Exam 1
Exam Page 1
Define each of the following:
a) Element.
An element is described as "the individual and unique entry in a data set about which data has been collected, analyzed and presented in a same manner
...
Exam 1
Exam Page 1
Define each of the following:
a) Element.
An element is described as "the individual and unique entry in a data set about which data has been collected, analyzed and presented in a same manner to differentiate" (Module 1).
b) Variable.
A variable is defined as a "particular measurable attribute that the researcher believes is needed to describe the element in their study" (Module 1).
c) Data.
Data (or the plural of datumn) is defined as things (such as numerical information, people, geographical areas,etc.) about which information can be collected and then analyzed.
Answer Key
Exam Page 2
Explain the difference between population and sample.
"The entire number of items in a large group" would be defined as the population. (Module 1) The sample is then taken from the population by a researcher and is studied.The sample taken from the population is, in fact, the subset of the population. You need the population to get the sample and without the population, there can be no sample.
Instructor Comments
Very good definitions.
Answer Key
Exam Page 3
Look at the following data and see if you can identify any outliers: 65 71 55 69 3 77 67 70 246 61 277
3, 246, 277
Instructor Comments
Very good.
Answer Key
Exam Page 4
The following pie chart shows the percentages of total items sold in a month in a certain fast food restaurant.
A total of 4900 fast food items were sold during the month. How many were burgers?
How many were french fries?
4900(.32)=1568
32% or 1,568 burgers were sold during the month.
4900(.18)=882
18% or 882 french fries were sold during the month.
Instructor Comments
Very good.
Answer Key
Exam 2
Exam Page 1
During an hour at a fast food restaurant, the following types of sandwiches are ordered:
Turkey Hamburger Cheeseburger Fish Hamburger Turkey Fish Chicken Fish Chicken Turkey Fish Hamburger Fish
Cheeseburger Fish Cheeseburger Hamburger Fish Fish Cheeseburger Hamburger Fish Turkey Turkey Chicken Fish Chicken Cheeseburger Fish Turkey Fish Fish Hamburger Fish Fish Turkey Chicken Hamburger Fish Cheeseburger Chicken Chicken Turkey Fish Chicken
Hamburger Chicken Fish Chicken
a) Make a frequency distribution for this data.
Types of Frequency Sandwiches Turkey 8
Chicken 10
Cheeseburger 6
Fish 18
Hamburger 8
Total 50
b) Make a relative frequency distribution for this data. Include relative percentages on this table.
Types of Frequency Relative Relative Sandwiches Frequency Percentage
Turkey 8 (8/50)= .16 (.16)100= 16%
Chicken 10 (10/50)= .20 (.20)100= 20%
Cheeseburger 12%
Fish
6
18 (6/50)= (18/50)=
.12
.36
(.12)100=
(.36)100=
36%
Hamburger 8 (8/50)= .16 (.16)100= 16%
Total 50 1 100%
Exam Page 2
Consider the following data:
422 389 414 401 466 421 399 387 450 407 392 410
440 417 490
Find the 20th percentile of this data. 387,389,392,399,401,407,410,414,417,421,422,440,450,466,490
i=(p n = (20) *15= 3
100 ) 100
i=3
392 is the 20th percentile of this data.
Exam Page 3
Consider the following data:
{29, 20, 24, 18, 32, 21}
a) Find the sample mean of this data.
x* = ∑xi
n
x*=(29+20+24+18+32+21) =144=24
6 6
b) Find the range of this data.
