Q12 In Exercises 11鈥�14, use the po... [FREE SOLUTION]

Chapter 6: Q12 (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 Median Repeat Exercise 11 using medians instead of means.

Short Answer

Expert verified

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

Sample	Sample median
(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 medians, the following table is obtained:

Sample median	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 median is not equal to the mean of the sampling distribution of the sample median.

c. Since the population median is not equal to the mean of the sample medians, it can be said that the sample medians do not target the value of the population median.

Since the mean value of the sampling distribution of the sample median is not equal to the population median, the sample median cannot be considered as a good estimator of the population median.

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 medians

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)

Note that for a sample of size 2, the median has the following formula:

$M e d i a n = \frac{{(\frac{n}{2})}^{t h} o b s + {(\frac{n}{2} + 1)}^{t h} o b s}{2} = \frac{{(\frac{2}{2})}^{t h} o b s + {(\frac{2}{2} + 1)}^{t h} o b s}{2} = \frac{1^{s t} o b s + 2^{n d} o b s}{2}$

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

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

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

Sample median	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 median and mean of the sample medians

The population median is computed as shown below:

$Population Median = \frac{{(\frac{n}{2})}^{t h} o b s e r v a t i o n + {(\frac{n}{2} + 1)}^{t h} o b s e r v a t i o n}{2} = \frac{{(\frac{4}{2})}^{t h} o b s e r v a t i o n + {(\frac{4}{2} + 1)}^{t h} o b s e r v a t i o n}{2} = \frac{2^{n d} o b s + 3^{r d} o b s}{2} = \frac{36 + 41}{2} = 38.5$

Thus, the population median is equal to 38.5.

The mean of the sample medians is computed below:

$Mean of Sample Medians = \frac{Sample {Median}_{1} + Sample {Median}_{2} + . . . . . + Sample {Median}_{16}}{16} = \frac{34 + 35 + . . . . + 51}{16} = 40.5$

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

Here, the population median (38.5) is not equal to the mean of the sampling distribution of the sample median (40.5).

Good estimator

Since the population median is not equal to the mean of the sample medians, it can be said that the sample medians do not target the value of the population median.

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 median (40.5) is not equal to the population median (38.5).

Thus, the sample median cannot be considered as a good estimator of the population median.

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	Mean	St.Dev.	Distribution
Males	23.5 in	1.1 in	Normal
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Short Answer

Step by step solution

Given information

Sampling distribution of sample medians

Population median and mean of the sample medians

Good estimator

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