Construct A Confidence Interval At The Given Level For P

Construct A Confidence Interval At The Given Level For The Population

Construct a confidence interval at the given level for the population mean based on the sample mean. Each question is worth 10 points. Make sure to answer the question at the end of each item. Use the formula: sample mean ± margin of error, where the margin of error = z_c * (standard deviation / √n).

Paper For Above instruction

Introduction

Confidence intervals are a fundamental statistical tool used to estimate the range of likely values for a population parameter, such as the mean, based on sample data. They provide a measure of uncertainty around the sample estimate and are crucial in decision-making processes across various fields including sports management, education, healthcare, and consumer research. This paper aims to construct confidence intervals at specified confidence levels for different scenarios, interpret their significance, and discuss their practical applications.

Scenario 1: Baseball Stadium Manager

A manager of a baseball stadium seeks to estimate the average age of all baseball fans with a 90% confidence level. From a random sample of 40 fans, the mean age is 32.5 years with a standard deviation of 2.3 years. Using the formula for the confidence interval:

CI = x̄ ± z_c * (s / √n)

Where:

• x̄ = 32.5
• s = 2.3
• n = 40
• z_c for 90% confidence ≈ 1.645

Calculating the margin of error:

Margin of error = 1.645 (2.3 / √40) ≈ 1.645 0.364 ≈ 0.599

Constructed confidence interval: (32.5 - 0.599, 32.5 + 0.599) ≈ (31.901, 33.099)

The manager finds this information useful to understand the typical ages of fans and tailor marketing or ticketing strategies accordingly.

Scenario 2: High School Principal

A sample of 42 computers has an average price of $875 with a standard deviation of $54. The principal wants a 95% confidence interval for the average computer price.

• x̄ = 875
• s = 54
• n = 42
• z_c for 95% confidence ≈ 1.96

Margin of error = 1.96 (54 / √42) ≈ 1.96 8.34 ≈ 16.34

Confidence interval: (875 - 16.34, 875 + 16.34) ≈ ($858.66, $891.34)

This interval helps the principal understand the typical market price for computers, aiding budget planning or negotiations.

Scenario 3: School Nurse

Survey of 30 girls shows an average weight of 134 pounds with a standard deviation of 12 pounds. The nurse aims for a 99% confidence interval.

• x̄ = 134
• s = 12
• n = 30
• z_c for 99% confidence ≈ 2.576

Margin of error = 2.576 (12 / √30) ≈ 2.576 2.19 ≈ 5.64

Confidence interval: (134 - 5.64, 134 + 5.64) ≈ (128.36, 139.64)

The nurse can use this information to monitor health standards or plan nutritional interventions for the population of girls.

Scenario 4: Cats’ Weights

a. For a 90% confidence level:

• x̄ = 9.8
• s = 2.4
• n = 36
• z_c for 90% ≈ 1.645

Margin of error = 1.645 (2.4 / √36) = 1.645 0.4 ≈ 0.658

Interval: (9.8 - 0.658, 9.8 + 0.658) ≈ (9.142, 10.458)

b. For a 95% confidence level:

• z_c for 95% ≈ 1.96

Margin of error = 1.96 (2.4 / √36) = 1.96 0.4 ≈ 0.784

Interval: (9.8 - 0.784, 9.8 + 0.784) ≈ (9.016, 10.584)

c. The intervals expand as the confidence level increases because the z_c value increases, leading to a larger margin of error. This reflects greater uncertainty, but also higher confidence that the true mean weight is within the interval. Simply put, higher confidence levels produce wider intervals, balancing certainty with precision.

Scenario 5: Consumer Purchasing Tires

A sample of 30 companies shows an average tire set price of $245 with a standard deviation of $38. The consumer wants a 95% confidence interval.

• x̄ = 245
• s = 38
• n = 30
• z_c for 95% ≈ 1.96

Margin of error = 1.96 (38 / √30) ≈ 1.96 6.93 ≈ 13.58

Interval: ($245 - 13.58, $245 + 13.58) ≈ ($231.42, $258.58)

This interval allows the consumer to assess the typical market price for a set of tires, aiding in making an informed purchase decision. If she samples 50 companies instead, the standard error decreases because:

New standard error = 38 / √50 ≈ 5.37

New margin of error = 1.96 * 5.37 ≈ 10.52

New confidence interval: ($245 - 10.52, $245 + 10.52) ≈ ($234.48, $255.52)

With a larger sample size, the interval narrows, providing a more precise estimate of the average tire price, thus increasing the reliability of the estimate.

Conclusion

Constructing confidence intervals provides valuable insights into the likely range of a population parameter, facilitating better decision-making across diverse contexts. The width of the intervals depends on the confidence level and the variability in the data, with higher confidence levels producing wider intervals. Larger sample sizes tend to reduce the margin of error, yielding more precise estimates. Practitioners can utilize these intervals to inform strategies—from marketing to healthcare to purchasing decisions—by understanding the underlying statistical estimates and their associated uncertainties.

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