Practice Test 3 Bus2023 Spring09 Solutions

8/19/2019 Practice Test 3 Bus2023 Spring09 Solutions

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Practice Test 3 –Bus 2023

Directions: For each question find the answer that is the best solution provided. There isonly one correct answer.

1. A random variable that can assume only a finite number of values is referred to as a(n)a. infinite sequence b. finite sequencec. discrete random variabled. discrete probability function

ANS!"# c

$. The number of customers that enter a store durin% one day is an e&ample of a. a continuous random variable b. a discrete random variablec. either a continuous or a discrete random variable' dependin% on the number of the

customersd. either a continuous or a discrete random variable' dependin% on the %ender of thecustomers

ANS!"# b

. An e&periment consists of measurin% the speed of automobiles on a hi%hway by the useof radar equipment. The random variable in this e&periment is speed' measured in miles per hour. This random variable is aa. discrete random variable b. continuous random variablec. comple& random variable

d. None of these alternatives is correct.ANS!"# b

. A measure of the avera%e value of a random variable is called a(n)a. variance b. standard deviationc. e&pected valued. None of the alternative answers is correct.

ANS!"# c Note# So we can write !(&) * +

,. The e&pected value for a binomial probability distribution is

a. !( x) * pn(1 - n) b. !( x) * p(1 - p)c. !( x) * np

d. !( x) * np(1 - p)ANS!"# c

1



. The variance for the binomial probability distribution isa. /ar( x) * p(1 - p) b. /ar( x) * np

c. /ar( x) * n(1 - p)d. /ar( x) * np(1 - p)

ANS!"# d

0. Assume that you have a binomial e&periment with p * . and a sample si2e of ,. Thevariance of this distribution is

a. $ b. 1$c. .d. Not enou%h information is %iven to answer this question.

ANS!"# b"ecall that variance of the binomial is /ar( x) * np(1 - p) * ,(.)(.) * 1$

3. Twenty percent of the students in a class of 1 are plannin% to %o to %raduate school.The standard deviation of this binomial distribution isa. $ b. 1c. d. $

ANS!"# c"ecall that variance of the binomial is /ar( x) * np(1 - p) * 1(.$)(.3) * 1Std(&) * *

!&hibit ,-3The student body of a lar%e university consists of 4 female students. A random sample of 3students is selected.

5. "efer to !&hibit ,-3. hat is the random variable in this e&periment6a. the 4 of female students b. the random sample of 3 studentsc. the number of female students out of 3d. the student body si2e

ANS!"# c

1. "efer to !&hibit ,-3. hat is the probability that amon% the students in the samplee&actly two are female6

a. .35 b. .$5c. .1d. .0

ANS!"# c

"ecall that 7( 8 * &) * (n

x ) p& (1-p)(n-&) 7( 8 * $) * (3

$ ) ).$ (.)(3-$) * .1

$



11. "efer to !&hibit ,-3. hat is the probability that amon% the students in the sample atleast 0 are female6

a. .1 b. .35

c. .13d. .35ANS!"# a Note# This is the same set up as above but now you need 7 (8 90) * 7 (0) : 7(3)

1$. "efer to !&hibit ,-3. hat is the probability that amon% the students in the sample atleast are male6

a. .1 b. .05c. .0d. .55

ANS!"# d Note# Same set up' but now the percenta%e of inters is not females' but males. 7 (8 9) * 7() :7 (0) : 7(3). ;a<e sure to use p*. since the probability of interest is males.

1. A numerical measure from a sample' such as a sample mean' is <nown asa. a statistic b. a parameter c. the mean deviationd. the central limit theorem

ANS!"# a

1. A subset of a population selected to represent the population is aa. subset b. samplec. small populationd. None of the alternative answers is correct.

ANS!"# b

1,. A simple random sample of , observations from a population containin% elementswas ta<en' and the followin% values were obtained.