{18,20,21,24,29,32}
Range is 14
(32-18)=14
c) Find the sample standard deviation of this data.
s2=∑(xi-x)2 = (18-24)2 + (20-24)2+(21-24)2+(24-24)2+(29-24)2+(32-24)2 = 36+16+9+0+25+64=
150=30
n-1 6-1 5 5
s=√s2 = √30 = 5.477
d) Find the coefficient of variation. cov=standard deviation*100=5.477*100 =22.82
mean 24
Exam Page 4
Suppose that you have a set of data that has a mean of 65 and a standard deviation of 10.
a) Is the point 75 above, below, or the same as the mean. How many standard deviations is 75 from the mean.
x*65
z=x-u= 75-65=1
o 10
z=1
The point 75 is above the mean (because it is a positive number), meaning that the data point is one standard deviation above the mean.
b) Is the point 85 above, below, or the same as the mean. How many standard deviations is 85 from the mean.
x*65
z=x-u= 85-65=2
o 10
z=2
The point 85 is above the mean (because it is a positive number), meaning that the data point is two standard deviations above the mean.
c) Is the point 57.5 above, below, or the same as the mean. How many standard deviations is 57.5 from the mean.
x*65
z=x-u= 57.5-65=-0.75
o 10
z=-0.75
The point 57.5 is below the mean (because it is a negative number), meaning that the data point is .75 standard deviations below the mean.
d) Is the point 107 above, below, or the same as the mean. How many standard deviations is 107 from the mean.
x*65
z=x-u= 107-65=4.2
o 10
z=4.2
The point 107 is above the mean (because it is a positive number), meaning that the data point is
4.2 standard deviations above the mean.
Exam Page 5
Consider the following set of data:
{22, 14, 35, 49, 8, 18, 30, 44}
a) Find the median.
{8,14,18,22,30,35,44,49}
Median=22+30=26
26
b) Find the mode of this set.
{8,14,18,22,30,35,44,49}
No mode (no number appears more than once).
Exam 3
Exam Page 1
Suppose A and B are two events with probabilities:
P(A)=.35,P(Bc )=.45,P(A∩B)=.25.
Find the following:
a) P(A∪B).
P(AUB)=P(A)+P(B)-P(AnB)
P(Bc)=> P(B)=1-P(Bc)=1-.45=.55 P(AUB)=.35+.55-.25=0.65
b) P(Ac ).
P(Ac)=> P(A)=1-P(Ac)=>
P(Ac)=1-P(A)= 1-.35=.65
c) P(B).
P(Bc)=> P(B)=1-P(Bc)
P(B)=1-.45=.55
Exam Page 2
Suppose you are going to make a password that consists of 4 characters chosen from
{2,7,8,c,f,k,t,z}. How many different passwords can you make if you cannot use any character more than once in each password?
n=8 (only 8 characters to choose from) r=4 (four character passwords
P(n,r)= n!=
(n-r)!
P(8,4)= 8!= 8! =8(7)(6)(5)=1680 different passwords
8-4! 4!
Exam Page 3
Find the answer to each of the following by first reducing the fractions as much as possible: a) P(490,2)=
P(n,r)=n!
(n-r)!
P(490,2)= 490! = 490!= 490(498)=244,020
(490-2)! 488!
b) C(670,665)=
C(n,r)= n!
r!(n-r)!
C(670,665)= 670! = 670! = 670(669)(668)(667)(666)=1,108,399,190,634 665!(670-665)! 665!5! 5(4)(3)(2)(1)
-1.0 points
Instructor Comments
In (a) you multiplied by 498 instead of 489.
Answer Key
Exam Page 4
Suppose A and B are two events with probabilities:
P(Ac )=.20,P(B)=.30,P(A∩B)=.20.
a) What is (A│B) ?
P(AlB)= P(AnB) = .20 =.666
P(B) .30
b) What is (B│A) ?
P(Ac)= P(A)=1-P(Ac)=1-.20=.80 P(BlA)= P(AnB) = .20 =.25
P(A) .80
Exam Page 5
In a manufacturing plant, three machines A, B, and C produce 45 %, 35 %, and 20 %, respectively, of the total parts production. The company's quality control department determined that 1.5 % of the parts produced by machine A, 2 % of the parts produced by machine B, and 1 % of the parts produced by machine C are defective. If a part is selected at random and found to be defective, what is the probability that it was produced by machine A?
