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ABNORMAL P
Subject
Mathematics
Date
Jun 9, 2024
Type
docx
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27
Uploaded by CorporalIbexMaster1220
Define each of the following:
a) Element.
a)
The element of a data set is simply the individual and unique entry in a data set about which data has been collected, analyzed and presented in the same manner.
b)
Variable.
b)
A variable is a particular, measurable attribute that the researcher believes
is needed to describe the element in their study.
2024 MATH-110 COMBINATION OF EXAMS QUESTIONS WITH 100% SOLVED ANSWERS 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
Explain the difference between population and sample. Population is the entire number of items in a large group.
A sample is representative group from the population.
Look at the following data and see if you can identify any outliers:
65 71 55 69 3 77 67 70 246 61 277
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
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
The outliers are:
3 246 277
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?
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?
Burgers : 4900(.32) = 1568
French Fries : 4900(.18) = 882
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
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
FishCheeseburger
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
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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)= .2
0
(.20)100
=
20%
Cheeseburg
er 12%
Fish
6
18
(6/50)= (18/50)
=
.12
.36
(.12)100=
(.36)100
=
36%
Hamburger
8
(8/50)=
.1
6
(.16)100
=
16%
Total
50
1
100%
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.
Consider the following data:
{29, 20, 24, 18, 32, 21}
a)
Find the sample mean of this data.
x* = ∑x
i
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.
s
2
=∑(x
i
-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=√s
2
= √30 = 5.477
d)
Find the coefficient of variation. cov=standard
deviation
*100=5.477
*100 =22.82
mean
24
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.
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).
Suppose A and B are two events with probabilities:
P(A)=.35,P(B
c
)=.45,P(A∩B)=.25.
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Find the following:
a)P(A
∪
B).
P(AUB)=P(A)+P(B)-P(AnB)
P(B
c
)=> P(B)=1-P(B
c
)=1-.45=.55 P(AUB)=.35+.55-.25=0.65
b)P(A
c
).
P(A
c
)=> P(A)=1-P(A
c
)=>
P(Ac)=1-P(A)= 1-.35=.65
c)P(B).
P(B
c
)=> P(B)=1-P(B
c
)
P(B)=1-.45=.55
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!
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
Find the answer to each of the following by first reducing the fractions as much as possible: a) P(490,2)=
b) C(670,665)=
In (a) you multiplied by 498 instead of 489.
Answer Key
Suppose A and B are two events with probabilities:
P(A
c
)=.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(A
c
)= P(A)=1-P(A
c
)=1-.20=.80 P(BlA)= P(AnB)
= .20
=.25
P(A)
.80
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
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
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
Since we do not want to count all of the possible orderings, we use combinations.
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
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
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!
p
x
(1-p)
(n-x)
= x!(n-x)!
f(x)= 8!
.08
6
(1-.08)
(8-6)
=.000006209
6!(8-6)!
For seven failures:
n=8, x=7, p=.08
f(x)= n!
p
x
(1-p)
(n-x)
= x!(n-x)!
f(x)= 8!
.08
7
(1-.08)
(7-6)
=.000000155
7!(8-7)!
For eight failures:
n=8, x=8, p=.08
f(x)= n!
p
x
(1-p)
(n-x)
= x!(n-x)!
f(x)= 8!
.08
8
(1-.08)
(8-8)
=.000000002
8!(8-8)!
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A factory has eight safety systems. During an emergency, the probability of any one of the safety systems failing is .08. What is the prob
The probability is:
.000006209+.000000155+.000000002=.000006366
Instructor Comments
The numbers are so small that with rounding your are fine.
Answer Key
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<
Z<
.86)=P(Z<
.86)-P(Z<
-.09)=.80511-.46414=.34097
A company manufactures a large number of rods. The lengths of the rods are normally distributed with a mean length of 3.7 inches and a standard deviation of .35 inches. If you choose a rod at random, what is the probability that the rod you chose will be:
a)
Less than 3.5 inches?
u= 3.7, o=.35, x=3.5 z=x-u
= 3.5-
3.7
=-.57
o
.35
P(Z<
-.57)=.28434
b)
Greater than 3.5 inches?
u= 3.7, o=.35, x=3.5
z=x-u
= 3.5-3.7
=-.57
o
.35
P(Z>
-.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(-.86<
Z<
.86)=P(Z<
.86)-P(Z<
-.86)=.80511-.19489=.61022
A life insurance
salesperson expects to sell between zero and five insurance policies per day. The probability of these is given as
follows:Policies
Sold
Per Day
Probability, f(x)
0
.04
1
.11
2
.23
3
.26
4
.19
5
.17
Find the expected number of insurance policies that the salesperson will sell per day. Also, find the variance and standard deviation of this data.