1$ 13 15 $ $1

A point estimate of the population mean isa. b. 13c. $d. 1

ANS!"# b



Note# Since this is a sample that is supposed to be representative of the population its mean is an

appro&imate of the population mean. So we %et that ! (

=

x ) * +.1. The samplin% error is the

a. same as the standard error b. absolute value of the difference between an unbiased point estimate and the

correspondin% population parameter c. error caused by selectin% a bad sampled. standard deviation multiplied by the sample si2e

ANS!"# b

10. The probability distribution of all possible values of the sample mean is called thea. central probability distribution b. samplin% distribution of the sample meanc. random variationd. standard error

ANS!"# b

13. From a population of $ elements' the standard deviation is <nown to be 1. A sampleof 5 elements is selected. >t is determined that the sample mean is ,. The standard error of themean is

a. b. $c. %reater than $d. less than $

ANS!"# d

"ecall that ? =

x * ? @ n * 1 @ 5 * 1 @ 0 * $

15. From a population of , elements' a sample of $$, elements is selected. >t is <nownthat the variance of the population is 5. The standard error of the mean is appro&imately

a. 1.1$$ b. $c. d. 1.30

ANS!"# d Note# Same process as above but now a finite population

? =

x *)1@()( −− N n N

(? @ n ) *)1,@()$$,,( −−

( @ $$, ) * 1.30

$. As the sample si2e increases' thea. standard deviation of the population decreases b. population mean increasesc. standard error of the mean decreasesd. standard error of the mean increases

ANS!"# c



$1. "andom samples of si2e are ta<en from an infinite population whose mean andstandard deviation are $ and 1,' respectively. The distribution of the population is un<nown.The mean and the standard error of the distribution of sample mean are

a. and 1,

b. $ and 1,c. $ and .10d. $ and $.,

ANS!"# d

"ecall that the mean of the samplin% distribution and population are the same. e also <now ? =

x *

? @ n * 1, @ . * 1, @ * $.,

$$. A theorem that allows us to use the normal probability distribution to appro&imate thesamplin% distribution of sample means and sample proportions whenever the sample si2e is lar%eis <nown as the

a. appro&imation theorem b. normal probability theoremc. central limit theoremd. central normality theorem

ANS!"# c

$. A population has a mean of 13 and a standard deviation of $. A sample of observations will be ta<en. The probability that the mean from that sample will be between 13and 13 is

a. .1,5 b. .313,

c. .1d. .00$

ANS!"# aTo ma<e a probability statement we need to simply calculate the 2-values for each.

B13 * (8 C + ) @ ? =

x where ?

=

x * ? @ n

* $ @.

* $@3 *

* (13 C 13) @ * 1

B13 * (8 C + ) @ ? =

x * (13 C 13) @ * $

7 ( 1 D B D $ ) * .1,5

$. A population has a mean of 3 and a standard deviation of 1$. A sample of

observations will be ta<en. The probability that the sample mean will be between 3., and 33.5is

a. .0 b. .0$c. .5,11

,



d. None of the alternative ANS!"S is correct.ANS!"# c-same format at above.$,. An estimate of a population parameter that provides an interval believed to contain thevalue of the parameter is <nown as the

a. confidence level b. interval estimatec. parameter valued. population estimate

ANS!"# b

$. The confidence associated with an interval estimate is called thea. level of si%nificance b. de%ree of associationc. confidence leveld. precision

ANS!"# c

$0. The 2 value for a 50.34 confidence interval estimation isa. $.$ b. 1.5c. $.d. $.$5

ANS!"# d-simply use the 2-table and find the 2 value that %ives the middle 50.34

$3. >t is <nown that the variance of a population equals 1'5. A random sample of 1$1 has been ta<en from the population. There is a .5, probability that the sample mean will provide amar%in of error of

a. 0.3 or less b. 1. or lessc. .5 or lessd. 1'5 or less

ANS!"# a"ecall that the mar%in of error is the difference between the sample value and population

parameter and it is bounded by BE@$ ? =

x * (1.5( ) @ ( ) D 0.3

!&hibit 3-1>n order to estimate the avera%e time spent on the computer terminals per student at a localuniversity' data were collected from a sample of 31 business students over a one-wee< period.Assume the population standard deviation is 1.$ hours.

$5. "efer to !&hibit 3-1. The standard error of the mean isa. 0.,



b. .1c. .1d. .1

ANS!"# d

"ecall that ?

=

x

* ? @n

* 1.$@5 * .1

. "efer to !&hibit 3-1. ith a .5, probability' the mar%in of error is appro&imatelya. .$ b. 1.5c. .$1d. 1.

