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Unformatted text preview: Ofﬂine HW #6 C .50 summed i, The diameter of 3.5 inch diskettes is normally distributed. Period‘ u rol in nectors randoml select a
FEE sample of 23 diskettes. If the mean diameter of the diskettes i too lare or too sm.lthe diskette punch is shu
down for adjustment; othenNise, the punching process continues. ' e ast sample showed a mean and standard 8 'W deviation of 3.47 and 0.08 inches, respectively. Using 0 = 0.01, the appropriate decision is : . do not reject the null hypothesis and do not shut down the punch u g 11‘ ; 3‘s #0 H 1 ,‘ 1 3 5'
o 0 ~ 0 reject the null hypothesis and shut down the punch I z 3}”, II: : 0,08
reject the null hypothesis and do not shut down the punch * : “K " , 33"" :5 s _ ' .1“
I f .03 ,r" '
do not reject the null hypothesis and shut down the punch Vs.” I" I ‘3 % m
H “RES!“ ““Ft<':ﬁof «171‘: 3., l”,
1.%ﬁ ‘~‘ .1 4i 3.:QH ' ’11s .0‘ ,3). 1 . The diameter of 3.5 inch diskettes 'is normally distributed Periodically, quality control inspectors randomly select a
sample of 18 diskettes. If the mean diameter of the diskettes is too large or too small the diskette punch is shut
996.! down for adjustment; othenNise, the punching process continues. The last sample showed a mean and standard
ﬂ ”l3 deviation of 3.55 and 0.08 inches, respectively. Using a = 0.05, the appropriate decision is: NW 74: 3.§S,m=.°3, M»: m ., . ‘ t g x .R0 35:“ 3‘50
reject the null hypothesis and do not shut down the punch W! W 3 3" C53. do not reject the null hypothesis and shut down the punch ' 2 lid do not reject the null hypothesis and do not shut down the punch MIR“: ['1‘ \F t."# 08 1.71; U, reject the null hypothesis and shut down the punch Jug”. . 015,11
C_____———————— .2..____.__.«_*.r\ 5
tum '3 mic 3‘“ 2" 3. A market researcher is testing consumer opinions about a new product. Consumers are asked to rate the product on
a scale of from O to 10 as to whether or not they would purchase the product if it were available (where O = 9% absolutely would not purchase, 10 = absolutely would purchase). Past studies have shown that this scale is
4‘. W normally distributed in the oulation. The market researcher will not give the go ahead for further development of
the DVOdUCt UNESSW is greater than 6.00. A sample of eight scores are obtained as follows: 7, 8, S, .  ", 8. The observed t statistic is:
(F we sum «in: 14,, we
1.81 '
_ “5: K : (“0 W. A”. K>QOQ(HM£ “IHOIIXTQ'TIO I". >40 .
, 4.83 . . ~\" ~
~ FEQx;*X 2 i. ‘)t(# A; : {.452 H. A company believes that it controls more than 40% of the total market share for one of its products. To prove this
belief, a random sample of 16S purchases of this product are contacted. It is found that 70 of the 16S purchased this company's brand of the product. If a researcher wants to conduct a statistical test for this problem, the
alternative hypothesis would be: ,: Nounnut MTG. 0F Hakim SHﬁRE h I h CM‘TKORLE“ 31 Conn!"
t e popu ation proportion is greatert an 0.40.
o H: ”w "m“: ,qo ir vanssmug
the population mean is equal to 0.40. ‘ ~ g k ’ u: MINE DEﬂoﬂm
1;. the population proportion is not equal to 0.40. the population proportion is less than 0.40. «an»: (outset: MRI. mm 40'].
0V NRRKCT 3“” S. A company believes that it controls more than 40% of the total market share for one of its products. To prove this belief, a random sample of 165 purchases of this product is taken. It is found that 70 of the 165 purchased this
company's brand of the product. If a researcher wants to conduct a statistical test for this problem, the observed RSV STR'USTK valueis: 76
A _ .
. 0.63 (“~ ‘ “~3 , Q ‘ ”.5 ’ 0.933 f'. 1.03 _, A
,, LL .4»! mm:
' 0'35 2 '  = 01.19
C (M (.c A . A bank inspector monitors the default rate on personal loans at the AFB member banks. One standard that she
examines is that no more than 5% of personal loans should be in default. On each Friday, the default rate is
calculated for a sample of 500 personal loans. Last Friday's sample contained 30 defaulted loans. The bank inspector's alternate hypothesis is: “005 Q: i’ol‘ULn'rioiI N16. or Manual. LoniIs N 057mm
p=.05 H”: ’Y . .05 vs. H“. «>135 1P Wt Rtxéd' H“
p¢.05 Wt HRVE DEHOI‘STRR‘WO THIlT no“ THM 5'7.