E(x)=u=∑xf(x)=
=0(.04)+1(.11)+2(.23)+3(.26)+4(.19)+5(.17)=2.96
The insurance salesperson can expect to sell 2.96 insurance policies per day. Variance:
Var(x)=o
2
=∑(x-u)
2
f(x)=
=(0-2.96)
2
(.04)+(1-2.96)
2
(.11)+(2-2.96)
2
(.23)+(3-2.96)
2
(.26)+(4-2.96)
2
(.19)+(5-
2.96)
2
(.17)=1.8984
Standard Deviation:
o=√o
2
= o=√o
2
=√1.8984=1.3778
An archer is shooting arrows at a target. She hits the target 55% of the time. If she takes 14 shots at the target, what is the probability that she will hit the target exactly 7 times?
n=14, x=7, p=.55
f(x)= 14!
.55
7
(1-.55)
(14-7)
=.286719357
An archer is shooting arrows at a target. She hits the target 55% of the time. If she takes 14 shots at the target, what is the probability t
14!(14-7)!
-2.0 points
Instructor Comments
You can see in the Answer Key where you are off in the denominator, which throws your answer off.
Answer Key
Suppose that you take a sample of size 26 from a population that is not normally distributed. Can the sampling distribution of x
x
be approximated by a normal probability distribution?
The sampling distribution of x
x
cannot be approximated by a normal probability distribution because if the population isn't normally distributed, the sample size has to be at least 30.
Suppose that you are attempting to estimate the annual income of 1200 families. In order to use the infinite standard deviation formula, what sample size, n, should you use?
n
<
0.05= N
n
<
0.05=
1200
n <
0.05(1200)
n <
60
Sample size must be less than 60.
Suppose that in a large hospital system, that the average (mean) time that it takes for a nurse to take the temperature and blood pressure of a patient is 175 seconds with a standard deviation of 50 seconds. What is the probability that 35 nurses selected at random will have a mean time of 160 seconds or less to take the temperature and blood pressure of a patient?
u=175, o=50, n=35, x
x
=160
o
x
x
= o
√n
o
x
x
= 50
= 8.4515
√35 z=
x
x
-u
=
o
x
x
z=
x
x
-u
=160-175
= -1.77
o
x
x
8.4515
P(Z<-1.77)=.03836
Suppose that in a very large city 7.2 % of the people have more than two jobs. Suppose that you take a
random sample of 80 people in that city, what is the probability that 7 % or less of the 80 have more
than two jobs?
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Suppose that in a very large city 7.2 % of the people have more than two jobs. Suppose that you take a random sample of 80 people in Now, we find the z-score:
We want P(Z< - .07). From the standard normal table, we find:
P(Z< -.07)=.47210.
So there is a .47210 probability that the percentage of the sample that have more than two jobs is less than 7 %.
.
-2.0 points
Instructor Comments
In your calculation of the standard deviation, you didn't get the square root. That throws the rest of your
calculations off.
Answer Key
A new drug is introduced that is supposed to reduce fevers. Tests are done with the drug. The drug is given to 70 people who have fevers. It is found that the mean time that it takes for the fever to get back
to normal for this test group is 400 minutes with a standard deviation of 85 minutes. Find the 95% confidence interval for the mean time that the drug will take to reduce all fevers for all people.
Case 1: Large population and large sample size n=70 x
x
=400 s=85 z=1.96
x
x
-z
s
<u< x
x
+z
s
√n
√n
400-1.96 85
<u<400+1.96 85
√70
√70
380.0875<u<419.9125
A certain school has 380 male students. The school nurse would like to know how many calories the male students consume per day. So, she samples 65 male students and finds that the mean calorie consumption of the 65 is 1950 calories per day with a standard deviation of 200 calories per day. Find the 95 % confidence interval for mean calorie intake of all the male students in the school.