ANS!"# a

;ar%in of !rror * BE@$ ? =

x * 1.5.1.$

1. "efer to !&hibit 3-1. >f the sample mean is 5 hours' then the 5,4 confidence interval isa. 0. to 11.5 hours b. 0. to 1. hoursc. 0.3 to 1.$ hoursd. 3.0 to 5.$ hours

ANS!"# d

"ecall that

=

x G 2E@$ ?. So Hust apply the formula to %et 3.0 to 5.$. This means we are 5,4 surethe true population mean falls within these values.

$. >f we chan%e a 5,4 confidence interval estimate to a 554 confidence interval estimate'

we can e&pect thea. width of the confidence interval to increase b. width of the confidence interval to decreasec. width of the confidence interval to remain the samed. sample si2e to increase

. >n hypothesis testin%' the hypothesis tentatively assumed to be true isa. the alternative hypothesis b. the null hypothesisc. either the null or the alternatived. None of the other answers are correct.

ANS!"# b

. Iour investment e&ecutive claims that the avera%e yearly rate of return on the stoc<s sherecommends is at least 1.4. Iou plan on ta<in% a sample to test her claim. Thecorrect set of hypotheses is

a. J# µ D 1.4 Ja# µ ≥ 1.4

b. J# µ ≤ 1.4 Ja# µ K 1.4

c. J# µ K 1.4 Ja# µ ≤ 1.4

0



d. J# µ ≥ 1.4 Ja# µ D 1.4

ANS!"# d

,. A soft drin< fillin% machine' when in perfect adHustment' fills the bottles with 1$ ouncesof soft drin<. Any overfillin% or underfillin% results in the shutdown andreadHustment of the machine. To determine whether or not the machine is properly adHusted' the correct set of hypotheses is

a. J# µ D 1$ Ja# µ ≤ 1$

b. J# µ ≤ 1$ Ja# µ K 1$

c. J# µ ≠ 1$ Ja# µ * 1$

d. J# µ * 1$ Ja# µ ≠ 1$

ANS!"# d

. >f a hypothesis test leads to the reHection of the null hypothesis' a

a. Type >> error must have been committed b. Type >> error may have been committedc. Type > error must have been committedd. Type > error may have been committed

ANS!"# d

0. The probability of ma<in% a Type > error is denoted by

a. α

b. β

c. 1 - α

d. 1 - βANS!"# a

3. A one-tailed test is aa. hypothesis test in which reHection re%ion is in both tails of the samplin%

distribution b. hypothesis test in which reHection re%ion is in one tail of the samplin% distributionc. hypothesis test in which reHection re%ion is only in the left tail of the samplin%

distributiond. hypothesis test in which reHection re%ion is only in the ri%ht tail of the samplin%

distribution

ANS!"# b

Exhibit 9-1

n * & * $. s * 1$ J# µ ≤ $

Ja# µ K $

5. "efer to !&hibit 5-1. The test statistic equalsa. $. b. .3

3



c. -$.d. -.3

ANS!"# aSince we have that the standard deviations is un<nown but we have nD it is a

2-stat * ( C +) @ ?

=

x

* ($-$.) @ (1$@ ) * . @ 1$ * $.

. "efer to !&hibit 5-1. The p-value isa. .,10 b. .$1c. .10d. $.1

ANS!"# c

1. "efer to !&hibit 5-1. >f the test is done at a ., level of si%nificance' the null hypothesisshould

a. not be reHected b. be reHectedc. Not enou%h information is %iven to answer this question.d. None of the other answers are correct.

ANS!"# b

5



Short Answer:

Question 1: Binoia! Distribution

a. Thirty-two percent of the students in a mana%ement class are %raduate students. Arandom sample of , students is selected. Lsin% the binomial probability function' determine the probability that the sample contains e&actly $ %raduate students6

b. A production process produces $4 defective parts. A sample of , parts from the

production is selected. hat is the probability that the sample contains e&actly two defective parts6 Lse the binomial probability function and show your computations to answer thisquestion.

c. hen a particular machine is functionin% properly' 34 of the items produced are non-defective. >f three items are e&amined' what is the probability that one is defective6 Lse the binomial probability function to answer this question.

ANS!"#

"ecall that for each we use 7( 8 * &) * (n

x ) p& (1-p)(n-&)

a. let p * .$ and n * ,

then 7( 8 * $) * (,$ ) .$$ (1-.$)(,-$) * .$$

b. Similarly n * ,' but not p * .$

7( 8 * $) * (,

$ ) .$$ (1-.$)(,-$) * .03c. Now we have n * $ with p * .3

7( 8 * 1) * ((

1 ) .31 (1-.3)(-1) * .3

1



Questions 2:

A random sample of 1 credit sales in a department store showed an avera%e sale of M1$..From past data' it is <nown that the standard deviation of the population is M..

a. etermine the standard error of the mean. b. ith a .5, probability' determine the mar%in of error.c. hat is the 5,4 confidence interval of the population mean6

ANS!"S#

a. "ecall that ?

