% confidence interval estimate for the mean weight of the carry-on luggage. Question 4 A statistician employed by a consumer testing organization reports that at 95% confidence he has determined that the true average content of the Uncola soft drinks is between 11.7 to 12.3 ounces. He further reports that his sample revealed an average content of 12 ounces, but he forgot to report the size of the sample he had selected. Assuming the standard deviation of the population is 1.28, determine the size of the sample.

Paper For Above instruction

Statistical inference plays a vital role in decision-making processes across various domains, enabling researchers and professionals to draw conclusions about populations based on sample data. This paper addresses four specific problems involving confidence intervals and related calculations to understand population parameters with a specified level of confidence. Through detailed explanations and calculations, we explore methodologies for estimating proportions, means with known and unknown standard deviations, and determining sample sizes necessary for desired confidence levels.

The first problem involves estimating the proportion of voters opposing a policy based on a sample. The second problem deals with constructing a confidence interval for the average income of airline pilots, highlighting the importance of selecting the correct t-value for the desired confidence level. The third problem evaluates the mean weight of carry-on luggage, employing different confidence levels and known standard deviations. The final problem involves determining the sample size required to estimate a population mean within a specified confidence interval, given limited information about the sample.

Analysis and Solutions

Problem 1: Confidence Interval for Population Proportion

A polling sample consisted of 600 voters, with 210 opposed to the container bill. The sample proportion (p̂) is calculated as:

p̂ = 210 / 600 = 0.35

To estimate the true proportion (p) with 92% confidence, we utilize the formula for a confidence interval for a population proportion:

CI = p̂ ± Z_α/2 * √[p̂(1 - p̂) / n]

Where:

- n = 600

- p̂ = 0.35

- Z_α/2 corresponds to the critical Z-value for 92% confidence. From standard normal tables, Z_α/2 ≈ 1.75.

Calculations:

S.e. = √[0.35 * 0.65 / 600] ≈ √[0.2275 / 600] ≈ √0.000379 ≈ 0.0195

Lower limit = 0.35 - 1.75 * 0.0195 ≈ 0.35 - 0.0341 ≈ 0.3159

Upper limit = 0.35 + 1.75 * 0.0195 ≈ 0.35 + 0.0341 ≈ 0.3841

Hence, the 92% confidence interval for the proportion of voters opposing the bill is approximately (0.316, 0.384).

Problem 2: Confidence Interval for the Mean Income of Pilots

Given:

- Sample size (n) = 87

- Sample mean (x̄) = $99,400

- Standard deviation (σ) = $12,000

Since the population standard deviation is known, and the sample size is reasonably large, the confidence interval uses the z-distribution:

CI = x̄ ± Z_α/2 * (σ / √n)

For a 95% confidence level, Z_α/2 ≈ 1.96.

Calculations:

Margin of error = 1.96 (12,000 / √87) ≈ 1.96 (12,000 / 9.33) ≈ 1.96 * 1285.8 ≈ 2,520

Lower bound = 99,400 - 2,520 ≈ \$96,880

Upper bound = 99,400 + 2,520 ≈ \$101,920

Thus, the 95% confidence interval for the average yearly income of pilots is approximately (\$96,880, \$101,920).

Regarding the t-value for a 95% confidence interval in small samples when the population standard deviation is known, we typically do not use t; instead, z-value suffices. However, if the population standard deviation were unknown, the t-distribution with degrees of freedom n-1 would be used, and t ≈ 2.00 for df=86.

Problem 3: Confidence Intervals for Carry-On Luggage Weight

Given:

- Sample size (n) = 25

- Sample mean (x̄) = 18 pounds

- Known population standard deviation (σ) = 7.5 pounds

The formulas for confidence intervals with known σ are:

CI = x̄ ± Z_α/2 * (σ / √n)

Calculations:

1. For 97% confidence (Z_α/2 ≈ 2.17):

Margin of error = 2.17 (7.5 / √25) = 2.17 (7.5 / 5) = 2.17 * 1.5 ≈ 3.26

CI = 18 ± 3.26 ⇒ (14.74, 21.26) pounds

2. For 95% confidence (Z_α/2 ≈ 1.96):

Margin of error = 1.96 (7.5 / 5) = 1.96 1.5 ≈ 2.94

CI = 18 ± 2.94 ⇒ (15.06, 20.94) pounds

These intervals provide estimates for the average weight of carry-on luggage at different confidence levels.

Problem 4: Determining Sample Size Based on Confidence Interval

The confidence interval for the mean content of soft drinks is between 11.7 and 12.3 ounces with 95% confidence; the sample mean (x̄) is 12 ounces, and the population standard deviation (σ) is 1.28.

The confidence interval formula:

CI = x̄ ± Z_α/2 * (σ / √n)

Given the interval width:

Width = 12.3 - 11.7 = 0.6

Half-width (margin of error, E):

E = 0.3

Solving for n:

n = (Z_α/2 * σ / E)^2

For a 95% confidence level, Z_α/2 ≈ 1.96.

Calculations:

n = (1.96 1.28 / 0.3)^2 ≈ (1.96 4.267)^2 ≈ (8.36)^2 ≈ 69.9

Rounding up, the sample size needed is approximately 70.

Conclusion

This analysis illustrates essential applications of confidence intervals in real-world scenarios, from estimating proportions and means to calculating necessary sample sizes. Proper understanding and calculation of these intervals enable more accurate and credible decision-making in research, policy, and industry practices, emphasizing the importance of selecting appropriate confidence levels, statistical distributions, and sample sizes based on the specific context and known parameters.

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