exam iii tian xia

PSYC2311/MATH1380 – Elementary StatisticsExam 3 – April 14th, 2016 – Chapters 8 - 11

Name: Tian XiaDate: 04/14/2016

Instructions:Answer each question completely. For short answer and essay questions, use complete sentences and standard paragraph form. Incomplete sentences and sentence fragments will be ignored. While it is not necessary for students to reference course material or external research for essay questions, students should support their responses with conclusions from course material and external research. In other words, justify your responses.

Students must show their work for arithmetic problems.

Students are bound by the Baptist University of the Americas Honor Code. Any violations will be dealt as described in the Student Handbook. The professor will assign zeros to students' work involved in cheating and the lowering of the overall grade of the course. Individual grades for cheating are not dropped from the overall average. Furthermore, cheating includes making known to other students in other sections the nature of the exam. Also, cheating includes sharing any exam material and/or responses with others. Under no circumstances are students to communicate with other students, tutors, faculty, or other individuals regarding questions and answers on the exam.

Let the peace of Christ rule in your hearts, since as members of one body you were called to peace. And be thankful. Let the word of Christ dwell in you richly as you teach and admonish one another with all wisdom, and as you sing psalms, hymns and spiritual songs with gratitude in your hearts to God. And whatever you do, whether in word or deed, do it all in the name of the Lord Jesus, giving thanks to God the Father through him. – Colossians 3:15-17

Hypothesis TestingExplain the difference between a Type I and Type II error in hypothesis testing. Provide an example of a null hypothesis, and explain using common language, of committing a Type I error (5 points).

Type I Error: Rejection of the null hypothesis when it is true.Type II Error: Failure to reject a null hypothesis that is false.

Example. H0: The drug has no effect on the patients.

If the significance level is .1, which is a high α level, and you testing value may fall into rejection region. As a result, you will reject the null hypothesis by saying that the drug has effect on the patients when the null hypothesis that the drug has no effect on the patients is actually true.

It is that adopting a .10 α level leaves some room for mistaking a chance difference for a real difference. By lowering the α value, reduces the probability of this kind of mistake but may fall into Type II error.

List the steps used in hypothesis testing. Taking these steps into consideration, why can’t the hypothesis be proved true? (5 points).

1. Recognize two possibilities for the population mean: The null hypothesis, H0


The alternative hypothesis H1

2. Assume for the time being that H0 is true.3. Use a sampling distribution that shows the differences between sample means and the null

hypothesis mean when H0 is true.4. Gather sample data from the population. Calculate a mean, x̄15. Calculate the difference between the sample mean and the null hypothesis mean.6. Using the sampling distribution, determine the probability of the difference that was actually

observed (or one more extreme). 7. If the probability is small, reject the null hypothesis; if the probability is large, conclude that the

data are consistent with the null hypothesis.

As agree to use a sample, some uncertainty/error about the results is introduced.

In what ways is the t distribution similar to the standard normal distribution? In what ways is the t distribution difference from the standard normal distribution? (5 points).

(1) It is symmetrical, bell-shaped, and similar to the standard normal curve.(2) It differs from the standard normal curve, in that it has an additional parameter, called degrees

of freedom, which changes its shape.

The average undergraduate cost for tuition, fees, room and board for all institutions last year was $19,410. A random sample of costs this year for 40 institutions of higher learning indicated that the average cost was $22,098, with a standard deviation of $6,050. At a 99% confidence level, is there sufficient evidence to conclude that the cost of attendance for undergraduate education has increased? (source: New York Times Almanac, 2007). State the null hypothesis, the alternative hypothesis, the critical value and statistic used. You do not need to show the specific calculations, but you must show the values calculated to make comparisons and arrive at a conclusion. (15 points).

H0: The cost of attendance for undergraduate education has not increased, => H0=19,410H1: The cost has increasedBecause the standard distribution of the population is unknown, we use t statistics.Assume H0 is true, t= (x̄-µx̄)/ Sx̄=(x̄-µx̄)/(S/√n)= (22098-19410)/(6050/√40)=2.81df=n-1=39, α=0.01, => the critical value= 2.426<2.81


So we should reject the Null Hypothesis, which means there is sufficient evidence to conclude that the cost of attendance for undergraduate education has increased.

A large university in Mexico City reports that the average salary of parents of an entering class is $91,600 (converted from Pesos). To see how this compares to Baptist University of the Americas, president Maciel 28 randomly selected families and finds there average income is $88,500 (converted from local currencies). If the standard deviation for salaries is $10,000, can president Maciel conclude there is a difference? At a confidence level of 90%, is he correct? (15 points).H0: There is no a difference => H0=91600H1: There is a difference

The sample size is less than 30 and the standard deviation of the population is unknown, so we use t-distribution.t=(x̄-µ0)/ Sx̄=(x̄-µ0)/( S/√n)= (88500-91600)/ (10000/√28)=-1.64df=n-1=27, α=.10, so the critical value= -1.314>-1.64

With a confidence level of 90%, he is correct, there is a difference. We should reject null hypothesis.

