CHAPTER 13 REVIEW



CHAPTER 13 REVIEW

1. The following are the approximate percentages for the different blood types among white Americans: A: 40%; B: 11%; AB: 4%; O: 45%. A random sample of 1000 black Americans yielded the following blood type data: A: 270; B: 200; AB: 40; O: 490. Does this sample indicate that the distribution of blood types among black Americans differs from that of white Americans?

ANS:

H0: ρA = 0.40, ρB = 0.11, ρAB = 0.04, ρO = 0.45

Ha: the actual population proportions are different from those in H0.

Blood type Observed values Expected Values

A 270 (0.4)(1000) = 400

B 200 (0.11)(1000) = 110

AB 40 (0.04)(1000) = 40

O 490 (0.45)(1000) = 450

[pic]

df = 4 – 1 = 3

From the χ2 table, we see that 119.44 is larger than any value for 3 df. Hence, P value < 0.0005. [The TI-83 gives a P value of 1.02 x 10-25 (χ2cdf(119.44, 1E99, 3).]

Because P value < 0.05, we reject the null and conclude that not all the proportions are as hypothesized in the null. We have very strong evidence that the proportions of the various blood types among black Americans differ from the proportions among white Americans.

2. The statistics teacher, Ms. Uong, used her calculator to simulate rolling a die 96 times and storing the results in a list L1. She did this by entering MATH PRB randInt(1, 6, 96) ( L1. Next she stored the list (STAT SortA (L1)). She then counted the number of each face value. The results were as follows:

Face value Observed

1 19

2 15

3 10

4 14

5 17

6 21

Does it appear that the teacher’s calculator is simulating a fair die?

ANS:

H0: p0 = p2 = p3 = p4 = p5 = p6 = 1/6

Ha: Not all of the proportions are equal to 1/6.

Face value Observed Expected

1 19 (1/6)(96) = 16

2 15 (1/6)(96) = 16

3 10 (1/6)(96) = 16

4 14 (1/6)(96) = 16

5 17 (1/6)(96) = 16

6 21 (1/6)(96) = 16

χ2 = 4.75, df = 6 – 1 = 5 ( P-value > 0.25 (table) or P value = 0.45 (calculator [χ2cdf(4.75, 1E99, 5)])

Because the P value is > 0.05, we have no evidence to reject the null. We cannot conclude that the calculator is failing to simulate a fair die.

3. A study of 150 cities wanted to determine if crime rate is related to outdoor temperature. The results of the study are summarized in the following table:

Crime Rate

| |Below |Normal |Above |

|Below |12 |8 |5 |

|Normal |35 |41 |24 |

|Above |4 |7 |14 |

Do these data provide evidence, at the 0.02 level of significance, that the crime rate is related to the temperature at the time of the crime?

ANS:

H0: The crime rate is independent of temperature.

Ha: The crime rate is not independent of temperature.

Crime Rate

| |Below |Normal |Above |Total |

|Below |12 |8 |5 |25 |

| |(8.5) |(9.33) |(7.17) | |

|Normal |35 |41 |24 |100 |

| |(34) |(37.33) |(28.67) | |

|Above |4 |7 |14 |25 |

| |(8.5) |(9.33) |(7.17) | |

|Total |51 |56 |43 |150 |

[pic]

df = (3 – 1)(3 – 1) = 4 ( 0.01 < P value < 0.02 (table). [P value = 0.012 (calculator).]

Because P value < 0.02, we reject H0. We have strong evidence that the number of crimes committed is related to the temperature at the time of the crime.

4. A study yields a chi-square statistic value of 20 (χ2 = 20). What is the P value of the test if

(a) the study was a goodness-of-fit test with n = 12?

(b) the study was a test of independence between two categorical variables, the row variable with 3 values and the column variable with 4 values?

ANS:

(a) n = 12 ( df = 12 – 1 = 11 ( 0.025 < P value < 0.05

[Using the calculator: χ2cdf(20, 1E99, 11) = 0.045.]

(b) r = 3, c = 4 ( df = (3 – 1)(4 – 1) = 6 ( 0.0025 < P value < 0.005

[Using the calculator: χ2cdf(20, 1E99, 6) = 0.0028.]