Machine A=P(A) Machine B=P(B) Machine C=P(C)
Def=P(Def)
P(A)=.45, P(B)=.35, P(C)=.20
P(DeflA)=.015, P(DeflB)=.02. P(DeflC)=.01
P(AlDef)= P(A)*P(DeflA) = .45*.015
=0.4286
(P(A)*P(DeflA))+(P(B)*P(DeflB))+(P(C)*P(DeflC)) .45*.015+.35*.02+.20*.01
Exam Page 6
The probability that a certain type of battery in a smoke alarm will last 4 years or more is .65. The probability that a battery will last 7 years or more is .10. Suppose that the battery is 4 years old and is still working, what is the probability that the battery will last at least 7 years?
P(E)=.65
P(F)=.10
(smoke alarm lasting more than 4 years) = E (smoke alarm lasting more than 7 years) = F
If the smoke alarm has lasted 7 years, it would have lasted for 4 years already, so P(EnF)=.10 P(EnF)=.10
P(FlE)=P(EnF)=.10=.1538 P(E) .65
Exam Page 7
Suppose that 5 out of 11 people are to be chosen to go on a mission trip. In how many ways can these 5 be chosen if the order in which they are chosen is not important.
n=11 r=5
C(n,r)= n!
r!(n-r)!
C(11,5)= 11! = 11(10)(9)(8)(7)(6)=2772
5!6! 5(4)(3)(2)(1)
-1.0 points
Instructor Comments
You didn't need the 6 on top. It cancelled as part of the 6! on the bottom.
Answer Key
Exam 4
Exam Page 1
A factory has eight safety systems. During an emergency, the probability of any one of the safety systems failing is .08. What is the probability that six or more safety systems will fail during an emergency?
For six failures:
n=8, x=6, p=.08
f(x)= n! px(1-p)(n-x) = x!(n-x)!
f(x)= 8! .086(1-.08)(8-6) =.000006209
6!(8-6)!
For seven failures:
n=8, x=7, p=.08
f(x)= n! px(1-p)(n-x) = x!(n-x)!
f(x)= 8! .087(1-.08)(7-6) =.000000155
7!(8-7)!
For eight failures:
n=8, x=8, p=.08
f(x)= n! px(1-p)(n-x) = x!(n-x)!
f(x)= 8! .088(1-.08)(8-8) =.000000002
8!(8-8)!
The probability is:
.000006209+.000000155+.000000002=.000006366
Instructor Comments
The numbers are so small that with rounding your are fine.
Answer Key
Exam Page 2
Find each of the following probabilities:
a. Find P(Z ≤ 1.27) .
P(Z<1.27)=.89796
b. Find P(Z ≥ -.73) .
P(Z>-.73)=1-P(Z<-.73)=1-.23270=.7673
c. Find P(-.09 ≤ Z ≤ .86).
P(-.09-.57)=1-P(Z<-.57)=1-.28434=.71566
c) Between 3.4 inches and 4 inches?
u= 3.7, o=.35, x=3.4 x=4
z=x-u= 3.4-3.7=-.86 z=x-u= 4-3.7=.86
o .35 o .35
P(-.865,000 hits.
Type I Error: Reject the null hypothesis that the mean number of website hits is 5000 per month when the mean number of website hits is not greater than 5000 per month.
Type II Error: Do not reject the null hypothesis that the mean number of website hits is 5000 per month when the mean number of website hits is greater than 5000 per month.
Exam Page 2
Suppose that we have a problem for which the null and alternative hypothesis are given by: H0: μ=959.
H1:μ <959.
Is this a right-tailed test, left-tailed test, or two-tailed test. Find the z value based on a level of significance of .13.
This is a left tailed test. H0:u=959.
H1:u<959.
Left tailed test: P(Z.05.
Right tailed test; P(Z>z)=a P(Z>z)=.03 z=1.88
a=.03, p=.05, n=130 x-=11
z-score:
u=np u=np=130(.05)=6.5
os=√np(1-p)
os=√130(.05)(1-.05)=2.48495
z score=x--u= 11-6.5 =1.8109
os 2.48495
We do not reject the null hypothesis (z score is less than 1.88).
