Problem 10
Question
Geben sei eine normalverteilte Zufallsstichprobe vom Umfang \(n=20\), aus deren Realisation sich die Werte \(\sum_{i=1}^{n} x_{i}=100\) und \(S^{2}=10\) ergeben. Geben Sie ein Konfidenzintervall für den Erwartungswert zum Konfidenzniveau \(0.9\) an.
Step-by-Step Solution
Verified Answer
The 90% confidence interval for the mean is \([3.777, 6.223]\).
1Step 1: Understand the Problem
We need to calculate a confidence interval for the population mean \(\mu\) using a normally distributed random sample. Given are \(n=20\), \(\sum_{i=1}^{n} x_{i}=100\), and the sample variance \(S^{2}=10\). The confidence level is 90%.
2Step 2: Calculate the Sample Mean
The sample mean \(\bar{x}\) is calculated by dividing the sum of the sample values by the number of observations. \[\bar{x} = \frac{\sum_{i=1}^{n} x_{i}}{n} = \frac{100}{20} = 5\]
3Step 3: Determine the T-Critical Value
Since \(n\) is 20 and not large, we use the t-distribution. The degrees of freedom are \(n-1=19\). For a 90% confidence interval, we need \(\alpha=0.1\), so we find \(t_{\alpha/2,19}\) from a t-table or calculator. This value is approximately 1.729.
4Step 4: Calculate the Standard Error
The standard error of the mean is calculated using the sample variance \(S^{2}\): \[\text{SE} = \sqrt{\frac{S^2}{n}} = \sqrt{\frac{10}{20}} = \sqrt{0.5} \approx 0.707\]
5Step 5: Construct the Confidence Interval
The confidence interval is given by: \[\bar{x} \pm t_{\alpha/2,19} \times \text{SE} = 5 \pm 1.729 \times 0.707\] This calculates to \[5 \pm 1.223\] Therefore, the confidence interval is \([3.777, 6.223]\).
Key Concepts
Normal DistributionT-DistributionStandard ErrorSample Mean
Normal Distribution
A normal distribution is an important concept in statistics, often referred to as a bell curve because of its distinctive shape. It's symmetric and describes how values of a random variable are likely to be distributed around the mean. In normal distribution, the mean, median, and mode are all equal.
Some key properties include:
Some key properties include:
- 68% of the data falls within one standard deviation from the mean.
- 95% is within two standard deviations, and
- 99.7% is within three standard deviations.
T-Distribution
The t-distribution, also known as Student's t-distribution, is used when the sample size is small and the population standard deviation is unknown. It resembles the normal distribution but has fatter tails, which account for the additional variability expected with smaller samples.
Here’s how it works:
Here’s how it works:
- The t-distribution becomes more like the normal distribution as sample size increases.
- Degrees of freedom (df), usually defined as the number of observations minus one, influence its shape.
- For our problem with a sample size of 20, we use df = 19.
Standard Error
The standard error (SE) is a measure of how much the sample mean is likely to vary from the true population mean. It's calculated using the formula \( \text{SE} = \sqrt{\frac{S^2}{n}} \), where \( S^2 \) is the sample variance and \( n \) is the sample size.
The standard error gives insight into the precision of the sample mean as an estimate of the population mean:
The standard error gives insight into the precision of the sample mean as an estimate of the population mean:
- Smaller SE indicates more reliable estimates.
- Larger samples typically have smaller SE, increasing precision.
Sample Mean
The sample mean, usually denoted as \( \bar{x} \), is the average of all numbers in a sample. It's calculated by dividing the sum of sample values by the number of observations. The sample mean serves as an unbiased estimator of the population mean under the assumption of random sampling.
In practice:
In practice:
- Use \( \bar{x} = \frac{\sum x_i}{n} \) to find the sample mean.
- Acts as the center point in confidence interval calculations.
- Helps identify the distribution's central tendency.