you are asked to estimate the average weight of an adult male passenger for a study to maximize the number of seats in a certain airplane. you collect random weights of 1000 men passengers from a population of 100,000. you find that the average man in the sample weighs 180 pounds and the standard deviation of the sample is 30 pounds. what is the 95% confidence interval?
95% = mean ± 1.96 SEm
SEm = SD/√n
I'll let you do he calculations.
To calculate the 95% confidence interval for the population's average weight of adult male passengers, you can follow these steps:
Step 1: Determine your sample mean and sample standard deviation.
- Sample mean (x̄) = 180 pounds (given)
- Sample standard deviation (σ) = 30 pounds (given)
Step 2: Determine your sample size.
- Sample size (n) = 1000 men passengers (given)
Step 3: Determine the confidence level and find the critical value.
- The confidence level is 95%, which corresponds to an alpha level (α) of 0.05.
- Since the sample size is large (n > 30), you can use a Z-distribution.
- The critical value for a 95% confidence interval is ±1.96 (which can be found in the Z-score table or using a calculator).
Step 4: Calculate the margin of error.
- Margin of error (E) = (Critical value) * (Sample standard deviation / √Sample size)
- E = 1.96 * (30 / √1000) ≈ 1.84 pounds (rounded to two decimal places)
Step 5: Calculate the lower and upper bounds of the confidence interval.
- Lower bound = Sample mean - Margin of error
- Lower bound = 180 - 1.84 ≈ 178.16 pounds (rounded to two decimal places)
- Upper bound = Sample mean + Margin of error
- Upper bound = 180 + 1.84 ≈ 181.84 pounds (rounded to two decimal places)
Therefore, the 95% confidence interval for the average weight of adult male passengers is approximately 178.16 to 181.84 pounds.