Aczel Solution 005

CHAPTER 1

Chapter 05 - Sampling and Sampling Distributions

Chapter 5

Sampling and Sampling distributions

5-1. Parameters are numerical measures of populations. Sample statistics are numerical measures of samples. An estimator is a sample statistic used for estimating a population parameter.

5-2.

= 97.9225(estimate of )

s = 51.8303

(estimate of )

s2 = 2,686.38(estimate of 2the population variance)

5-3.

= x/n = 5/12 = 0.41667

(5 out of 12 accounts are over $100.)

5-4.

= 2121.667

s = 1737.714

Basic Statistics from Raw Data

Measures of Central tendency

Mean2121.6667Median

Measures of Dispersion

If the data is of a

SamplePopulation

Variance3019651.52

St. Dev.1737.71445

5-5. average price = 4.367standard deviation = 0.3486

Basic Statistics from Raw Data

Measures of Central tendency

Mean4.3676471Median

Measures of Dispersion

If the data is of a

SamplePopulation

Variance0.12154412

St. Dev.0.34863178

5-6. = x/n = 11/18 = 0.6111, where x = the number of users of the product.

5-7. We need 25 elements from a population of 950 elements. Use the rows of Table 5-1, the rightmost 3 digits of each group starting in row 1 (left to right). So we skip any such 3-digit number that is either > 950 or that has been generated earlier in this list, giving us a list of 25 different numbers in the desired range. The chosen numbers are:

480, 11, 536, 647, 646, 179, 194, 368, 573, 595, 393, 198, 402, 130, 360, 527, 265, 809, 830,

167, 93, 243, 680, 856, 376.

5-8. We will use again Table 5-1, using columns this time. We will use right-hand columns, first 4 digits from the right (going down the column):

4,1943,4024,8303,5371,305.

5-9 We will use Table 5-1, sets of 2 columns using all 5 digits from column 1 and the first 3 digits from column 2, continuing by reading down in these columns. Then we will continue to the set: column 3 and first 3 digits column 4. We skip any numbers that are > 40,000,000. The resulting voter numbers are:

10,480,150

22,368,465

24,130,483

37,570,399

1,536,020.

5-10. There are 7 x 24 x 60 minutes in one week: (7)(24)(60) = 10,080 minutes. We will use Table 5-1

Start in the first row and go across the row, then to the next row (left to right using all 5 digits

in each set), discarding any of the resulting 5-digit numbers that are > 10,080. The resulting

minute numbers are:

1,5362,0116,2437,8566,1216,907

5-11.A sampling distribution is the probability distribution of a sample statistic. The sampling distribution is useful in determining the accuracy of estimation results.

5-12. Only if the population is itself normal.

5-13.E = 125SE 20/ = 8.944

5-14. The fact that, in the limit, the population distribution does not matter. Thus the theorem is very

general.

5-15. When the population distribution is unknown.

5-16. The Central Limit Theorem does not apply.

5-17. is binomial. Since np = 1.2, the Central Limit Theorem does not apply and we cannot use the

normal distribution.

5-18. = 1,247

= 10,000n = 100

P( < 1,230) = P = P(Z < 1.7) = .5 .4554 = 0.0446

Sampling Distribution of Sample Mean

Population Distribution

MeanStdev

1247100

Sample SizeSampling Distribution of X-bar

n100MeanStdev

124710

P(X 3.6) = P = P(Z > 1.333) = 0.0912



MeanStdev

3.41.5


n100MeanStdev

3.40.15

xP(X>x)

3.60.0912

5-21. P(12 < < 15) = P

= P(5.5 < Z < 9.5) = 2 (.5) = 1.000 (approximately)

(Use template: Sampling Distribution.xls, sheet: x-bar)

Sampling Distribution of Sample MeanPopulation Distribution

MeanStdevIs the population normal?