Case 3: Finite population
N=300 n=65 x
x
=1950 s=200 z=1.96
x
x
-z
s √N-
n
<u< x
x
+z s
√N-n
√n √N-1
√n √N-1
A certain school has 380 male students. The school nurse would like to know how many calories the male students consume per day. So
The population is finite. So, we should use Case 3: Finite population.
Use:
In the statement of the problem, we are given: N=380 n=65 x ¯
=1950 s=200
For a 95% confidence level, table 6.1 gives z=1.96
1905.67 < μ< 1994.33
1950-1.96 200
√ 300-65
<u<1950+1.96 200
√
300-65
√65
√300-1
√65 √ 300-1
1906.895<u<1993.105
-2.0 points
Instructor Comments
1950-1.96 200
√ 300
-65
<u<1950+1.96 200
√
300-65
√65
√
300
-1
√65 √ 300-1
300 should be 380.
Answer Key
In a large city, the city supervisor wants to find the average number of aluminum cans that each family recycles per month. So, she surveys 21 families and finds that these 21 families recycle an average of 160 cans per month with a standard deviation of 45 cans per month. Find the 90 % confidence interval for the mean number of cans that all of the families in the city recycle per month.
Case 2: Large population and small sample size
n=21 x
x
=160 s=45 t=1.725
x
x
±
t s
√n 160±1.725 45
√21 160±16.9392
143.0608<u<176.9392
df=n-1=21-1=20
A shipment of 400 new blood pressure monitors have arrived. Tests are done on 55 of the new monitors and it is found that 10 of the 5
We have a finite population, so we will use Case 2:
The proportion of the sample that are defective is 10/55 = .182 so we set P=.182. As we mentioned previously, we estimate p by P. So, p
.182± .08
So the proportion of the total that are defective is between .102 and .262.
A doctor has a large number of patients and would like to know if his patients prefer to fill in forms electronically or prefer to hand write their forms. So, he surveys 95 patients and finds that 53 prefer electronic forms while 42 prefer hand written forms. Find the 95% confidence limit for the proportion of all patients that prefer the electronic forms.
Case 1:
P/p=p/n=53/95=.56 n=95 z=1.96
P± z √p(1-p)
√n
.56± 1.96 √.56(1-.56)
√95
.56±.1
.46 to .66
A shipment of 400 new blood pressure monitors have arrived. Tests are done on 55 of the new
monitors and it is found that 10 of the 55 give incorrect blood pressure readings. Find the 90%
confidence interval for the proportion of all the monitors that give incorrect readings.
Case 2:
P/p=p/n=10/55=.182 n=55 N= 400 z=1.645
P± z √p(1-p)
√N-n
√n
√N-1
.182± 1.645 √.182(1-.182)
√400-55
√55
√400-1
.182±.06191
0.12009 to 0.24391
-2.0 points
Instructor Comments
You have a math error in your calculation. You can see in the Answer key that the value past the
± should be 0.079587 (or round to 0.08).
Answer Key
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a)
Define the null and alternative hypothesis for the following. Also, explain what it would mean to make a Type I error and explain what it would mean to make a Type II error.
The newspaper in a certain city had a circulation of 18,000 per day in 2010. You believe that the newspaper’s circulation is different than 18,000 today.
H
0
=18,000 circulations. H
1
≠18,000 circulations.
Type I Error: Reject the null hypothesis that the mean number of newspaper circulations is 18,000 per day when the mean number of newspaper circulations is 18,000 per day.
Type II Error: Do not reject the null hypothesis that the mean number of newspaper circulations is
18,000 per day when the mean number of newspaper circulations is different from 18,000 per day.
b)
Define the null and alternative hypothesis for the following. Also, explain what it would mean to make a Type I error and explain what it would mean to make a Type II error.
A certain website had 5000 hits per month a year ago. You believe that the number of hits per month is more than that today.
H
0
=5,000 hits. H
1
>5,000 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.
Suppose that we have a problem for which the null and alternative hypothesis are given by: H
0
: μ=959.
H
1
: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. H
0
:u=959.
H
1
:u<959.