Analysis of Difference between Means for Large SampleA survey of 1000 students nationwide showed an average ACT score of 21.4, A survey of 500 students in Texas revealed an average ACT score of 20.8. If the standard deviation for the 1000 students across the nation and the 500 students in Texas is 3, can we conclude that the ACT scores in Texas are below the national average? Use a confidence level of 95% to determine your conclusion. (10 points).

The sample size is greater than 30 (1000 & 500). So we can use z statistics.

H0: the ACT scores in Texas are not below the national average, => µ1=µ2

H1: The act scores in Texas are below the national average.

Assume H0 is true,

Z= [(x̄1- x̄2) - (µ1- µ2)]/(⍴12/N1+⍴2

2/N2), and since µ1=µ2

Z= (x̄1- x̄2)/(⍴12/N1+⍴2

2/N2) = (21.4-20.8)/ (9/1000+9/500) =0.6/0.164316767=3.651483717


With a confidence level of 95%, the critical value is 1.65.

So the value of the z statistic is outside the critical region which means the probability of the average act scores for the school in Texas being the same as the average scores for the country is less than 5%. Hence, we reject the null hypothesis. We conclude that the ACT scores in Texas are below the national average.

Analysis of Variance – One Way ANOVAWhat are the assumptions for ANOVA? (5 points).-Each group sample is drawn from a normally distributed population-All populations have a common variance-All samples are drawn independently of each other-Within each sample, the observations are sampled randomly and independently of each other-Factor effects are additive

List the specific steps to complete an one-way ANOVA, including how to find the critical value, calculate the F statistic, and draw a conclusion. (5 points).

1. Calculate the means2. Calculate over all mean.3. Calculate SST and the its degree of freedom4. Calculate SSW and the its degree of freedom5. Calculate SSB (=SST-SSW) and the its degree of freedom6. Recognize two possibilities for the population mean: The null hypothesis, H0

The alternative hypothesis H1

7. Assume the null hypothesis is true: the population means are the same for all groups.8. Calculate F statistics value with the formula: F= (SSB/dfssb)/(SSW/dfssw)9. Go to F table and find the critical value based on the dfssb and dfssw.10. Check if the calculated F value falls into rejection area or not and make conclusion to

retain or reject the null hypothesis.


The per student cost (in thousands of dollars, converted from Pesos) for cyber charter school tuition for school districts in three areas of Mexico are shown below. At a confidence level of 95%, is there a difference in the means? (20 points).

Area I Area II Area III 6.2 7.5 5.89.3 8.2 6.46.8 8.5 5.66.1 8.2 7.16.7 7.0 3.06.9 9.3 3.5

x̄ 7.0 8.12 5.23

Xgrand = (6.2+9.3+6.8+6.1+6.7+6.9+7.5+8.2+8.5+8.2+7.0+9.3+5.8+6.4+5.6+7.1+3.0+3.5)/18=6.78

SST= (6.2-6.78)2+(9.3-6.78)2+(6.8-6.78)2+(6.1-6.78)2+(6.7-6.78)2+(6.9-6.78)2+(7.5-6.78)2+(8.2-6.78)2+(8.5-6.78)2+(8.2-6.78)2+(7.0-6.78)2+(9.3-6.78) 2+ (5.8-6.78) 2+(6.4-6.78) 2+(5.6-6.78) 2+(7.1-6.78)2+(3.0-6.78)2+(3.5-6.78)2=48.73

dfsst=18-1=17

SSW= (6.2-7)2+(9.3-7)2+(6.8-7)2+(6.1-7)2+(6.7-7)2+(6.9-7)2+(7.5-8.12)2+(8.2-8.12)2+(8.5-8.12)2+(8.2-8.12)2+(7.0-8.12)2+(9.3-8.12) 2+ (5.8-5.23) 2+(6.4-5.23) 2+(5.6-5.23) 2+(7.1-5.23)2+(3.0-5.23)2+(3.5-5.23)2

=23.36dfssw=(m-1)*n=5*3=15

SSB=SST-SSW=48.73-23.36=25.37dfssb=3-1=2

H0: there is no difference between the means => µI= µII=µIII,

H1: There is difference between the means.

Assume H0 is true, F statstic=(SSB/dfssb)/(SSW/dfssw) = (25.37/2)/(23.36/15)=12.685/1.56=8.145

With a confidence level of 95%, the critical value is 3.6823.


The score is outside the critical region, so we reject the null hypothesis. There is a differences between the means.

Three random samples of times (in minutes) that commuters are stuck in traffic are shown below. Is there a difference in the average times among the three cities? Use a 95% confidence level to draw a conclusion.

San Antonio Monterrey Santiago 59 54 5362 52 5658 55 5463 58 4961 53 52

____________________________________Mean 60.6 54.4 52.8

Xgrand =55.93

SST= 234.9333333, df=m*n-1=14SSW=65.2, df=(m-1) * n=12SSB=SST-SSW=234.9333333-65.2=169.73, df=2

H0: there is no difference in the average times among the three cities, => µI= µII=µIII,

H1: There is

Assume H0 is true.

