Finite Population Correction Factor Quiz

PDF page 328: We saw that the sample size has an important effect on the variance and thus the standard deviation of the sampling distribution. Also of interest is the proportion of the total population that has been sampled. We have assumed that the population is extremely large and that we have sampled a small part of the population. As the population becomes smaller and we sample a larger number of observations the sample observations are not independent of each other. To correct for the impact of this, the Finite Correction Factor can be used to adjust the variance of the sampling distribution. It is appropriate when more than 5% of the population is being sampled and the population has a known population size. There are cases when the population is known, and therefore the correction factor must be applied. The issue arises for both the sampling distribution of the means and the sampling distribution of proportions. The Finite Population Correction Factor for the variance of the means shown in the standardizing formula is: ̄

PDF page 328: We saw that the sample size has an important effect on the variance and thus the standard deviation of the sampling distribution. Also of interest is the proportion of the total population that has been sampled. We have assumed that the population is extremely large and that we have sampled a small part of the population. As the population becomes smaller and we sample a larger number of observations the sample observations are not independent of each other. To correct for the impact of this, the Finite Correction Factor can be used to adjust the variance of the sampling distribution. It is appropriate when more than 5% of the population is being sampled and the population has a known population size. There are cases when the population is known, and therefore the correction factor must be applied. The issue arises for both the sampling distribution of the means and the sampling distribution of proportions. The Finite Population Correction Factor for the variance of the means shown in the standardizing formula is: ̄

PDF page 330: Average a number that describes the central tendency of the data; there are a number of specialized averages, including the arithmetic mean, weighted mean, median, mode, and geometric mean. Central Limit Theorem Given a random variable with known mean μ and known standard deviation, σ, we are – sampling with size n, and we are interested in two new RVs: the sample mean, X . If the size (n) of the sample is – σ sufficiently large, then X ~ N(μ, ). If the size (n) of the sample is sufficiently large, then the distribution of the n sample means will approximate a normal distributions regardless of the shape of the population. The mean of the σ , sample means will equal the population mean. The standard deviation of the distribution of the sample means, n is called the standard error of the mean. Finite Population Correction Factor adjusts the variance of the sampling distribution if the population is known and more than 5% of the population is being sampled. Mean a number that measures the central tendency; a common name for mean is "average." The term "mean" is a – shortened form of "arithmetic mean." By definition, the mean for a sample (denoted by x ) is – Sum of all values in the sample x =

PDF page 328: It is learned that the population of White German Shepherds in the USA is 4,000 dogs, and the mean weight for German Shepherds is 75.45 pounds. It is also learned that the population standard deviation is 10.37 pounds. If the sample size is 100 dogs, then find the probability that a sample will have a mean that differs from the true probability mean by less than 2 pounds.

PDF page 329: N = 4000, n = 100, σ = 10.37, μ = 75.45, x − μ = ± 2

PDF page 328: It is learned that the population of White German Shepherds in the USA is 4,000 dogs, and the mean weight for German Shepherds is 75.45 pounds. It is also learned that the population standard deviation is 10.37 pounds. If the sample size is 100 dogs, then find the probability that a sample will have a mean that differs from the true probability mean by less than 2 pounds.

PDF page 329: N = 4000, n = 100, σ = 10.37, μ = 75.45, x − μ = ± 2

PDF page 328: It is learned that the population of White German Shepherds in the USA is 4,000 dogs, and the mean weight for German Shepherds is 75.45 pounds. It is also learned that the population standard deviation is 10.37 pounds. If the sample size is 100 dogs, then find the probability that a sample will have a mean that differs from the true probability mean by less than 2 pounds.

PDF page 329: N = 4000, n = 100, σ = 10.37, μ = 75.45, x − μ = ± 2

PDF page 328: It is learned that the population of White German Shepherds in the USA is 4,000 dogs, and the mean weight for German Shepherds is 75.45 pounds. It is also learned that the population standard deviation is 10.37 pounds. If the sample size is 100 dogs, then find the probability that a sample will have a mean that differs from the true probability mean by less than 2 pounds.

PDF page 329: N = 4000, n = 100, σ = 10.37, μ = 75.45, x − μ = ± 2

PDF page 329: When a customer places an order with Rudy's On-Line Office Supplies, a computerized accounting information system (AIS) automatically checks to see if the customer has exceeded his or her credit limit. Past records indicate that the probability of customers exceeding their credit limit is .06. Suppose that on a given day, 3,000 orders are placed in total. If we randomly select 360 orders, what is the probability that between 10 and 20 customers will exceed their credit limit?

PDF page 329: N = 3000, n = 360, p = 0.06

PDF page 329: When a customer places an order with Rudy's On-Line Office Supplies, a computerized accounting information system (AIS) automatically checks to see if the customer has exceeded his or her credit limit. Past records indicate that the probability of customers exceeding their credit limit is .06. Suppose that on a given day, 3,000 orders are placed in total. If we randomly select 360 orders, what is the probability that between 10 and 20 customers will exceed their credit limit?

PDF page 329: N = 3000, n = 360, p = 0.06

PDF page 329: When a customer places an order with Rudy's On-Line Office Supplies, a computerized accounting information system (AIS) automatically checks to see if the customer has exceeded his or her credit limit. Past records indicate that the probability of customers exceeding their credit limit is .06. Suppose that on a given day, 3,000 orders are placed in total. If we randomly select 360 orders, what is the probability that between 10 and 20 customers will exceed their credit limit?

PDF page 329: N = 3000, n = 360, p = 0.06

PDF page 329: f ⎝Z⎠ = 0.4744 * 2 = 0.9488

PDF page 329: = p −2.74 ∠ z ∠ −0.38 = 0.4969 − 0.1480 = 0.3489

PDF page 328: It is learned that the population of White German Shepherds in the USA is 4,000 dogs, and the mean weight for German Shepherds is 75.45 pounds. It is also learned that the population standard deviation is 10.37 pounds. If the sample size is 100 dogs, then find the probability that a sample will have a mean that differs from the true probability mean by less than 2 pounds.

PDF page 329: N = 4000, n = 100, σ = 10.37, μ = 75.45, x − μ = ± 2