Left tailed test: P(Z<z)=a P(Z<z)=.13
z=-1.125
x
x
It is recommended that pregnant women over eighteen years old get 85 milligrams of vitamin C each day. A doctor is concerned that her pregnant patients are not getting enough vitamin C. So, she collects data on 52 of her patients and finds that the mean vitamin intake of these 52 patients is 76 milligrams per day with a standard deviation of 22 milligrams per day. Based on a level of significance of α =
.001, test the hypothesis.
H
0
:u=85. H
1
:u<85.
Left tailed test:
P(Z<z)=a P(Z<z)=.001 z=-
3.09
a=.001, u=85, o=22, x
-
=76 n=52
z-score:
o -
= o
= 22
=3.051
√n
√52
z=x
-
-u= 76-85
=-2.94985
o -
3.051
Do not reject the null hypothesis (z-score is greater than -3.09).
A mayor claims that the unemployment rate in his city is 5 %. Many people think that the unemployment
rate is higher. So, 130 residents of the city are contacted and it is found that 11 of them are unemployed. Can the mayor’s claim be supported to a level of significance of α = .03, test the hypothesis.
H
0
:p=.05.
H
1
:p>.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
o
s
=√np(1-p)
o
s
=√130(.05)(1-.05)=2.48495
z score=x
-
-u= 11-6.5
=1.8109
o
s 2.48495
We do not reject the null hypothesis (z score is less than 1.88).
Suppose we have independent random samples of size n1 = 570 and n2 = 675. The proportions of success in the two samples are p1= .4
From table 6.1, we see that 99% confidence corresponds to z=2.58. Notice that the sample sizes are each greater than 30, so we may us
Suppose we have independent random samples of size n
1
= 570 and n
2
= 675. The proportions of success in the two samples are p
1
= .41 and p
2
= .53. Find the 99% confidence interval for the difference in the two population proportions.
n
1 = 570, n
2 = 675, p
1
= .41, p
2 = .53, z=2.58 P
1
-
P
2
±z√p
1
(1-p
1
)
+ p
2
(1-p
2
)
n
1
n
2
.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
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1=
1
2
1
2
1
2
1
1
So,
the interval
is (-.1927,-.0473).
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 (µ
1
- µ
2
).
b)
Can you be 95% confident that there is a difference in the means of the two populations? a) z=1.96
n
1
=77, n
2
=72, x- 32
, x- =26 s
1
=4.1 s
2
=6.3
(x
-
-x
-
)-z√s
2
+s
2
+<u -u <(x
-
-x
-
2)+z√s
2
+s
2
n
1
n
2
n1
n2
(32-26)-1.96√4.1
2 + 6.3
2 +<u
1
-u
2
< (32-26)+1.96√4.1
2 + 6.3
2 77
72
77
72
6±1.7194
4.2806<u
1
-u
2
<7.7194
b) Yes we can be 95% confident due to the fact that both values cover positive values (confidence intervals are both positive).
A head librarian supervises a number of libraries in a large county. He wants to know if full-time library workers and part-time library workers re-shelve books at the same rate. So, he checks the records of 42 full-time library workers and finds that they re-shelve an average of 113 books per hour with a standard deviation of 8.1 books per hour. The records of 42 part-time library show that they re-
shelve an average of 110 books per hour with a standard deviation of 10.1 books per hour.
Using a level of significance of α=.05, is there enough evidence to indicate a difference in the mean number of books re-shelved by full-time workers compared to part-time workers?
H
0
:u
1
-u
2
=0 H
1
:u
1
-u
2
≠0
2
1
Two Tailed Test: a=.05 P(Z<z)=a/2 P(Z>z)=a/2
P(Z<z)=.05/2=.025 P(Z>z)=a/2=.05/2=.025
z=-1.96 and z=1.96
Z-Score:
z=(x
-
-x
-
)-0
√s
1
2
+ s
2
√ n
1
n
2
z=(113-110)-0
√8.1
2
+ 10.1
2
√ 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.
