TASK LIST NO. 7: Point and Interval Estimation
Task 1
Find the confidence interval for the unknown mean value \(\mu\) of a population, in the case where \(\sigma\) is known, based on an \(n\)-element simple sample \(X_1, ..., X_n\).
Data: From a population with a standard deviation \(\sigma=0.14\), a sample of \(n=100\) elements was taken (e.g., voltage measurements on a processor). The sample mean was \(\bar{x}=2.5\). Determine the 95% confidence interval for the mean (assume \(1-\alpha = 0.95\), which gives \(u_{\alpha} = 1.96\)).
Task 2
The durability of 10 randomly selected structural elements was measured (or e.g., battery life of 10 laptops). The following results were obtained: \(383, 284, 339, 340, 305, 386, 378, 335, 344, 346\).
Assuming that the distribution of the feature is normal, determine the 95% confidence interval for the mean durability.
Hint: Since \(n=10\) is small (\(n<30\)) and we do not know \(\sigma\), calculate \(s\) from the sample and use the Student's t-distribution.
Task 3
To determine the electron charge, 26 measurements were made using the Millikan method. A mean of \(\bar{x} = 1.574 \cdot 10^{-19}\) and a standard deviation of \(s = 0.043 \cdot 10^{-19}\) were obtained.
Determine the confidence interval for the mean charge at a confidence level of \(0.99\).
Task 4
A 300-element sample was taken from a population of cotton fibers and their lengths were measured. The mean \(\bar{x}=27.43\) mm and variance \(s^2=51.598\) were calculated.
Find the 95% realization of the confidence interval for the unknown average fiber length.
Hint: With such a large \(n\) (\(n=300\)), the Student's t-distribution is practically identical to the normal distribution, so the statistic \(u_{\alpha}\) can be used.
Task 5
Measurements of sea depth (or e.g., network latency) are made in a certain specific location.
How many independent measurements must be made to assume with a confidence level of \(0.95\) that the absolute error of estimating the mean will not exceed \(10\) m, if the error distribution is normal with a variance of \(\sigma^2 = 180\) \(m^2\)?
Task 6
Among 120 randomly selected employees of a certain plant, 17 did not meet the work performance standard (in IT: 17 out of 120 servers did not meet SLA requirements).
Determine the 95% realization of the confidence interval for the fraction \(p\) of employees not meeting the standard in the entire plant.
Task 7
15 measurements were made of the time to repair yarn breaks on looms. The sample variance was calculated as \(s^2 = 134.2\). Assuming that this time has a normal distribution, determine the 90% confidence interval for the variance \(\sigma^2\) and the standard deviation \(\sigma\). The sample variance ( s^2 ) is calculated using division by ( n-1 ).
Hint: Use the chi-square (\(\chi^2\)) distribution tables.
Task 8
Two samples were drawn for a certain feature with a normal distribution:
- Sample 1: \(n=25\), mean \(\bar{x}=15\), deviation \(s=5\).
- Sample 2: \(n=100\), mean \(\bar{x}=15\), deviation \(s=5\).
Calculate the lengths of the 95% confidence intervals for both samples. How does a fourfold increase in sample size affect precision (interval width)?
Task 9
A simple sample of size \(n=5\) is given: \(\{2, 4, 6, 8, 10\}\). Calculate the value of the unbiased estimator of the expected value (\(\bar{x}\)) and the unbiased estimator of the variance (\(s^2\)).
Explain why we divide by \(n-1\) instead of \(n\) for variance.
Task 10
The percentage starch content was examined in 80 potatoes. The sample mean was \(\bar{x}=17.525\%\), and the standard deviation was \(s=1.84\%\).
Assuming a confidence level of 0.95, estimate the average starch content in the entire batch.