Q14 In Exercises 11鈥�14, use the po... [FREE SOLUTION]

Chapter 6: Q14 (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 Variance Repeat Exercise 11 using variances instead of means.

Short Answer

Expert verified

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

Sample	Sample variance
(34,34)	0
(34,36)	2
(34,41)	24.5
(34,51)	144.5
(36,34)	2
(36,36)	0
(36,41)	12.5
(36,51)	112.5
(41,34)	24.5
(41,36)	12.5
(41,41)	0
(41,51)	50
(51,34)	144.5
(51,36)	112.5
(51,41)	50
(51,51)	0

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

Sample variance	Probability
0	$\frac{4}{16}$
2	$\frac{2}{16}$
12.5	$\frac{2}{16}$
24.5	$\frac{2}{16}$
50	$\frac{2}{16}$
112.5	$\frac{2}{16}$
144.5	$\frac{2}{16}$

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

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

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

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 variances

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 sample variance has the following formula:

$s^{2} = \frac{{鈭�}_{i = 1}^{n} {(x_{i} - \overset{炉}{x})}^{2}}{n - 1}$ where

is the ith sample value

is the sample mean

n is the sample size

The sample means of all the 16 samples are tabulated below:

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)	${\overset{炉}{x}}_{5} = \frac{36 + 34}{2} = 37$
(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)	${\overset{炉}{x}}_{16} = \frac{51 + 51}{2} = 51$

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

Sample	Sample variance
(34,34)	$s_{1}^{2} = \frac{{(34 - 34)}^{2} + {(34 - 34)}^{2}}{2 - 1} = 0$
(34,36)	$s_{2}^{2} = \frac{{(34 - 35)}^{2} + {(36 - 35)}^{2}}{2 - 1} = 2$
(34,41)	$s_{3}^{2} = \frac{{(34 - 37.5)}^{2} + {(41 - 37.5)}^{2}}{2 - 1} = 24.5$
(34,51)	$s_{4}^{2} = \frac{{(34 - 42.5)}^{2} + {(51 - 42.5)}^{2}}{2 - 1} = 144.5$
(36,34)	$s_{5}^{2} = \frac{{(36 - 35)}^{2} + {(34 - 35)}^{2}}{2 - 1} = 2$
(36,36)	$s_{6}^{2} = \frac{{(36 - 36)}^{2} + {(36 - 36)}^{2}}{2 - 1} = 0$
(36,41)	$s_{7}^{2} = \frac{{(36 - 38.5)}^{2} + {(41 - 38.5)}^{2}}{2 - 1} = 12.5$
(36,51)	$s_{8}^{2} = \frac{{(36 - 43.5)}^{2} + {(51 - 43.5)}^{2}}{2 - 1} = 112.5$
(41,34)	$s_{9}^{2} = \frac{{(41 - 37.5)}^{2} + {(34 - 37.5)}^{2}}{2 - 1} = 24.5$
(41,36)	$s_{10}^{2} = \frac{{(41 - 38.5)}^{2} + {(36 - 38.5)}^{2}}{2 - 1} = 12.5$
(41,41)	$s_{11}^{2} = \frac{{(41 - 41)}^{2} + {(41 - 41)}^{2}}{2 - 1} = 0$
(41,51)	$s_{12}^{2} = \frac{{(41 - 46)}^{2} + {(51 - 46)}^{2}}{2 - 1} = 50$
(51,34)	$s_{13}^{2} = \frac{{(51 - 42.5)}^{2} + {(34 - 42.5)}^{2}}{2 - 1} = 144.5$
(51,36)	$s_{14}^{2} = \frac{{(51 - 43.5)}^{2} + {(36 - 43.5)}^{2}}{2 - 1} = 112.5$
(51,41)	$s_{15}^{2} = \frac{{(51 - 46)}^{2} + {(41 - 46)}^{2}}{2 - 1} = 50$
(51,51)	$s_{16}^{2} = \frac{{(51 - 51)}^{2} + {(51 - 51)}^{2}}{2 - 1} = 0$

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

Sample variance	Probability
0	$\frac{4}{16}$
2	$\frac{2}{16}$
12.5	$\frac{2}{16}$
24.5	$\frac{2}{16}$
50	$\frac{2}{16}$
112.5	$\frac{2}{16}$
144.5	$\frac{2}{16}$

Population variance and mean of the sample variances

The population mean is computed below:

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

The population variance is computed as shown below:

$蟽^{2} = \frac{{鈭�}_{i = 1}^{N} {(x_{i} - 渭)}^{2}}{N} = \frac{{(34 - 40.5)}^{2} + {(36 - 40.5)}^{2} + {(41 - 40.5)}^{2} + {(51 - 40.5)}^{2}}{4} = 43.25$

Thus, the population variance is equal to 43.25.

The mean of the sample variances is computed below:

${\overset{炉}{s}}^{2} = \frac{s_{1}^{2} + s_{2}^{2} + . . . . . + s_{16}^{2}}{16} = \frac{0 + 2 + . . . . + 0}{16} = 43.25$

Thus, the mean of the sampling distribution of the sample variance is equal to 43.25.

Here, the population variance (43.25) is equal to the mean of the sampling distribution of the sample variance (43.25).

Good estimator

Since the population varianceis equal to the mean of the sample variances, it can be said that the sample variancestarget the value of the population variance.

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 variance (43.25) is equal to the population variance (43.25).

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

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Short Answer

Step by step solution

Given information

Sampling distribution of sample variances

Population variance and mean of the sample variances

Good estimator

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