What is sampling error?
It is the difference between the sample mean and the population mean Could the value of the sampling error be zero?
Yes it is possible to have a zero sampling error. However, it is very low probability that this could happen. If it were zero, what would this mean?
This means that the population is uniform and the sample mean and the population mean are equal.
List the reasons for sampling. Give an example of each reason for sampling.
1. Contacting whole population is time consuming. If the population is California residents, it will take a long time to send everyone a survey and then process the results. 2. Contacting whole population is costly. Same example of California residents, it will be very costly to send by mail a survey to all residents and then process millions of responses. 3. Checking all population is physically impossible. If the population is infinite like the water at California shores, its is impossible to check the bacteria levels for all the water on California shores. 4. Some tests are destructive to the population. Like testing for an epidemic of e-coli bacteria in lettuce, we can’t take every lettuce produced in a farm and damage it while testing for the bacteria. The whole crop will be damaged. 5. Sample results are adequate. Valid and reliable samples provide adequate results that are very close to the population results. For example, checking the price of rice in the retailers in US, it is sufficient to get a retailer from each region to get a good idea of the price of rice in the whole nation.
Population is defined as including all items with the characteristic one wishes to understand. Because there is seldom enough time or money to gather information from everyone or everything in a population, the goal is to find a representative sample (or subset) of that population. For example, a researcher might study the success rate of a new ‘quit smoking’ program on a sample group ...
Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $110,000. This distribution follows the normal distribution with a standard deviation of $40,000.
A. If we select a random sample of 50 households, what is the standard error of the mean?
σ/√n = 40000/√50 = 5656.85
B. What is the expected shape of the distribution of the sample mean? “The central limit theorem states that, for large random samples, the shape of the sampling distribution of the sample mean is close to the normal probability distribution” “Most statisticians consider a sample of 30 or more to be large enough for the central limit theorem to be employed”
From the text book, since the sample size is 50 which is greater than 30 we can apply the central limit theorem and it should be normally distributed.
C. What is the likelihood of selecting a sample with a mean of at least $112,000?
Z = (sample mean – population mean) / standard error of mean Z = (X – μ) / σ/√n, Where X is the sample mean, μ is the population mean Z = (112000 –110000)/5656.85) = 0.3535
P(Z) = P (0-1.76777) = 0.5+ P (–1.7677 < z < 0) = 0.5 =0.5 + 0.4616 = 0.9616.
E. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000.
P (100000 < X < 112000) = P(Z > -1.7677) – P(Z >0.3535) = 0.9616 – 0.3632 =0.5984.
A state meat inspector in Iowa has been given the assignment of estimating the mean net weight of packages of ground chuck labeled “3 pounds.” Of course, he realizes that the weights cannot be precisely 3 pounds. A sample of 36 packages reveals the mean weight to be 3.01 pounds, with a standard deviation of 0.03 pounds.
The Standard deviation is a measure of the variation (or spread) of a data set. For a variable x, the standard deviation of all possible observations for the entire population is called the population standard deviation or standard deviation of the variable x. It is denoted σx or, when no confusion will arise, simply σ. Suppose that we want to obtain information about a population standard ...
a. What is the estimated population mean? 3 pounds
b. Determine a 95 percent confidence interval for the population mean. Since sample is big (>30), we can use the formula of X – (z)σ/√n