In Exercises 11–14, use the population of {34, 36, 41, 51} of the amounts of caffeine (mg/12oz) in Coca-Cola Zero, Diet Pepsi, Dr Pepper, and Mellow Yello Zero.

Assume that  random samples of size n = 2 are selected with replacement.

Sampling Distribution of the Sample Mean

a. After identifying the 16 different possible samples, find the mean of each sample, then construct a table representing the sampling distribution of the sample mean. In the table, combine values of the sample mean that are the same. (Hint: See Table 6-3 in Example 2 on page 258.)

b. Compare the mean of the population {34, 36, 41, 51} to the mean of the sampling distribution of the sample mean.

c. Do the sample means target the value of the population mean? In general, do sample means make good estimators of population means? Why or why not?

Short Answer

Expert verified

a. The following table represents the sampling distribution of the sample means:

Sample

Sample mean

(34,34)

34

(34,36)

35

(34,41)

37.5

(34,51)

42.5

(36,34)

35

(36,36)

36

(36,41)

38.5

(36,51)

43.5

(41,34)

37.5

(41,36)

38.5

(41,41)

41

(41,51)

46

(51,34)

42.5

(51,36)

43.5

(51,41)

46

(51,51)

51

Combining all the same values of means, the following table is obtained:

Sample mean

Probability

34

116

35

216

36

116

37.5

216

38.5

216

41

116

42.5

216

43.5

216

46

216

51

116

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

c. Since the population mean is equal to the mean of the sample means, it can be said that the sample means target the value of the population mean.

Since the mean value of the sampling distribution of the sample mean is equal to the population mean, the sample mean can be considered as a good estimator of the population mean.

Step by step solution

01

Given information

A population of the amounts of caffeine in 3 different drink brands is provided.

Samples of size equal to 2 are extracted from this population with replacement.

02

Sampling distribution of sample means

a.

All possible samples of size 2 selected with replacement are tabulated below:

(34,34)

(36,34)

(41,34)

(51,34)

(34,36)

(36,36)

(41,36)

(51,36)

(34,41)

(36,41)

(41,41)

(51,41)

(34,51)

(36,51)

(41,51)

(51,51)

The mean has the following formula:

Mean=i=1nxin where

xi denotes the ith observation of the sample

n is the sample size

The following table shows all possible samples of size equal to 2 and the corresponding sample means

Sample

Sample mean

(34,34)

x¯1=34+342=34

(34,36)

x¯2=34+362=35

(34,41)

x¯3=34+412=37.5

(34,51)

x¯4=34+512=42.5

(36,34)

x¯5=36+342=35

(36,36)

x¯6=36+362=36

(36,41)

x¯7=36+412=38.5

(36,51)

x¯8=36+512=43.5

(41,34)

x¯9=41+342=37.5

(41,36)

x¯10=41+362=38.5

(41,41)

x¯11=41+412=41

(41,51)

x¯12=41+512=46

(51,34)

x¯13=51+342=42.5

(51,36)

x¯14=51+362=43.5

(51,41)

x¯15=51+412=46

(51,51)

x¯16=51+512=51

Combining the values of means that are the same, the following probability values are obtained:

Sample mean

Probability

34

116

35

216

36

116

37.5

216

38.5

216

41

116

42.5

216

43.5

216

46

216

51

116

03

Population mean and mean of the sample means

b.

The population mean is computed as shown below:

μ=34+36+41+514=40.5

Thus, the population mean is equal to 40.5.

The mean of the sample means is computed below:

MeanofSampleMeans=x¯1+x¯2+.....+x¯1616=34+35+....+5116=40.5

Thus, the mean of the sampling distribution of the sample mean is equal to 40.5.

Here, the population mean is equal to the mean of the sampling distribution of the sample mean.

04

Good estimator

c.

Since the population mean is equal to the mean of the sample means, it can be said that the sample means target the value of the population mean.

A good estimator is a sample statistic whose sampling distribution has a mean value equal to the population parameter.

The mean value of the sampling distribution of the sample mean (40.5) is equal to the population mean (40.5).

Thus, the sample mean can be considered as a good estimator of the population mean.

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Most popular questions from this chapter

In Exercises 7–10, use the same population of {4, 5, 9} that was used in Examples 2 and 5. As in Examples 2 and 5, assume that samples of size n = 2 are randomly selected with replacement.

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