5.1.1. Mean and standard deviation of

The mean and standard deviation calculated for the sampling distribution of are called the mean and standard deviation of and denoted by and respectively.

Let us calculate the mean of the 10 values of listed in Table 5.3.

Alternatively, we can calculate the mean of the sampling distribution of listed in Table 5.5 as

=

.

Now let us calculate the mean of population: the annual salaries of all five employees:

The mean of the sampling distribution of always equal to the mean of the population.

Mean of the sampling distribution:

The mean of the sampling distribution of is equal to the mean of the population.

Hence

= .

Hence, if we take all possible samples (of the same size) from a population and calculate their means, the mean of all these sample means will be the same as the population mean.

The sample mean is called an estimator of the population mean . When the expected value (or mean) of a sample statistic is equal to the value of the corresponding population mean, that sample statistic is said to be an unbiased estimator. For the sample mean , = . Hence, is an unbiased estimator of .

Let us talk about population standard deviation and standard deviation of the sampling distribution of .

First, let us find standard deviation of salaries of five employees.

We will use

to obtain standard deviation of population.

Now let us calculate the value of .

We will use formula

.

We obtain

As we see, the standard deviation of is not equal to the standard deviation of the population distribution. The standard deviation of is equal to the standard deviation of the population divided by the square root of the sample size.

That is,

We also call as a standard error of .

We use the above formula for standard error if the sample size is a small in comparison to the population size. The sample size is considered to be small compared to the population size if the sample size is equal to or less than 5% of the population size, that is, if

If this condition does not satisfied, we use the following formula to

calculate .

The term is often called a finite population correction factor.

In most practical applications, the sample size is usually small compared to the population size.

Consequently, in most cases the formula used for calculating is

5.1.2. Central limit theorem

Theorem: Whatever the population, the distribution of is approximately

normal when n is large. In random sampling from the population with mean and standard deviation , when n is large, the distribution of is approximately normal with mean and standard deviation . Consequently,

i s approximately

Remark:

The sample size is usually considered to be large if .

The central limit theorem states that the distribution of is normally distributed and can be used to compute a random variable, Z, with mean 0 and variance 1:

.

Example:

The amount of telephone bills for all households in a large city have a distribution that is skewed to the right with a mean of $30 and a standard deviation equal to $7. Calculate the mean and standard deviation of and describe the shape of its sampling distribution when the sample size is

a) n =35; b) n =70

Solution:

Although the population distribution is not normal, in each case the sample size is large .

Hence the central limit theorem can be applied.

a) Let be the mean value of telephone bills paid by a sample of 35 households. Then the sampling distribution of is approximately normal with

and .

Figure 5.3 shows the population distribution and sampling distribution of .

b) Let be the mean value of telephone bills paid by a sample of 70 households. Then the sampling distribution of is approximately normal with

and .

Figure 5.4 shows the population distribution and sampling distribution of .

`

Example:

Consider a population with mean 75 and standard deviation of 11.

a) If a random sample of size 64 is selected, what is the probability that the sample mean will be between 73 and 78?

b) If a random sample of size 80 is selected, what is the probability that the sample mean will be between 68 and 83?

Solution:

We have and Since n =64 is large, the central limit theorem tells us that the distribution of is approximately normal.

a) To calculate we convert to the standardized variable

.

The Z –values corresponding to 73 and 78 are

For 73:

For 78:

Consequently,

b) We now have n =80.

.

Example:

The prices of all houses in a large city have a probability distribution with a mean of $80 000 and a standard deviation of $15 000. Let be the mean price of a sample of 200 houses selected from this city.

a) What is the probability that the mean price obtained from this sample will be within $2 000 of the population mean?

b) What is the probability that the mean price obtained from this sample will be more than the population mean by $1 500 or more?

Solution:

The sampling distribution of is approximately normal because the sample size is large (n >30).

a) We need to find the probability

.

b) The probability that the mean price obtained from the sample of 200 houses will be more than the population mean by $1 500 or more is written as

.

Exercises

1. A large population has mean 90 and standard deviation 8. Calculate and for a random sample of size

a) n =9 and b) n =36.

2. A large population has a standard deviation 10. What is the standard deviation of for a random sample of size

a) n =25 ; b) n =100; and c) n =400

3. Consider a large population with and

Assuming , find the mean and standard deviation of the sample

mean for the sample size of

a) 18; b) 90

4. A population of N=5000 has a . In each of the following cases which formula you use to calculate and why?

Using the appropriate formula, calculate for each of these cases.

a) n =300; b) n =200; c) n =500; d) n =100.

5. For a population with and

a) For a sample selected from this population, and .

Find the sample size. Assume .

b) For a sample selected from this population, and . Find the sample size. Assume .

6. The following data show the number of automobiles owned in a population of five families:

2; 1; 0; 2; 3

a) List the ten possible samples of size 3 for this population.

(Use sampling without replacement).

b) Using the ten values, compute the mean and variance of .

c) Compute the mean and variance of the population. Compare your results to those in part b. Interpret your findings.

7. For a population, , and , find the Z values for each of the following for n =49.

a) ; b) ; c)

8. A population has a mean of 58 and a standard deviation of 12. Assuming

, find the following probabilities for a sample size of 50.

a) ; b)

9. Let X be a continuous random variable that has a distribution with and . Assuming , find the probability that the sample mean for a random sample of 64 taken from this population will be

a) less than 82.3;

b) more than 86.7.

10. The amount of electric bills for all households in a city have a skewed probability distribution with a mean of $65 and a standard deviation of $25. Find the probability that the sample mean amount of electric bills for a random sample of 75 households selected from this city will be

a) more than $70;

b) between $58 and $63;

c) within $6 of the population mean;

d) more than the population mean by at least $5?

11. The balances of all saving accounts at a local bank has a mean equal $45 360 and standard deviation equal to $5 900. Find the probability that the mean of a sample of 80 saving accounts selected from this bank will be

a) less than $46 500;

b) between $40 000 and $49 200;

c) within $1 800 of the population mean;

d) lower than the population mean by $1 200 or more.

12. The mean and standard deviation of the population are 55 and 7, respectively. Sample of 40 observations is selected randomly from this population.

a) What is the probability that sample mean will be between 54 and 56?

b) Find the shortest interval centered at 55, where will lie with

probability 0.95.

13. Consider a random sample of size n =100 from a population that has a standard deviation of .

a) Find the probability that the sample mean will lie within 2 units of the population mean-that is, .

b) Find the number k so that .

c) What is the probability that will differ from by more than 4 units?

Answers

1. a) =90; =2.67; b) =90; =1.33; 2. a) 2; b) 1; c) 0.5;

3. a) =60; =2.828; b) =60; =1.265;

4. a) =1.120; b) =1.414; c) =0.849; d) =2.000; 5. a) n =25;

b) n =64; 7. a) Z =2.44; b) Z =-7.25; c) Z=-3.65; 8. a) 0.1530; b) 0.7611;

9. a) 0.0001; b) 0.9505; 10. a) 0.0418; b) 0.2373; c) 0.9624; d) 0.0418;

11.a) 0.9582; b) 1.0; c) 0.9936; d) 0.0344;12. a) 0.6318; b) [52.82; 57.18];

13. a) 0.6826; b) k=3.29; c) 0.0456.