5 – 7 The following data were collected while conducting a chi-square test for independence:

Preference

| |Brand A |Brand B |Brand C |

|Male |16 |22 |15 |

|Female |18* |30 |28 |

5. What null hypothesis is being tested?

ANS:

H0: Gender and preference are independent in the population.

6. What is the expected value of the cell marked with the *?

ANS:

Preference

| |Brand A |Brand B |Brand C |Total |

|Male |16 |22 |15 |53 |

|Female |18* |30 |28 |76 |

| |(20.03) | | | |

|Total |34 |52 |43 |129 |

The expected value = [pic]

7. How many degrees of freedom are involved in the test?

ANS:

df = (2 – 1)(3 – 1) = 2

8. A χ2 goodness-of-fit is performed on a random sample of 360 individuals to see if the number of birthdays each month is proportional to the number of days in the month. χ2 is determined to be 23.5. The P value for this test is

(a) 0.001 < P < 0.005

(b) 0.02 < P < 0.025

(c) 0.025 < P < 0.05

(d) 0.01 < P < 0.02

(e) 0.05 < P < 0.10

ANS: D

Because n = 12 months, df = 12 – 1 = 11. Reading from the χ2 table, we have 0.01 < P < 0.02. [Using the calculator, χ2cdf(23.5, 1E99, 11) = 0.0150

9. For the following two-way table, compute the value of χ2.

| |C |D |

|A |15 |25 |

|B |10 |30 |

(a) 2.63

(b) 1.22

(c) 1.89

(d) 2.04

(e) 1.45

ANS: E

| |C |D |Total |

|A |15 |25 |40 |

| |(12.5) |(27.5) | |

|B |10 |30 |40 |

| |(12.5) |(27.5) | |

|Total |25 |55 |80 |

[pic]

10. An AP Statistics students noted that the probability distribution for a binomial random variable with n = 4 and p = 0.3 is approximately given by:

|n |P |

|0 |0.24 |

|1 |0.41 |

|2 |0.27 |

|3 |0.08 |

|4 |0.01 |

The student decides to test the randBin function on her calculator by putting 500 values into a list from this function (randBin(4, 0.3, 500) ( L1) and counting the number of each outcome. She obtained

|n |Observed |

|0 |110 |

|1 |190 |

|2 |160 |

|3 |36 |

|4 |4 |

Do these data indicate that the “randBin” function on the calculator is failing to correctly generate the correct quantities of 0, 1, 2, and 3 from this distribution?

ANS:

Ho: p0 = 0.24, p1 = 0.41, p2 = 0.27, p3 = 0.07, p4 = 0.01

Ha: Not all of the proportions stated in Ho are correct.

|n |Expected |

|0 |0.24(500) = 120 |

|1 |0.41(500) = 205 |

|2 |0.27(500) = 135 |

|3 |0.08(500) = 40 |

|4 |0.01(500) = 5 |

[pic]

[On the calculator, χ2cdf(7.16, 1E99, 4) = 0.128

The P-value is greater than 0.05, which is too high to reject the null. We do not have strong evidence that the calculator is not performing as it should.

11. A chi-square test for the homogeneity of proportions is conducted on three populations and one categorical variable that has four values. Computation of the chi-square statistic yields χ2 = 17.2. Is this finding significant and the 0.01 level of significance?

ANS:

df = 4 – 1 = 3

χ2cdf(17.2, 1E99, 3) = 0.009

The finding is significant at the 0.01 level of significance because 0.009 < 0.01.

12. Which of the following best described the difference between a test for independence and a test for homogeneity of proportions?

(a) There is no difference because they both produce the same value of the chi-square test statistic.

(b) A test for independence has one population and two categorical variables, whereas a test for homogeneity of proportions has more than one population and only one categorical variable.

(c) A test for homogeneity of proportions has one population and two categorical variables; whereas, a test for independence has more than one population and only one categorical variable.

(d) A test for independence uses count data when calculating chi-square and a test for homogeneity uses percentages or proportions when calculating chi-square.

ANS: B

13. Restaurants in two parts of a major city were compared on customer satisfaction to see if location influences customer satisfaction. A random sample of 38 patrons from the Big Steak Restaurant in the eastern part of town and another random sample of 36 patrons from the Big Steak Restaurant on the western side of town were interviewed for the study. The restaurants are under the same management, and the researcher established that they are virtually identical in terms of décor and service. The results are presented in the following table.