=

x

* ? @

n

* @ 1 *

b. "ecall that the ;ar%in of !rror * BE@$ ? =

x * 1.5 * 0.3

c.

=

x G BE@$ ? =

x 1$ :@- 0.3 11$.1 to 1$0.3

11





Question ":

The price of a particular brand of Heans has a mean of M0.55 (Hust assume 3) and a standarddeviation of M0. A sample of 5 pairs of Heans is selected. Answer the followin% questions.a. hat is the probability that the sample of Heans will have a mean price less than M6 b. hat is the probability that the sample of Heans will have a mean price between M3 andM56c. hat is the probability that the sample of Heans will have a mean price within M of the

population mean6

ANS!"S: #ere we si$!% &in' the (-state &or each one: 2-stat * ( C +) @ ? =

x

Note that ? =

x * ? @ n

* 0 @5

* 1a. 7 ( 8 K ) * 7 ( B D $ ) * .50003 b. 7 ( 3 D 8 D 5 ) * 7 ( D B D 1) * .50c. 7 (, D 8 1 ) * 7 (- D B D ) .550

1



Question ):

The avera%e %asoline price of one of the maHor oil companies has been M1. per %allon.Pecause of shorta%es in production of crude oil' it is believed that there has been a si%nificantincrease in the avera%e price. >n order to test this belief' we randomly selected a sample of of the companyQs %as stations and determined that the avera%e price for the stations in the sample

was M1.1. Assume that the standard deviation of the population (σ) is M.1$.a. State the null and the alternative hypotheses.

b. Test the claim at ∝ * .,.

c. hat is the p-value associated with the above sample results6

ANS!"S#

Note that ? =

x * ? @ n

* .1$ @.

* .$

a. J# µ ≤ 1

Ja# µ > 1

>t is a one tailed test because we thin< that the results of the shorta%e is anincrease.

b. z * ( C +) @ ? =

x * (1.1 C 1) @ .$ * ,R therefore' reHect J' there is sufficient

evidence at ∝ * ., to conclude that there has been an increase in the avera%e

price. e <now this for two reasons.The critical value we compare our B to is 1.5. >f it is si%nificantly lar%er than1.5 then we thin< the price is much hi%her.e can also use the p-value method. >f our p-value of our test stat is smaller than

∝ * .,' then we reHect Jo and conclude there is a statistically si%nificant result

and in this case suspect the price of %as has increased. The p-value of B * , isD .1. So this is much less than .,

c. almost 2ero D .1 because it is smaller than the last computed table value

1



Question *:

a. At a local university' a sample of 5 evenin% students was selected in order to determinewhether the avera%e a%e of the evenin% students is si%nificantly different from $1. The avera%e

a%e of the students in the sample was $ with a standard deviation of .,. etermine whether or not the avera%e a%e of the evenin% students is si%nificantly different from $1. Lse a .1 level of si%nificance.

ANS!"#

Note that ? =

x * ? @ n

* ., @5

* .,

J# µ * $1

Ja# µ * $1 z *( C +) @ ?

=

x * ($ C $1) @ ., * R therefore' reHect J' there is

sufficient evidence at ∝ * .1 to conclude that the avera%e a%e of the evenin%

students is si%nificantly different from $1

b. >n order to determine the avera%e price of hotel rooms in Atlanta' a sample of hotels wasselected. >t was determined that the avera%e price of the rooms in the sample was M11$ withstandard deviation of M1. Lse a ., level of si%nificance and determine whether or not theavera%e room price is si%nificantly different from M13.,.

ANS!"# Same method as above. So ma<e sure to find ?

=

x .J# µ * 13.,

Ja# µ * 13., z * ( C +) @ ? =

x * (11$ C 13.,) @ $ * 1.0,R therefore' do not reHect J'

there is not sufficient evidence at ∝ * ., to conclude that that the avera%e room

price is si%nificantly different from M13.,.

1,

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Practice Test 3 Bus2023 Spring09 Solutions