Assume That A Population Is Normally Distributed With A Mean
Assume That A Population Is Normally Distributed With A Mean Of 100 An
Consider a population that is normally distributed with a mean of 100 and a standard deviation of 15. To determine whether observing a sample mean of 115 or more is unusual, we utilize the properties of the sampling distribution and the Central Limit Theorem (CLT). The question explores how the sample size influences the likelihood of such an extreme sample mean.
When the sample size is small, such as n = 3, the standard error (SE) of the mean is calculated by dividing the population standard deviation by the square root of the sample size: SE = 15 / √3 ≈ 8.66. The z-score corresponding to a sample mean of 115 is (115 - 100) / 8.66 ≈ 1.73. Using standard normal distribution tables, a z-score of 1.73 corresponds to a probability of about 8.4%. Since this probability is relatively high, a sample mean of 115 or more in a small sample of size 3 would not be considered very unusual.
Increasing the sample size to n = 20 dramatically reduces the standard error to SE = 15 / √20 ≈ 3.35. The z-score now becomes (115 - 100) / 3.35 ≈ 4.48. The probability associated with such a z-score is practically zero (
Similarly, examining the probability of obtaining a sample mean of 105 or more in a sample of size 20, the z-score is (105 - 100) / 3.35 ≈ 1.49, corresponding to a probability of approximately 6.8%. This is also relatively unlikely, suggesting that such an event, while not impossible, is somewhat uncommon given the population characteristics. It illustrates why the CLT is fundamental: as sample size increases, the sampling distribution of the mean becomes approximately normal with a smaller standard error, enabling more precise probability assessments.
The CLT justifies using normal distribution calculations for sample means regardless of the original population distribution, provided the sample size is sufficiently large or the population is normal. It ensures that the distribution of the sample mean tends to a normal shape, which facilitates the calculation of probabilities and understanding of sampling variability.
Population Errors of Errors and Sampling Analysis
Given four typists with errors per page: Stephanie (3), Adam (2), Leslie (1), and Shay (4), we first calculate the population mean errors per page:
μ = (3 + 2 + 1 + 4) / 4 = 10 / 4 = 2.5 errors per page.
Next, the population standard deviation σ is computed using the formula for the population standard deviation:
σ = √[( (3 - 2.5)² + (2 - 2.5)² + (1 - 2.5)² + (4 - 2.5)² ) / 4]
= √[(0.25 + 0.25 + 2.25 + 2.25) / 4] = √[5 / 4] = √1.25 ≈ 1.12 errors per page.
For sampling with replacement, the total number of possible samples of size 2 is 4² = 16. Each combination can be listed, and the mean errors per page for each sample can be calculated. For example:
- Stephanie & Stephanie: (3 + 3) / 2 = 3
- Stephanie & Adam: (3 + 2) / 2 = 2.5
- Stephanie & Leslie: (3 + 1) / 2 = 2
- Stephanie & Shay: (3 + 4) / 2 = 3.5
- Adam & Adam: (2 + 2) / 2 = 2
Calculating all 16 sample means, the average of these sample means will equal the population mean of 2.5 errors per page. This conforms to the property that the mean of the sampling distribution of sample means equals the population mean.
The key distinction between the population standard deviation and the standard error of the mean is that the former describes variability within the entire population, while the latter describes the variability expected among sample means, which decreases as sample size increases. The standard error quantifies the precision of the sample mean estimate, illustrating how sample size influences the reliability of statistical inference.
References
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