Patron’s Ratings of Restaurants

| |Excellent |Good |Fair |Poor |

|Eastern |10 |12 |11 |5 |

|Western |6 |15 |7 |8 |

Do these data provide good evidence that location influences customer satisfaction?

ANS:

Ho: The proportions of patrons from each side of town that rate the restaurant Excellent, Good, Fair, and Poor are same for eastern and western.

Ha: Not all the proportions are the same.

| |Excellent |Good |Fair |Poor |Total |

|Eastern |10 |12 |11 |5 |38 |

| |(8.22) |(13.86) |(9.24) |(6.68) | |

|Western |6 |15 |7 |8 |36 |

| |(7.78) |(13.14) |(8.76) |(6.32) | |

|Total |16 |27 |18 |13 |74 |

[pic](from table)

[Use calculator, χ2cdf(2.86, 1E99, 3) = 0.41]

The P-value is large. We cannot reject the null. These data do not provide us with evidence that location influences customer satisfaction.

14. The number of defects from a manufacturing process by day of the week are as follows:

| |Monday |Tuesday |Wednesday |Thursday |Friday |

|Number |36 |23 |26 |25 |40 |

The manufacturing is concerned that the number of defects is greater on Monday and Friday. Test, at the 0.05 level of significance, the claim that the proportion of defects is the same each day of the week.

ANS:

Ho: p1 = p2 = p3 = p4 = p5 (the proportion of defects is the same each day of the week.)

Ha: At least one proportion is not equal to the others (the proportion of defects is not the same each day of the week.)

The number of expected defects is the same for each day of the week. Because there were a total of 150 defects during the week, we would expect, if the null is true, to have 30 defects each day.

| |Monday |Tuesday |Wednesday |Thursday |Friday |

|Observed |36 |23 |26 |25 |40 |

|Expected |30 |30 |30 |30 |30 |

[pic](from table.)

[Use calculator, χ2cdf(7.533, 1E99, 4) = 0.11]

The P value is too large to reject the null. We do not have strong evidence that there are more defects produced on Monday and Friday than on other days of the week.

15. A study was done on opinions concerning the legalization on marijuana at Mile High College. One hundred fifty-seven respondents were randomly selected from a large pool of faculty, students, and parents at the college. Respondents were given a choice of favoring the legalization of marijuana, opposing the legalization of marijuana, or favoring making marijuana a legal but controlled substance. The results of the survey were as follows.

| |Favor Organization |Oppose Legalization |Favor Legalization |

| | | |with Control |

|Students |17 |9 |6 |

|Faculty |33 |40 |27 |

|Parents |5 |8 |12 |

Do these data support, at the 0.05 level, the contention that they type of respondent (student, faculty, or parent) is related to the opinion toward legalization? Is this a test of independence or a test of homogeneity of proportions?

ANS:

Because we have a single proportion from which we drew our sample and we are asking if two variables are related within that population, this is a chi-square test for independence.

Ho: Type of respondent and opinion toward the legalization of marijuana are independent.

Ha: Type of respondent and opinion toward the legalization of marijuana are not independent.

| |Favor Organization |Oppose Legalization |Favor Legalization |Total |

| | | |with Control | |

|Students |17 |9 |6 |32 |

| |(11.21) |(11.62) |(9.17) | |

|Faculty |33 |40 |27 |100 |

| |(35.03) |(36.31) |(28.66) | |

|Parents |5 |8 |12 |25 |

| |(18.76) |(36.31) |(7.16) | |

|Total |55 |57 |45 |157 |

[pic](from table).

[Use calculator, χ2cdf(10.27, 1E99, 4) = 0.036]

Because P-value < 0.05, we reject Ho. We have evidence that the type of respondent is related to opinion concerning the legalization of marijuana.

16. In a recent year, at the 6 p.m. time slot, television channels 2, 3, 4, and 5 captured the entire audience with 30%, 25%, 20%, and 25%, respectively. During the first week of the next season, 500 viewers are interviewed.

a) If viewer preferences have not changed, what number of persons is expected to watch each channel?

b) Suppose that the actual observed numbers are as follows:

|2 |3 |4 |5 |

|139 |138 |112 |111 |

Do these numbers indicate a change? Are the differences significant?

