Exam III Tian Xia

PSYC2311/MATH1380 – Elementary Statistics
Exam 3 – April 14th, 2016 – Chapters 8 - 11
Name: Tian Xia
Date: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 materialor 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 showtheir 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 Testing
Explain 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, x̄ 1
5. 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=19,410
H1: The cost has increased
Because 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)=2.81
df=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=91600
H1: 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)=-1.64
df=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 ofDifference between Means for Large Sample
A 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= [(x̄ 1- x̄ 2) - (µ1- µ2)]/(⍴1
2
/N1+⍴2
2
/N2),and since µ1=µ2
Z= (x̄ 1- x̄ 2)/(⍴1
2
/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 ofVariance – One Way ANOVA
What 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 means
2. Calculate over all mean.
3. Calculate SST and the its degree of freedom
4. Calculate SSW and the its degree of freedom
5. Calculate SSB (=SST-SSW) and the its degree of freedom
6. 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 5.8
9.3 8.2 6.4
6.8 8.5 5.6
6.1 8.2 7.1
6.7 7.0 3.0
6.9 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
=23.36
dfssw=(m-1)*n=5*3=15
SSB=SST-SSW=48.73-23.36=25.37
dfssb=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 53
62 52 56
58 55 54
63 58 49
61 53 52
____________________________________
Mean 60.6 54.4 52.8
Xgrand =55.93
SST= 234.9333333, df=m*n-1=14
SSW=65.2, df=(m-1) * n=12
SSB=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=15.619
The 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 ofVariance – Two-Way ANOVA
How 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： 2
b. Degree of freedom between for variable B： 1
c. Degree of freedom between for factor A x B ：2
d. 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).
Duration
Gender One Week Two weeks
Female 86, 92, 87, 88, 78, 95 78, 62, 56, 54, 65, 63
Male 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 effect
3. 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

So
1. 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 weeks
52 85
67 94
53 82
42 84
68 78
71 91
Mean Table
One Week Two Weeks Average
Male 58.83 85.67 72.25
Female 87.67 63 75.33
Average 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
Female
One week Two weeks
86 78
92 62
87 56
88 54
78 65
95 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 lost
Diet/Exercise Program A B
I 5, 6, 4 8, 10, 15
II 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 effect
2. Exercise program has effect
3. 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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