Exam 8
Exam Page 1
Suppose we have independent random samples of size n1 = 570 and n2 = 675. The proportions of success in the two samples are p1= .41 and p2 = .53. Find the 99% confidence interval for the difference in the two population proportions.
n1 = 570, n2 = 675, p1= .41, p2 = .53, z=2.58 P1-P2±z√p1(1-p1) + p2(1-p2)
n1 n2
.41-.53±2.58√.41(1-.41) + .53(1-.53)
570 675
-.12±2.58(.0281678)
-.12± .072673
The confidence interval is (-.047327,-.192673).
Instructor Comments
You have your confidence interval written backwards.
Answer Key
So, the interval is (-.1927,-.0473).
Exam Page 2
In certain hospital, nurses are required to constantly make rounds to check in on all of the patients. The nursing supervisor would like to know if there is a difference between the number of rounds completed per shift by the nurses on the day shift compared to the nurses on the night shift. So, the nursing supervisor checks the records of 77 day shift nurses and finds that they complete an average (a mean) of 32 rounds per shift with a standard deviation of 4.1 rounds per shift. The nursing supervisor also checks the records of 72 night shift nurses and finds that they complete an average (a mean) of 26 rounds per shift with a standard deviation of 6.3 rounds per shift.
a) Find the 95% confidence interval for estimating the difference in the population means (µ1 - µ2).
b) Can you be 95% confident that there is a difference in the means of the two populations? a) z=1.96
n1=77, n2=72, x- 32, x- =26 s1=4.1 s2=6.3
(x- -x- )-z√s 2 +s 2+z)=a/2
P(Zz)=a/2=.05/2=.025
z=-1.96 and z=1.96
Z-Score:
z=(x- -x- )-0
√s12 + s 2
√ n1 n2 z=(113-110)-0
√8.12 + 10.12
√ 42 42 z=3/1.9978 z=1.5017.
The z-score is between -1.96 and 1.96 so we do not reject the null hypothesis. There is no difference in the mean number of books re-shelved by full time and part time workers.
Exam Page 4
Consider the following dependent random samples
Observations 1 2 3 4 5 6
x-values 155 140 142 159 162 169
y-values 169 150 151 145 173 180
a) Determine the difference between each set of points, xi - yi
b) Do hypothesis testing to see if µd < 0 at the α = .05. a)
d-=d1+d2...dn=-14-10-9+14-11-11=-6.833
n 6
b)
H0:ud=0 H1:ud<0
Left Tailed test (do not forget to switch t to a negative)
n=6, DOF=n-1=6-1=5 t.05=2.571-->-2.571, d-=-6.833, sd=10.34
sd =√∑(d1-d-)2=10.34
√ n-1 t= d- (sd/√n)
t= -6.833 =-6.833/4.22=-1.619 (10.34/√6)
We do not reject that null hypothesis due to the fact that t is not less than -t.05 (because
-1.619>-2.571).
Exam Page 5
A new energy drink is supposed to improve a person’s time in the one mile run. The times, in seconds, of eight runners with and without the drink are given below:
Runner 1 2 3 4 5 6 7 8
x-time (before) 260 273 266 265 261 247 268 271
y-time (after) 263 270 280 279 266 242 263 259
Find the 98 % confidence interval for mean of the differences, µd.
DOF=n-1
DOF=8-1=7 df=7 so t=2.998 d-=-1.375 sd=9.365
d-=d1+d2...dn=-3+3-14-14-5+5+5+12 =-1.375
n 8
-5.0 points
Instructor Comments
You did not complete the problem.