ANS:

a) Expected:

|2 |3 |4 |5 |

|(0.30)(500) = 150 |(0.25)(500) = 125 |(0.20)(500) = 100 |(0.25)(500) = 125 |

b) Ho: The television audience is distributed over channels 2, 3, 4, and 5 with percentages 30%, 25%, 20%, and 25%, respectively.

Ha: The audience distribution is not 30%, 25%, 20%, and 25%, respectively.

[pic]

[pic]

Since 5.167 < 7.81, we do not have sufficient evidence to reject the null hypothesis.

17. A grocery store manager wishes to determine whether a certain product will sell equally well in any of five locations in the store. Five displays are set up, one in each location, and the resulting numbers of the product sold are noted.

|1 |2 |3 |4 |5 |

|43 |29 |52 |34 |48 |

Is there enough evidence that location makes a difference? Test at both the 5% and 10% significance levels.

ANS

Ho: Sales of the product are uniformly distributed over the five locations.

Ha: Sales are not uniformly distributed over the five locations.

A total of 43 + 29 + 52 + 34 + 48 = 206 units were sold. If location does not matter, we would expect 206/5 = 41.2 units sold per location (uniform distribution.)

Expected:

|1 |2 |3 |4 |5 |

|41.2 |41.2 |41.2 |41.2 |41.2 |

Thus,

[pic]

[pic]

Since 8.903 < 9.49, we do not have sufficient evidence to reject the null hypotheses at the 5% level. However, at the 10% level, 8.903 > 7.78, we do have sufficient evidence to reject the null hypothesis.

18. In a nationwide telephone poll of 1000 adults representing Democrats, Republicans, and Independent, respondents were asked if their confidence in the U.S. banking system had been shaken by the savings and loan crisis. The answers, cross-classified by party affiliation, are given in the following two-way table.

| |Yes |No |No Opinion |

|Democrats |175 |220 |55 |

|Republicans |150 |165 |35 |

|Independents |75 |105 |20 |

Test the null hypothesis that shaken confidence in the banking system is independent of party affiliation. Use a 10% significance level.

ANS:

Ho: Party affiliation and shaken confidence in the banking system are independent.

Ha: Party affiliation and shaken confidence in the banking system are not independent.

| |Yes |No |No Opinion |Total |

|Democrats |175 |220 |55 |450 |

| |(180) |(220.5) |(49.5) | |

|Republicans |150 |165 |35 |350 |

| |(140) |(171.5) |(38.5) | |

|Independents |75 |105 |20 |200 |

| |(80) |(98) |(22) | |

|Total |400 |490 |110 |1000 |

[pic]

Since 3.024 < 7.78, there is not sufficient evidence to reject the null hypothesis of independence. Thus, at the 10% significance level, we cannot claim a relationship between party affiliation and shaken confidence in the banking system.

19. To determine whether men with a combination of childhood abuse and a certain abnormal gene are more likely to commit violent crimes, a study is run on a simple random sample of 575 males in the 25 to 35 age group. The data are summarized in the following table:

| |Not abused, normal gene |Abused, normal gene |Not abused, abnormal gene |Abused, abnormal gene |

|Criminal behavior |48 |21 |32 |26 |

|Normal behavior |201 |79 |118 |50 |

Is there evidence of a relationship between the four categories (based on childhood abuse and abnormal genetics) and behavior (criminal versus normal)? Explain.

ANS:

| |Not abused, normal gene |Abused, normal gene |Not abused, abnormal gene |Abused, abnormal gene |Total |

|Criminal behavior |48 |21 |32 |26 |127 |

| |(55) |(22.1) |(33.1) |(16.8) | |

|Normal behavior |201 |79 |118 |50 |448 |

| |(194.0) |(77.9) |(116.9) |(59.2) | |

|Total |249 |100 |150 |76 |575 |

Ho: The four categories (based on childhood abuse and abnormal genetics) and behavior (criminal versus normal) are independent.

Ha: The four categories (based on childhood abuse and abnormal genetics) and behavior (criminal versus normal) are not independent.

[pic]

Since 7.752 < 7.81, the data do not provide some evidence to reject Ho and conclude that there is not evidence of a relationship between the four categories (based on childhood and abnormal genetics) and behavior (criminal versus normal).

