Consider two sample means distributions corresponding to the same x distribution. The first sample mean distribution is based on samples of size n=100 and the second is based on a sample of size n=225. Which sample mean distribution has the smaller standard error? Explain.
Consider two sample means distributions corresponding to the same x distribution. The first sample mean distribution is based on samples of size n=100 and the second is based on a sample of size n=225. Which sample mean distribution has the smaller standard error? Explain.
MATLAB: An Introduction with Applications
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![**Comparison of Sample Mean Distributions**
Consider two sample mean distributions corresponding to the same \( \text{x} \) distribution. The first sample mean distribution is based on samples of size \( n = 100 \) and the second is based on a sample of size \( n = 225 \). Which sample mean distribution has the smaller standard error? Explain.
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**Explanation:**
The standard error of the sample mean is calculated using the formula:
\[ \text{Standard Error} = \frac{\sigma}{\sqrt{n}} \]
where \( \sigma \) is the standard deviation of the population, and \( n \) is the sample size.
Given two sample sizes,
- \( n_1 = 100 \)
- \( n_2 = 225 \)
By applying the formula for each:
- For \( n_1 = 100 \):
\[ \text{Standard Error}_1 = \frac{\sigma}{\sqrt{100}} = \frac{\sigma}{10} \]
- For \( n_2 = 225 \):
\[ \text{Standard Error}_2 = \frac{\sigma}{\sqrt{225}} = \frac{\sigma}{15} \]
Comparing the denominators,
\[ \sqrt{100} = 10 \quad \text{and} \quad \sqrt{225} = 15 \]
Since the denominator for the second sample mean distribution (15) is larger than that of the first sample mean distribution (10), it follows that:
\[ \frac{\sigma}{15} < \frac{\sigma}{10} \]
Therefore, the sample mean distribution for \( n = 225 \) has the smaller standard error. In other words, as the sample size increases, the standard error of the sample mean decreases, leading to more precise estimates of the population mean. This illustrates the principle that larger samples tend to provide more accurate estimates of population parameters.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8bcef5a6-2125-4478-a5fc-16fefbacd832%2F982f8a9d-ce80-4a69-8d56-ea350bcb3a9e%2Fo8xkzzm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Comparison of Sample Mean Distributions**
Consider two sample mean distributions corresponding to the same \( \text{x} \) distribution. The first sample mean distribution is based on samples of size \( n = 100 \) and the second is based on a sample of size \( n = 225 \). Which sample mean distribution has the smaller standard error? Explain.
---
**Explanation:**
The standard error of the sample mean is calculated using the formula:
\[ \text{Standard Error} = \frac{\sigma}{\sqrt{n}} \]
where \( \sigma \) is the standard deviation of the population, and \( n \) is the sample size.
Given two sample sizes,
- \( n_1 = 100 \)
- \( n_2 = 225 \)
By applying the formula for each:
- For \( n_1 = 100 \):
\[ \text{Standard Error}_1 = \frac{\sigma}{\sqrt{100}} = \frac{\sigma}{10} \]
- For \( n_2 = 225 \):
\[ \text{Standard Error}_2 = \frac{\sigma}{\sqrt{225}} = \frac{\sigma}{15} \]
Comparing the denominators,
\[ \sqrt{100} = 10 \quad \text{and} \quad \sqrt{225} = 15 \]
Since the denominator for the second sample mean distribution (15) is larger than that of the first sample mean distribution (10), it follows that:
\[ \frac{\sigma}{15} < \frac{\sigma}{10} \]
Therefore, the sample mean distribution for \( n = 225 \) has the smaller standard error. In other words, as the sample size increases, the standard error of the sample mean decreases, leading to more precise estimates of the population mean. This illustrates the principle that larger samples tend to provide more accurate estimates of population parameters.
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