Answer Key
Runner 1 2 3 4 5 6 7 8
x-time (before) 260 273 266 265 261 247 268 271
y-time (after) 263 270 280 279 266 242 263 259
Exam 9
Exam Page 1
Suppose you have 14 data points and you calculate the sample correlation coefficient and find that r = - .58. Can you be 95% confident that a linear relation exists between the variables? If so, is the relation positive or negative? Justify you answer.
n=14, r=-.58, 95% confident? CVCC=.53241
We can be 95% confident that a negative linear relation exists (because the original r is negative) between the variables due to the fact that the absolute of r is │r│=.58 and is greater than .53241 (which we can find on the critical values chart).
Exam Page 2
Suppose you have 40 data points and you calculate the sample correlation coefficient and find that r = - .30. Can you be 95% confident that a linear relation exists between the variables? If so, is the relation positive or negative? Justify you answer.
n=40, r=-.30, 95% confident? CVCC=.31201
We can be 95% confident that a no relation exists between the variables due to the fact that the absolute of r is │r│=.30 and is less than .31201 (which we can find on the critical values chart).
Exam Page 3
Compute the sample correlation coefficient for the following data:
Can you be 95% confident that a linear relation exists between the variables? If so, is the relation positive or negative? Justify you answer.
n=5
x-=2+3+7+9+12=6.6
5
y-=3+6+10+14+17=10
5
s 2=∑(x-x-)2= (2-6.6)2+(3-6.6)2+(7-6.6)2+(9-6.6)2+(12-6.6)2=17.3
n-1 5-1
sx=√s 2=√17.3=4.16
s 2=∑(y-y-)2= (3-10)2+(6-10)2+(10-10)2+(14-10)2+(17-10)2=32.5
n-1 5-1
sy=√s 2=√32.5=5.7
r=∑(xi-x-)2(yi-y-)2
sx sy
n-1
r=(2-6.6)(3-10)+(3-6.6)(6-10)+(7-6.6)(10-10)+(9-6.6)(14-10)+(12-6.6)(17-10)
( 4.16) (5.7) (4.16) (5.7) (4.16) (5.7 ) ( 4.16) (5.7) ( 4.16 ) (5.7)=3.96074=.99.
5-1 4
n=5, Confidence=95%?, r=.99, CVCC=.87834
Absolute of r is │r│= .99 and is greater than .87834 so a positive linear relation exists.
Exam Page 4
Find the best fit line for the following data:
n=5
x-=1+2+5+7+9=4.8 5
y-=15+12+11+9+7=10.8
5
s 2=∑(x -x-)2= (1-4.8)2+(2-4.8)2+(5-4.8)2+(7-4.8)2+(9-4.8)2=11.2
n-1 5-1
sx=√s 2=√11.2=3.35
s 2=∑(y -y-)2= (15-10.8)2+(12-10.8)2+(11-10.8)2+(9-10.8)2+(7-10.8)2=9.2
n-1 5-1
sy=√s 2=√9.2=3.03
r=∑(xi-x-)2(yi-y-)2
sx sy
n-1
r=(1-4.8)(15-10.8)+(2-4.8)(12-10.8)+(5-4.8)(11-10.8)+(7-4.8)(9-10.8)+(9-4.8)(7-10.8)
( 3.35) (3.03) (3.35) (3.03) (3.35) (3.03 ) ( 3.35) (3.03) ( 3.35 ) (3.03)=-
3.8561=-.964.
5-1
4
n=5, Confidence=95%?, r=-.964, CVCC=.87834
Absolute of r is │r│= .964 and is greater than .87834 so a negative (because original r was negative) linear relation exists.
x-=4.8, y-=10.8, Sx=3.35, Sy=3.03, r=-.964
Slope:
m=r Sy= -.964 3.03=-.873
Sx 3.35
Y intercept:
b=y--mx-=10.8-(-.873)(4.8)=14.99
Line Equation: y=mx+b
y=-.873x+14.99
Exam 10
Exam Page 1
Find the value of X2 for 13 degrees of freedom and an area of .100 in the right tail of the chi- square distribution.
X2=19.812.