20. In a large city, a group of AP Statistics students work together on a project to determine which group of school employees has the greatest proportion who are satisfied with their jobs. In independent simple random sample of 100 teachers, 60 administrators, 45 custodians, and 55 secretaries, the numbers satisfied with their jobs were found to be 82, 38, 34, and 36, respectively. Is there evidence that the proportion of employees satisfied with their jobs is different in different school system job categories?

ANS:

Ho: The proportion of employees satisfied with their jobs is the same across the various school system job categories.

Ha: At least two of the job categories differ in the proportion of employees satisfied with their jobs.

The observed and expected counts are as follows:

| |Satisfied |Not satisfied |Total |

|Teachers |82 |18 |100 |

| |(73.1) |(26.9) | |

|Administrators |28 |22 |60 |

| |(43.8) |(16.2) | |

|Custodians |34 |11 |45 |

| |(32.9) |(12.1) | |

|Secretaries |36 |19 |55 |

| |(40.2) |(14.8) | |

|Total |190 |70 |260 |

[pic]

Since 8.640 > 7.81, there is sufficient evidence to reject Ho and we can conclude that there is evidence that the proportion of employees satisfied with their jobs is not the same across all the school system job categories.

21. The following table provides the responses of a group of 100 children shown three different toys and asked which one they liked the best. Based on the data, is there evidence that one of the three shows a difference in the preference between the boys and girls at the 5% significance level?

| |Toy A |Toy B |Toy C |

|Boys |25 |24 |11 |

|Girls |9 |25 |6 |

ANS:

Ho. There is no difference in the preference for toys between boys and girls.

Ha: There is a difference in the preference for toys between boys and girls.

| |Toy A |Toy B |Toy C |Total |

|Boys |25 |24 |11 |60 |

| |(20.4) |(29.4) |(10.2) | |

|Girls |9 |25 |6 |40 |

| |(13.6) |(19.6) |(6.8) | |

|Total |34 |49 |17 |100 |

[pic]

Since 5.23 < 7.81, we do not have evidence against the null hypothesis; therefore, boys and girls do not differ in their choice of toys.

22. In a study of exercise habits in men working in the health care profession in Chicago, researchers classified the 356 sampled employees according to the level of education they completed and their exercise habits. The researchers want to ascertain if there is an association between the level of education completed and exercise habits at the 5% significance level. The data from the study are provided in the table below.

| |College |Some College |HS |

|Regularly |51 |22 |43 |

|Occasionally |92 |21 |28 |

|Never |68 |9 |22 |

ANS:

Ho: There is no association between educational level completed and exercise habits for men working in the health care profession in Chicago.

Ha: There is an association between educational level completed and exercise habits for men working in the health care profession in Chicago.

| |College |Some College |HS |Total |

|Regularly |51 |22 |43 |116 |

| |(68.75) |(16.94) |(30.30) | |

|Occasionally |92 |21 |28 |141 |

| |(83.57) |(20.60) |(36.83) | |

|Never |68 |9 |22 |99 |

| |(58.68) |(14.46) |(25.86) | |

|Total |211 |52 |93 |356 |

[pic]

Since 18.5097 > 9.49, there is evidence to reject the null hypothesis; in other words, there is evidence of an association between level of education and exercise habits among men working in the health care profession in Chicago.

The next set of questions refers to the following situation:

A survey was conducted to investigate the severity of rodent problems in egg and poultry operations. A random sample of operators was selected, and the operators were classified according to the type of operation and the extent of the rodent population. A total of 78 egg operators and 53 turkey operators were classified and the summary information is:

[pic]

23. The value of the test statistic is:

(a) about 5.99

(b) about 9.71

(c) about 6.81

(d) about 5.64

(e) about 8.60

Solution: d

24. The expected count in the (egg, mild infestation) cell is:

(a) about 26.00

(b) about 33.33

(c) about 53.00

(d) about 31.55

(e) about 78.00

Solution: d

25. The approximate p-value is found to be:

(a) about .060

(b) about .014

(c) about .032

(d) about .008

(e) about .05

Solution: a

The next set of questions refers to the following situation

In the paper “Color Association of Male and Female Fourth-Grade School Children” (J. Psych., 1988, 383-8), children were asked to indicate what emotion they associated with the color red. The response and the sex of the child are noted and summarized below. The first number in each cell is the count, the second number is the row percent.

