Instructions
Topics: Sampling, and Sampling Distribution Sampling
Slide 5: Exercise 1and Exercise 2,
Sampling Distribution
Slide 8: Exercise 1
Slide 9: Exercises 2 and 3
Topic: Central Limit Theorem
Slide10: Example 4
Slide11: Exercises 4, 5, and 6.
Also solve the 3 exercises underneath:
Exercise I*. An inspector wishes to estimate the mean weekly wage of the several thousands of workers employed in a plant within plus or minus $20 and with 99% degree of confidence. From past experience, the inspector knows that the weekly wages of these workers are normally distributed with a standard deviation of $40. What is the minimum sample size required?
Topics: Confidence Interval andNormal Distribution
Slide14: Example 1, Example 2
Slide15: Exercise 2.
Exercise II. A team of efficiency experts intends to use the mean of the random sample of size n = 150 to estimate the average mechanical aptitude (measured in points) of assembly-line workers in a large industry (population mean µ). If the sample mean x ¯=69.5 and, based on experience, the efficiency experts can assume that σ = 6.2 for such data
calculate a 90% confidence interval *
calculate a 95% confidence interval
calculate a 99% confidence interval *
for the average electromechanical aptitude of assembly line workers (µ) in the given industry. We can assume that the sample size n is much smaller than the population N of workers (n/N < 5%).
Exercise III*. A random sample of n = 64 with a mean x ¯=50 and a standard deviation (s) of 20 is taken from a population of N = 800. Find an interval estimate for the population mean such that we are:
90% confidence interval that the interval includes the population mean.*
95% confidence interval that the interval includes the population mean.
99% confidence interval that the interval includes the population mean.*
What does the result of part a, b, and c tells us?
Hint. Notice that n > 30, thus the theoretical sampling distribution of the mean is approximately normal, i.e., we can use the z value of 1.96 from the standard normal distribution to construct the 95% confidence interval for the unknown population, and we can use the sample standard deviation s as an estimate for the unknown σ.
Hint. Since n/N is higher than 5%, we have to use the population correction factor for the standard error of the mean. That is we have to use the formula:
µ=x ¯±z s/√n √((N-n)/(N-1))