Exam Page 2
Find the value of X2 for 17 degrees of freedom and an area of .050 in the left tail of the chi- square distribution.
a=1-.050=.95 X2=8.672.
Exam Page 3
Find the value of X2 values that separate the middle 80 % from the rest of the distribution for 17 degrees of freedom.
a=1-.80=.20/2=.10 in each of the tails. Area to right of first X2=.80+.10=.90 X2=10.085.
Area to right of second X2=.10 X2=24.769.
Exam Page 4
Find the critical value of F for DOF=(6,18) and area in the right tail of .01. F=4.01
Numerator= 6, Denominator=18 <---Look up in F Distribution Table in the right tail of .01.
Exam Page 5
The mayor of a large city claims that 30 % of the families in the city earn more than $ 100,000 per year; 52 % earn between $ 30,000 and $ 100,000 (inclusive); 18 % earn less than $ 30,000 per year.
In order to test the mayor’s claim, 285 families from the city are surveyed and it is found that: 90 of the families earn more than $ 100,000 per year;
135 of the families earn between $ 30,000 and $ 100,000 per year (inclusive); 60 of the families earn less $ 30,000.
Test the mayor’s claim based on 5 % significance level.
H0: The mayor's claim is correct.
H1: The mayor's claim is not correct.
More than two possible outcomes: Multinomial experiment-we use the Chi-Square chart here. DOF=3-1=2, Level of Significance=5%=.05 Chi Square=5.991
Expected frequencies= Ei=npi n=285
Earn more than $100,000 per year E1= 285(.30)=85.5
Earn between $ 30,000 and $ 100,000 per year E2=285(.52)=148.2 Earn less than $ 30,0001941 to 1970 E3=285(.18)=51.3
Observed Frequencies:
Earn more than $100,000 per year=90
Earn between $ 30,000 and $ 100,000 per year=135 Earn less than $ 30,000=60
Test Statistics Calculation:
X2=∑(Oi-Ei-)2=(90-85.5)2+(135-148.2)2+(60-51.3)2= 2.888.
Ei 85.5 148.2 51.3
X2 is smaller than the critical value of 5.991 so we do not reject the null hypothesis.
Exam Page 6
A trucking company wants to find out if their drivers are still alert after driving long hours. So, they give a test for alertness to two groups of drivers. They give the test to 475 drivers who have just finished driving 4 hours or less and they give the test to 635 drivers who have just finished driving 8 hours or more. The results of the tests are given below.
Passed Failed
Drove 4 hours or less 365 110
Drove 8 hours or more 440 195
Is there is a relationship between hours of driving and alertness? (Do a test for independence.) Test at the .5 % level of significance.
H0: There is a relationship between hours of driving and alertness.
H1: There is not a relationship between hours of driving and alertness.
# of Rows= 2, # of Columns=2 DOF=(# of rows-1)(# of columns-1)=(2-1)(2-1)=1 LOS=.005
CSCV=7.879 Passed Failed Row Totals
Drove 4 hours or less 365 110 475
Drove 8 hours or more 440 195 635
Column Totals 805 305 1110
Expected Values Calculation:
Eij= (TRi)(TCj) Total number in sample
E for drove 4 hours or less and Passed E11=(475)(805)/1110=344.48 E for drove 4 hours or less and Failed E12=(475)(305)/1110=130.52
E for drove 8 hours or more and Passed E21=(635)(805)/1110=460.52 E for drove 8 hours or more and Failed E22=(635)(305)/1110=174.48
X2=∑ (Oij-Eij)2= (365-344.48)2+(110-130.52)2+(440-460.52)2+(195-174.48)2=7.776
Eij 344.48 130.52 460.52 174.48
X2 is less than the CV of 7.879 so we do not reject the null hypothesis.
-1.0 points
Instructor Comments
Your null and alternative hypotheses are switched. The calculations are correct.
Answer Key
Passed Failed
Drove 4 hours or less 365 110
Drove 8 hours or more 440 195
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