F-statistic=(SSB/dfssb) / (SSW/dfssw) = (169.73/2)/(65.2/12)=84.87/5.43=15.619The critical value is 3.8853


So we reject the null hypothesis. There is a difference in the average times among the three cities

What factor might have influenced the results of the study? (5 points).

Effect size, variability, and sample size.

Analysis of Variance – Two-Way ANOVAHow does the two-way ANOVA differ from the one-way ANOVA? (5 points).One- way ANOVA has one independent variable in the experiment, two-way ANOWA has two factors/ two independent variables.

Explain what is meant by the main effect and interaction effect? (5 points).

Main effect: Significance test of the deviations of the mean levels of one independent variable from the grand mean.

Interaction effect: When the effect of one independent variable on the dependent variable depends on the level of another independent variable.

In a two-way ANOVA, variable A has three levels (instances) and variable B has two levels (instances). There are five data values in each cell. Find each degree-of-freedom value: (10 points)

a. Degree of freedom between for variable A： 2b. Degree of freedom between for variable B： 1c. Degree of freedom between for factor A x B ：2d. Degree of freedom within：24

For the following exercise, perform the following steps: state the hypothesis, find the critical value for each F test, complete the summary table and find the test value, make the decision, and summarize the


results. Assume that all variables are normally distributed, that the samples are independent, and that the population variances are equal.

Two special training program in outdoor survival are available for international missionaries. One lasts one week and the other lasts two weeks. The International Mission Board wishes to test the effectiveness of the programs and to see if there are any gender differences. Six missionaries are randomly selected and assigned to each of the programs according to gender. After completing the program, each is given a written test on his or her knowledge of survival skills. The test consists of 100 questions. The scores of the groups are shown below. Use a confidence level of 90% and draw a conclusion, using a two-way ANOVA. (30 points).

DurationGender One Week Two weeksFemale 86, 92, 87, 88, 78, 95 78, 62, 56, 54, 65, 63Male 52, 67, 53, 42, 68, 71 85, 94, 82, 84,78, 91

H0: Neither gender nor duration has no effect on the score of the result of the written test on missionaries’ knowledge of survival skills.H1: 1. Gender has effect

2. Duration has effect3. The interaction of gender and duration has effect

Sum of Squares d.f. Mean Square F Score

Sum of Square of 1st factor (Gender)

56.9184 1 56.9184/1=56.9184 56.9184/68.27501=0.833663737

Sum of Square of 2nd factor (Duration)

7.0635 1 7.0635/1=7.0635 7.0635/68.27501=0.103456594

Sum of Square within(Error)

1365.5002 20 1365.5002/20=68.27501

Sum of Square of both factors

3978.4763 1 3978.4763/1=3978.4763 3978.4763/68.27501=58.27133969

Sum of Square Total

5407.9584 23

F Distribution F (1,20) = 0.833663737; p<.10 F Distribution F (1,20) = 0.103456594; p<.10 F Distribution F (1,20) =58.27133969; p<.10 The critical value is 2.97


So1. Gender has no effect on the score of the result of the written test on missionaries’ knowledge of survival skills.2. Duration has no effect on the score of the result of the written test on missionaries’ knowledge of survival skills.3. The interaction of gender and duration has significant effect on the score of the result of the written test on missionaries’ knowledge of survival skills.

(All the work showed as below)

Male

One week Two weeks52 8567 9453 8242 8468 7871 91

Mean Table One Week Two Weeks AverageMale 58.83 85.67 72.25Female 87.67 63 75.33Average 73.25 74.335 73.79

d.f. gender= Number of Raws-1=1; d.f. duration=Number of Columns-1=1; d.f. within= (n-1)*m=(6-1)*4=20; d.f. within=d.f.gender*d.f.duration=1*1=1;d.f.total =1+1+20+1=23

FemaleOne week Two weeks86 7892 6287 5688 5478 6595 63


BONUS QUESTION:A researcher conducted a study of two different diets and two different exercise programs. The randomly selected students were assigned to each group for one month. The values below indicate the amount of weight each student lostDiet/Exercise Program A BI 5, 6, 4 8, 10, 15II 3, 4, 8 12, 16, 11

What procedure should be used to draw a conclusion? (2 points)Two-way ANOVA

What are the independent variables? (2 points)Diets and exercise program

What are the dependent variables? (2 points)The amount of weight each student lost

How many levels does each variable contain? (2 points)Two

What the hypotheses for the study? (8 points)Null hypotheses H0: Neither the diets nor the exercise programs has effect on the amount of weight each student lost.Alternative hypothesis H1:

1. Diets has effect2. Exercise program has effect3. The diet and exercise program interaction has effect

What are the F values for the hypothesis? (15 points)

exam iii tian xia

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