8/18/2019 ps8_sol (1)

1/4

EF 507 PS-Chapter 8 FALL 2008

1. The amount of material used in making a custom sail for a sailboat is normallydistributed. For a random sample of 15 sails, you find that the mean amount materialis 912 square feet, with a standard deiation of !" square feet. #hich of the following

represents a 99$ confidence interal for the population mean amount of material usedin a custom sail%

&' 912 ±  "9.2(' 912 ±  "2.!)' 912 ± "".*+' 912 ±  "!.

ANSWER: A

2. -et   1 X  , 2 X  , * X  , and " X   be a random sample of obserations from a population

with mean  µ  and ariance 2σ  . )onsider the following estimator of  µ    1/θ   0 .15   1 X   

.*5   2 X 

  .2   * X 

  .*   " X 

. #hat is the ariance of   1/

θ  %&' .235

('   2.235σ 

)' .55   2σ +' .55

ANSWER: C

*. #hich of the following statements is false%

&' &n estimator /θ   that is a function of the sample data 1 2, ,....., n X X X   is called

an unbiased estimator of the population parameter /θ  4 that is , ( )/ E   θ θ =(' The statistic  X  is an unbiased estimator of the population mean  µ 

)' The sample statistic s is an unbiased estimator of the population parameter σ   when n 6 1' is used as a diisor in the calculation of the sample standarddeiation+' 7one of the aboe

ANSWER: C

". 8n deeloping an interal estimate for a population mean, the population standarddeiation σ   was assumed to be . The interal estimate was "5.2 ± 2.*!. ad σ 

equaled 1!, the interal estimate would be&' "5.2 ±  ".32

(' 55.2±

 2.*!)' "5.5 ±  9.""+' 91.!" ±  ".32ANSWER: A

5. The lower limit of a 95$ confidence interal for the population proportion  P  giena sample si:e n 0 1 and sample proportion / P 0 .!2 is equal to

&' .315(' .!99)' .525

+' .5"ANSWER:  C

1

8/18/2019 ps8_sol (1)

2/4

!. 8n order to estimate the aerage daily down time, a manufacturer randomly samples" days of production records and found a mean of 51.53! minutes and standarddeiation of 35.9 minutes. &n $ confidence interal is gien by&' 51.!35 ±  1.212.'(' 51.!35 ±  1.235.9')' 51.!35 ±  1.*"12.'+' 51.!35 ±  1.*".3'ANSWER: C

3. The +aytona (each Tourism )ommission is interested in the aerage amount ofmoney a typical college student spends per day during spring break. They surey *5students and find that the mean spending is ;!*.53 with a standard deiation of;13.*2.

a' +eelop a 95$ confidence interal for the population mean daily spending.

ANSWER:

1 < 2,

8/18/2019 ps8_sol (1)

3/4

QUESTIONS 10 AND 11 ARE ASED ON T!E FOLLOWIN"

INFOR#ATION:

& researcher is interested in determining the percentage of all households in the =.>.that hae more than one home computer. 8n a surey of "92 households, 23$indicated that they own more than one home computer.

1. +eelop a 9$ confidence interal for the proportion of all households in the=.>. with more than one computer.

ANSWER:   n 0 "92, / p 0.23, < 2 .5 z z α    = 01.!"5

( ) ( ) ( )< 2/ / /1 < .23 1.!"5 .23 .3* < "92 .23 .** p z p p nα ± − = ± = ± .

 ence, =)-0 .** and -)- 0 .2*3

11. The researcher reports a confidence interal of .2*12 to .*9! but neglects totell you the confidence leel. #hat is the confidence leel associated with this

interal%ANSWER: #idth0   ( )< 2   / /1 uppose that the amount of time teenagers spend on the internet is normally

distributed with a standard deiation of 1.5 hours. & sample of 1 teenagers isselected at random, and the sample mean computed as !.5 hours.

12. +etermine the 95$ confidence interal estimate of the population mean.

ANSWER:   < 2 < !.5 1.9!'1.5 < 1' !.5 .29" x z nα    σ ± × = ± = ± ⇒ !.2! @ µ @ !.39"

1*. 8nterpret what the interal estimate in Auestion 12 tells you.

ANSWER: 8f we repeatedly draw independent random samples of si:e 1 from the population of teenagers, and confidence interal for each of these samples aredetermined, then oer a ery large number of repeated trials, 95$ of these confidenceinterals will contain the alue of the true population mean amount of time teenagersspend on the internet.

QUESTIONS 1% T!ROU"! 17 ARE ASED ON T!E FOLLOWIN"

INFOR#ATION:

& furniture moer calculates the actual weight as a proportion of estimated weight for a sample of *1 recent Bobs. The sample mean is 1.1* and the sample standarddeiation is .1!.

1". )alculate a 95$ confidence interal for the population mean using t  tables.

*

8/18/2019 ps8_sol (1)

4/4

ANSWER: Cien n  0 *1, 1.1*, x   = and  s  0 .1!. To calculate a 95$ confidence

interal, we need the t  table alue for df 0 *. #ith .5,α   =  this alue is .25,* 2."2.t    =

The confidence interal is < 2 < 1.1* 2."2'.1! < *1' 1.1* .59 x t s nα ± × = ± = ±   or 1.31 1.19. µ < <

15. &ssume that the population standard deiation is known to be .1!. )alculate a95$ confidence interal for the population mean using the z- table.

ANSWER: 8f σ   is known to be .1!, we may use the z  table alue .25 1.9! z    =  rather 

than the t   table alue' in the 95$ confidence interal.

< 2< 1.1* 1.9!'.1! < *1' 1.1* .5! x z nα    σ ± × = ± = ±  or 1.3" 1.1!. µ < <

"