13.11.2


n36MeanStdev

13.10.2

x1P(x1x)

30.0000

5-26.n = 16

= 1.5

= 2

P( > 0) = P = P(Z > -3) = .5 + .4987 = 0.9987



MeanStdev

1.52


n16MeanStdev

1.50.5

xP(X>x)

00.9987

5-27. p = 1/7

P( < .10) = P = P(Z < (1.648) = 0.5 ( 0.4503 =

0.0497, a low probability. The sample size, along with np and n(1 p), are large enough here that the sample distribution (over all the different samples of 180 people in the population) of the proportion of people who get hospitalized during the year is going to be pretty close to normal. Therefore, any one such sample proportion will be close to the predicted mean 1/7 with reasonable probability, and 1/10 is far enough away from that mean given our estimated sample standard deviation that the probability of falling even farther away than that from the mean is small.

5-28.

= 700

= 100n = 60

P(680 720) = P

= 2TA(1.549) = 0.8786

5-29.p = = 0.35

= = 0.0213

P = P( < 0.30) + P( > 0.40)

= P + P

= 1 2TA(2.344) = 0.0190

5-30. Estimator B is better. It has a small bias, but its variance is small. This estimator is more likely to produce an estimate that is close to the parameter of interest.

5-31. I would use this estimator because consistency means as n the probability of getting close to the parameter increases. With a generous budget I can get a large sample size, which will make this probability high.

5-32. = 1,287

s 2 =

EMBED Equation.3 = 1,287 = 1,300

5-33. Advantage: uses all information in the data.

Disadvantage: may be too sensitive to the influence of outliers.

5-34. Depends also on efficiency and other factors. With respect to the bias:

A has bias = 1/nB has bias = 0.01

A is better than B when 1/n < 0.01, that is, when n > 1/0.01 = 100

5-35. Consistency is important because it means that as you get more data, your probability of getting closer to your target increases.

5-36. = 30, = 48, = 32. The three sample means are known. The df for deviations from the three sample means are:

df = + + 3 = 30 + 48 + 32 3 = 107

5-37. a) the mean is the best number to use.

mean =43.667

DeviationDeviation

Samplefrom meansquared

34-9.66793.45089

517.33353.77289

40-3.66713.44689

38-5.66732.11489

473.33311.10889

506.33340.10689

528.33369.43889

440.3330.110889

37-6.66744.44889

SSD =358

degrees of freedom = 8

MSD = SSD / df = 358 / 8 = 44.75

b) choose the means of the respective block of numbers: 40.75, 49.667, 40.5 minimized SSD = 195.917, df = 6, MSD = 32.65283

mean =40.7549.66740.5

DeviationDeviation


34-6.7545.5625

5110.25105.0625

40-0.750.5625

38-2.757.5625

47-2.6677.112889

500.3330.110889

522.3335.442889

443.512.25

37-3.512.25

SSD =195.9167

c) Each of the numbers themselves. SSD = 0. MSD indicates that the variance is zero, which is true since we are using each of the individual numbers to reduce SSD to zero.

d) SSD = 719, df = 9, MSD = 79.889

mean =50

DeviationDeviation


34-16256

5111

40-10100

38-12144

47-39

5000

5224

44-636

37-13169

SSD =719

5-38. No, because there are n 1 = 19 1 = 18 degrees of freedom for these checks once you know their mean. Since 17 is on less, there is a remaining degree of freedom and you cannot solve for the missing checks.

5-39. Yes. ( + ( + + )/19 = . Since 18 of the are known and so is , we can solve the equation for the unknown .

5-40. df = n-k

as k increases, df decreases, SSD decreases, MSD decreases

5-41. E() = = 1,065

V() = /n = 5002/100 = 2,500

5-42. = 1,000,000

Want SD() 25

SD() = = 1,000 /

1,000 / 25

1,000/25 = 40

n 1,600. The sample size must be at least 1,600.

5-43. = 53

= 10

n = 400

E() = = 53SE() = / = 10 / = 0.5



MeanStdev

5310


n400MeanStdev

530.5

5-44.p = 0.5

n = 120

SE() = = = 0.0456

5-45. E() = p = 0.2

SE() = = = 0.04216

5-46. P = 0.5 maximizes the variance of .Proof:

V() =

=

EMBED Equation.3 (p(p 2) = (1 2p)

Set the derivative to zero:

(1 2p) = 0

1 = 2p

p = 1/2

The assertion may also be demonstrated by trying different values of p.

5-47. P(0.72 < < 0.82) = P(10.95 < Z

Aczel Solution 005

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