How Likely Are These Sample Means for Chickadees?

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SUMMARY

The discussion focuses on calculating the probability of obtaining sample means for Chickadees (Poecile atricapillus) based on provided sample data. The biologist collected three samples with mean masses of 13.23 grams, 9.64 grams, and 11.14 grams, while the population mean mass is 10.87 grams with a standard deviation of 1.89 grams. The user correctly identifies that the probability of obtaining sample means between 9.64 and 13.23 grams is calculated using z-scores, ultimately determining P(9.64 < x̄ < 13.23) = 0.6347. The user realizes the need to apply the standard deviation of the means for accurate z-score calculations.

PREREQUISITES
  • Understanding of normal distribution and its properties
  • Knowledge of z-score calculations
  • Familiarity with sample means and population parameters
  • Basic statistics concepts, including standard deviation and probability
NEXT STEPS
  • Study the concept of standard error of the mean (SEM) and its calculation
  • Learn about hypothesis testing and confidence intervals in statistics
  • Explore the Central Limit Theorem and its implications for sample means
  • Investigate statistical software tools for performing probability calculations, such as R or Python's SciPy library
USEFUL FOR

Students in biology and statistics, researchers analyzing avian populations, and anyone interested in statistical methods for sampling and probability calculations.

jmm
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Homework Statement



A biologist interested in the mass of Chickadees (Poecile atricapillus) in North Glenmore Park collects the 3 samples shown below:
Sample 1: 9 individuals with a mean mass of 13.23 grams
Sample 2: 16 individuals with a mean mass of 9.64 grams
Sample 3: 13 individuals with a mean mass of 11.14 grams

b. The population of Chickadees in North Glenmore Park has a mean mass of 10.87 grams and a standard deviation of 1.89 grams. Assuming the population is normally distributed, what is the probability of obtaining samples such as these, with means between 9.64 and 13.23 grams, if each sample were comprised of 14 individuals?

Homework Equations



z = (xbar - µ) / ∂

The Attempt at a Solution



So I think the probability would be P(xbar < 13.23) - P(xbar < 9.64). For xbar = 13.23, z=1.25 and p = 0.8925, for xbar = 9.64, z = -0.65 and p = 0.2578. So P(9.64 < xbar < 13.23) = 0.8925 - 0.2578 = 0.6347. Am I on the right track with this or am I completely off? I'm confused why they mention that each sample is composed of 14 individuals.

Thank you.
 
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Never mind; I just figured out that I need to use the standard deviation of the means to calculate the z scores instead of the population standard deviations. Thanks anyways!
 

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