Q11 In Exercises 11鈥�14, use the po... [FREE SOLUTION]

Chapter 6: Q11 (page 264)

In Exercises 11鈥�14, use the population of {34, 36, 41, 51} of the amounts of caffeine (mg/12oz) in鈥侰oca-Cola鈥俍ero,鈥侱iet鈥侾epsi,鈥侱r鈥侾epper,鈥俛nd鈥侻ellow鈥俌ello鈥俍ero.
Assume鈥倀hat鈥� 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	$\frac{1}{16}$
35	$\frac{2}{16}$
36	$\frac{1}{16}$
37.5	$\frac{2}{16}$
38.5	$\frac{2}{16}$
41	$\frac{1}{16}$
42.5	$\frac{2}{16}$
43.5	$\frac{2}{16}$
46	$\frac{2}{16}$
51	$\frac{1}{16}$

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

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.

Sampling distribution of sample means

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:

$M e a n = \frac{{鈭�}_{i = 1}^{n} x_{i}}{n}$ where

$x_{i}$ 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)	${\overset{炉}{x}}_{1} = \frac{34 + 34}{2} = 34$
(34,36)	${\overset{炉}{x}}_{2} = \frac{34 + 36}{2} = 35$
(34,41)	${\overset{炉}{x}}_{3} = \frac{34 + 41}{2} = 37.5$
(34,51)	${\overset{炉}{x}}_{4} = \frac{34 + 51}{2} = 42.5$
(36,34)	${\overset{炉}{x}}_{5} = \frac{36 + 34}{2} = 35$
(36,36)	${\overset{炉}{x}}_{6} = \frac{36 + 36}{2} = 36$
(36,41)	${\overset{炉}{x}}_{7} = \frac{36 + 41}{2} = 38.5$
(36,51)	${\overset{炉}{x}}_{8} = \frac{36 + 51}{2} = 43.5$
(41,34)	${\overset{炉}{x}}_{9} = \frac{41 + 34}{2} = 37.5$
(41,36)	${\overset{炉}{x}}_{10} = \frac{41 + 36}{2} = 38.5$
(41,41)	${\overset{炉}{x}}_{11} = \frac{41 + 41}{2} = 41$
(41,51)	${\overset{炉}{x}}_{12} = \frac{41 + 51}{2} = 46$
(51,34)	${\overset{炉}{x}}_{13} = \frac{51 + 34}{2} = 42.5$
(51,36)	${\overset{炉}{x}}_{14} = \frac{51 + 36}{2} = 43.5$
(51,41)	${\overset{炉}{x}}_{15} = \frac{51 + 41}{2} = 46$
(51,51)	${\overset{炉}{x}}_{16} = \frac{51 + 51}{2} = 51$

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

Sample mean	Probability
34	$\frac{1}{16}$
35	$\frac{2}{16}$
36	$\frac{1}{16}$
37.5	$\frac{2}{16}$
38.5	$\frac{2}{16}$
41	$\frac{1}{16}$
42.5	$\frac{2}{16}$
43.5	$\frac{2}{16}$
46	$\frac{2}{16}$
51	$\frac{1}{16}$

Population mean and mean of the sample means

The population mean is computed as shown below:

$渭 = \frac{34 + 36 + 41 + 51}{4} = 40.5$

Thus, the population mean is equal to 40.5.

The mean of the sample means is computed below:

$Mean of Sample Means = \frac{{\overset{炉}{x}}_{1} + {\overset{炉}{x}}_{2} + . . . . . + {\overset{炉}{x}}_{16}}{16} = \frac{34 + 35 + . . . . + 51}{16} = 40.5$

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.

Good estimator

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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Short Answer

Step by step solution

Given information

Sampling distribution of sample means

Population mean and mean of the sample means

Good estimator

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