Three firms carry inventories that differ in size. Firm A's...

Three firms carry inventories that differ in size. Firm A's inventory contains 2000 items, firm $\mathrm{B}$ 's inventory contains 5000 items, and firm $\mathrm{C}$ 's inventory contains 10,000 items. The population standard deviation for the cost of the items in each firm's inventory is $\sigma=144 .$ A statistical consultant recommends that each firm take a sample of 50 items from its inventory to provide statistically valid estimates of the average cost per item. Employees of the small firm state that because it has the smallest population, it should be able to make the estimate from a much smaller sample than that required by the larger firms. However, the consultant states that to obtain the same standard error and thus the same precision in the sample results, all firms should use the same sample size regardless of population size. a. Using the finite population correction factor, compute the standard error for each of the three firms given a sample of size 50 . b. What is the probability that for each firm the sample mean $\bar{x}$ will be within ±25 of the population mean $\mu ?$

Three firms carry inventories that differ in size. Firm A's inventory contains 2000 items, firm $\mathrm{B}$ 's inventory contains 5000 items, and firm $\mathrm{C}$ 's inventory contains 10,000 items. The population standard deviation for the cost of the items in each firm's inventory is $\sigma=144 .$ A statistical consultant recommends that each firm take a sample of 50 items from its inventory to provide statistically valid estimates of the average cost per item. Employees of the small firm state that because it has the smallest population, it should be able to make the estimate from a much smaller sample than that required by the larger firms. However, the consultant states that to obtain the same standard error and thus the same precision in the sample results, all firms should use the same sample size regardless of population size.
a. Using the finite population correction factor, compute the standard error for each of the three firms given a sample of size 50 .
b. What is the probability that for each firm the sample mean $\bar{x}$ will be within ±25 of the population mean $\mu ?$

Key Concepts

Finite Population Correction (FPC)

In sampling from a finite population, the finite population correction factor is used to adjust the standard error of the sample mean to account for the fact that sampling without replacement reduces variability. The correction factor is calculated as ?((N - n) / (N - 1)), where N is the population size and n is the sample size. This becomes significant when the sample size is a non-negligible fraction of the population size.

Standard Error of the Sample Mean

The standard error of the sample mean quantifies the dispersion of the sampling distribution of the sample mean. It is derived from the population standard deviation and is adjusted for the sample size. When sampling from a finite population, it is further adjusted by the finite population correction factor, resulting in the formula: ?/?n multiplied by ?((N - n)/(N - 1)). This measure indicates the precision of the sample mean as an estimate of the population mean.

Central Limit Theorem and Normal Distribution

The Central Limit Theorem states that the distribution of the sample mean will tend to be normal or nearly normal provided the sample size is sufficiently large, regardless of the population's distribution. This property allows statisticians to use the normal distribution to approximate probabilities and perform hypothesis testing and confidence interval estimation for the sample mean.

Probability Calculation for Sample Mean Estimates

When using the standard error and normal approximation, the probability that the sample mean falls within a certain range of the population mean can be computed using z-scores and the standard normal distribution. This process involves determining the standardized value (z-score) corresponding to the specified interval and then using the cumulative distribution function of the normal distribution to find the associated probability.

Transcript

00:01 We have three different companies, a, b, and c, and this has a population size of 2 ,000.

00:09 This has a population size of 5 ,000, and this has a population size of 10 ,000.

00:16 And we know that the standard deviation for all three of these is said to be 144.

00:25 44 and we are asked to use a sample size of 50 for all three of these.

00:34 And we want to have in part a, we want to know what the standard error is going to be using the finite correction factor.

00:42 And so that is to take the capital n minus the little n divided by one less than the population size, times the standard deviation 144.

00:56 Over the square root of the sample size.

01:00 And then we have to find, and i will find this for each one, and then we want to find in part b what the value is for the mean plus or minus 25.

01:13 And so this standard air is square root of, and that is going to be 2000 minus 50 is going to be 1950.

01:27 Divided by 1999 and then times 144 divided by the square root of 50.

01:38 And so that standard error comes out to be approximately 20 .11.

01:44 And i'm going to store that as x in my calculator.

01:48 So i'm storing that as x.

01:50 And now we're going to use the normal cdf.

01:53 And we don't know what the mean is, but we know the differences are going to be going to end up being negative 25 and 25.

02:01 So we're just going to use negative 25, 25.

02:04 We'll assume the mean is zero.

02:06 So this is all relative.

02:07 And then we're going to use this value for our standard a -r.

02:11 So normal cdf, negative 25, 25, 0, and again, that x value...