Frequency|

Row Pct | anger | happy | love | pain | Total

-------------+---------+---------+-------+-------+

f | 27 | 19 | 39 | 17 | 102

| 26.47 | 18.63 | 38.24 | 16.67 |

-------------+---------+---------+-------+-------+

m | 34 | 12 | 38 | 28 | 112

| 30.36 | 10.71 | 33.93 | 25.00 |

-------------+---------+--------+--------+-------+

Total 61 31 77 45 214

26. Under a suitable null hypothesis, the expected frequency for the cell corresponding to Anger and Males is:

(a) 15.9

(b) 55.7

(c) 30.4

(d) 31.9

(e) 29.1

Solution: d

27. The null hypothesis will be rejected at α =0.05 if the test statistic exceeds:

(a) 3.84

(b) 5.99

(c) 7.81

(d) 9.49

(e) 14.07

Solution: c

28. The approximate p-value is:

(a) Between .100 and .900

(b) Between .050 and .100

(c) Between .025 and .050

(d) Between .010 and .025

(e) Between .005 and .010

Solution: a

29. A random sample of 100 members of a union are asked to respond to two questions: Question 1. Are you happy with your financial situation? Question 2. Do you approve of the Federal government’s economic policies?

The responses are:

Question 1.

Yes No | Total

Question Yes 22 48 | 70

2 No 12 18 | 30

Total 34 66 | 100

To test the null hypothesis that response to Question 1 is independent of response to Question 2 at 5% level, the expected frequency for the cell (Yes, Yes) and the critical value of the associated test statistic are:

(a) 23.8 and 1.96 respectively

(b) 10.2 and 3.84 respectively

(c) 23.8 and 3.84 respectively

(d) 23.8 and 7.81 respectively

(e) 10.2 and 7.81 respectively

Solution: c

30. A survey was conducted to investigate whether alcohol consumption and smoking are related. The following information was compiled for 600 individuals:

Smoker Non-smoker

Drinker 193 165

Non-drinker 89 153

Which of the following statements is true?

(a) The appropriate alternative hypothesis is A: Smoking and Alcohol Consumption are independent.

(b) The appropriate null hypothesis is H: Smoking and Alcohol Consumption are not independent.

(c) The calculated value of the test statistic is 3.84.

(d) The calculated value of the test statistic is 7.86.

(e) At level .01 we conclude that smoking and alcohol consumption are related.

Solution: e

31. Doctors’ practices have been categorized as to being Urban, Rural, or Intermediate. The number of doctors who prescribed tetracycline to at least one patient under the age of 8 were recorded for each of these practice areas. The results are:

Urban Intermediate Rural

Tetracycline 95 74 31

No tetracycline 126 84 30

If the county type of practice and the use of tetracycline are independent, then the expected number of rural doctors who prescribe tetracycline is:

(a) 31.0

(b) 27.7

(c) 1.37

(d) 51

(e) 62

Solution: b

32. For the problem outlined above, the critical value (table E) of the test statistic when the level of significance is α =0.05, is:

(a) 0.1026

(b) 7.3778

(c) 5.9915

(d) 12.5916

(e) 7.8147

Solution: c

The next set of questions refers to the following situation:

A study was conducted to determine if the fatality rate depends on the size of the automobile. The analysis of accidents is as follows (with some values hidden):

DEATH SIZE

FREQUENCY | m | s | L | TOTAL

-------------------+-----------+-------+-------+

no | 63 | 128 | 46 | 237

-------------------+-----------+--------+--------+

yes | 26 | 95 | 16 | 137

-------------------+-----------+-------+--------+

TOTAL 89 223 62 374

33. Under a suitable null hypothesis, the expected frequency for the cell corresponding to fatal type of accident and small size automobile is:

(a) 81.68

(b) 67.00

(c) 61.43

(d) 63.41

(e) 59.72

Solution: a

34. The null hypothesis will be rejected at α =0.05 if the test statistic exceeds:

(a) 12.59

(b) 7.81

(c) 5.99

(d) 3.84

(e) 9.49

Solution: c

35. The approximate p-value is:

(a) less than 0.005

(b) between 0.005 and 0.010

(c) between 0.010 and 0.025

(d) between 0.025 and 0.050

(e) between 0.050 and 0.100

Solution: c

-----------------------

Temperature

Temperature

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download