Consider the following dependent random samples
Observation
s
1
2
3
4
5
6
x-values
15
5
14
0
14
2
15
9
16
2
16
9
y-values
16
9
15
0
15
1
14
5
17
3
18
0
a)
Determine the difference between each set of points, x
i
- y
i
b)
Do hypothesis testing to see if µ
d
< 0 at the α = .05. a)
d
-
=d1+d2...dn
=-14-10-9+14-11-11
=-6.833
n
6
b)
H
0
:u
d
=0 H
1
:u
d
<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
, s
d
=10.34
s
d
=√∑(d1-d
-
)
2
=10.34
√
n-1 t= d
-
(s
d
/√n)
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 withou
Find the 98 % confidence interval for mean of the differences, µd.
Note that n=8. We will define , di = xi - yi. After doing the appropriate calculations, we find that d
=-1.375 sd = 9.365.
we look at the student’s t chart for 98% confidence (the 98% is found along the bottom row of the chart) and DOF=8-1=7 (the df=7 is fo
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).
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
26
6
26
5
261
247
26
8
27
1
y-time (after)
263
270
28
0
27
9
266
242
26
3
25
9
Find the 98 % confidence interval for mean of the differences, µ
d
.
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
26
8
27
1
y-time (after)
263
270
280 279
266 242
26
3
25
9
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x
x
y
y
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).
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).
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
s
x
=√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
s
y
=√s 2=√32.5=5.7
r=∑(x
i -x
-)2(y
i
-y-)2
s
x
s
y
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
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x
x
y
y
Absolute of r is │r│= .99 and is greater than .87834 so a positive linear relation exists.
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
s
x
=√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
s
y
=√s 2
=√9.2=3.03
r=∑(x
i
-x
-
)
2
(y
i
-y
-
)
2
s
x
s
y
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, S
x
=3.35, S
y
=3.03, r=-.964
Slope:
m=r S
y
= -.964 3.03
=-.873
S
x
3.35
Y intercept:
b=y
-
-mx
-
=10.8-(-.873)(4.8)=14.99
Line Equation: y=mx+b
y=-.873x+14.99
Find the value of X
2
for 13 degrees of freedom and an area of .100 in the right tail of the chi- square
distribution.
X
2
=19.812.
Find the value of X
2
for 17 degrees of freedom and an area of .050 in the left tail of the chi- square distribution.
a=1-.050=.95 X
2
=8.672.
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Find the value of X
2
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 X
2
=.80+.10=.90 X
2
=10.085.
Area to right of second X
2
=.10 X
2
=
24.769.
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.
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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.
H
0
: The mayor's claim is correct.
H
1
: 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= E
i
=np
i n=285
Earn more than $100,000 per year E
1
= 285(.30)=85.5
Earn between $ 30,000 and $ 100,000 per year E
2
=285(.52)=148.2 Earn less than $ 30,0001941 to 1970 E
3
=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:
X
2
=∑(O
i -E
i -)
2=(90-85.5
)2+(135-148.2
)2+(60-51.3
)2= 2.888.
E
i
85.5
148.2
51.3
X
2 is smaller than the critical value of 5.991 so we do not reject the null hypothesis.
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.
Passe
d
Faile
d
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.
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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 group
Is there is a relationship between hours of driving and alertness? (Do a test for independence.) Test at the .5 % level of significance.
H0: Driving hours and alertness are independent events. H1: Driving hours and alertness are not independent events.
H
0
: There is a relationship between hours of driving and alertness.
H
1
: 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:
E
ij
=
(TR
i
)(TC
j
)
Total number in sample
E for drove 4 hours or less and Passed E
11
=(475)(805)/1110=344.48 E for drove 4 hours or less and Failed E
12
=(475)(305)/1110=130.52
E for drove 8 hours or more and Passed E
21
=(635)(805)/1110=460.52 E for drove 8 hours or more and Failed E
22
=(635)(305)/1110=174.48
X2=∑ (O
ij
-E
ij
)
2= (365-344.48)
2+(110-130.52)
2+(440-460.52)
2+(195-174.48)
2=7.776
E
ij
344.48
130.52
460.52
174.48
X
2
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
Passe
d
Faile
d
Drove 4 hours or less
365
110
Drove 8 hours or more
440
195
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We have two rows and three columns, so # of Rows =2 and # of Columns=2. The degrees of freedom are given by:
DOF = (# of Rows-1)(# of Columns-1)=(2-1)(2-1)=1. Using this, along with .005 (for the .5 % level of significance) we find in the chi-squa
This value is less than the critical value of 7.879. So, we do not reject the null hypothesis.
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