«F Locus nut in OEFRJLT,
WHICH VIOLMIS ”flit STRNDMD " Ho none
THE“ ‘70. A bank inspector monitors the default rate on personal loans at the AFB member banks. One standard that she
examines is that no more than 5% of personal loans should be in default. On each Friday, the default rate is
calculated for a sample of 500 personal loans. Last Friday's sample contained 30 defaulted loans. If the bank
inspector uses hypothesis testing techniques to test this and if alpha is .10, then the critical value is: 1645 Rt‘sm Ho l? 2 91‘b5l.1,%1, (Tam aﬁ
1.28
1.645
t. 1.28 3. A bank inspector monitors the default rate on personal loans at the AFB member banks. One standard that she
examines is that no more than 5% of personal loans should be in default. On each Friday, the default rate is y. 3:,
calculated for a sample of 500 personal loans. Last Friday’s sample contained 30 defaulted loans. If the bank At”?
inspector uses hypothesis testing techniques to test this and if alpha is .10, then the observed value is: 0° ,. A
0.046 .. _ b ‘0‘. ‘_ .05, 3.06 —1.03
in o .03 (.95)
g 1.03 . 500
it; '0.046 = LOU. q. Management at a large company has recently implemented a more participatory management style. Management
believes that at least half of the company's employees feel that they are more a part of the decision making
process than before. To test this, the company's business researcher randomly polls 180 company employees. One
hundred and seven of those polled say that they feel they are more a part of the decision making process than
before. Suppose the researcher is using a 1% level of significance. The critical value is: 1645 'Q : Minnnon MTG. or wheat: Fé'tLNG Till? w: non: A mgr 0F
TlII mm“ Mun“ r8003: 1H»: strong 1.96 a 2.33 “0:,“ 3'“ "" ‘1‘”?th A 3 (i
‘ ,wi ~.50 ¢ 3‘ x "1 . _ w . )
.. 2575 2 . (Mu.3 'S ., 1.3L ,RESECY HglF 2 '2.“
J i550 lo . Management at a large company has recently implemented a more participatory management style. Management
believes that at least half of the company's employees feel that they are more a part of the decision making
process than before. To test this, the company's business researcher randomly polls 180 company employees. One
hundred and seven of those polled say that they feel they are more a part of the decision making process than
before. Suppose the researcher is using a 1% level of significance. The observed value is: . 2.53 565 RBWS
1.84 '
2.57 1.56 H. A graphical curve constructed by plotting beta values against various values of the alternative hypothesis is called:
Pareto chart BtACK , VHCES 33o  33\
power curve
p operating characteristic curve ogive curve H. The following data (in pounds), which were selected randomly from a normally distributed population of values,
represent measurements of a machine part that is supposed to weigh, on average, 8.3 pounds. 8.1 8.4 8.3 8.2 8.5 8.6 8.4 8.3 8.4 8.2
8.8 8.2 8.2 8.3 8.1 8.3 8.4 8.5 8.5 8.7 Use these data and a = .01 to test the hypothesis that the parts average 8.3 pounds. Test of mu 8.3 vs not = 8.3 II Variable N Mean StDev SE Mean 99% CI T P
Weights 20 8.3700 0.1895 0.0424 (8.2488, 8.4912) 1.65 0.115 . Us; THE Tm sranxric. ‘70 REACH a «Humor! Mm 14. it: K 4 4.00:," ”1.“! on 1: ’t.oo:’13’lcﬁ“3fﬂ"~1° “n‘ﬁ Ho
. “a 1n: Q Nam: To «Emu A concwtiwz 5mg ?:,n$ 7'°“>""‘ To Rﬂtc‘r H.
. us: “m :1 'To arm! A codcwxiw: SNCE $3 i: M Tu: cIJFniL'm Rm "0 B.One survey conducted by RHI Management Resources determined that the Lexus is the favorite luxury car for 25% of
CF05. Suppose a financial management association conducts its own survey of CFOs in an effort to determine
whether this figure is correct. They use an alpha of .05. Following is the MINITAB output with the results of the
survey. Discuss the ﬁndings, including the hypotheses, one or twotailed tests, sample statistics, and the
conclusion. Explain from the data why you reached the conclusion you did. K0 {4
‘l‘csmndi lint i is)? Pmimrlitm ”("1‘0“ '3 89¢ ‘f 7"” HSSOC ' B‘UEVCS VCR .
“““‘““7——£"“‘“‘~‘ ‘3 “Win an Lawn. Test of p : 025 vs )9 not : 0.35 Exact $1516 ’SIDEO H“.
Sample X .‘I Sample p 95?; CI Pry$111.19
1 79 384 0 206.729 (0.166390, 0.249653) 0.045 a use “a ,y nwé To REHCH a (cactusml 3
Since the p  value rounded to 3 decimal places is O. 045, then we reject the null hypothesis. ‘ USE THE “ST 379757“ TO REM“ a<ou<w$mk 17:5 3 :1, “ML,2 m M —_““_= r 1 003
I, on” ~15“ Mk \lL—J “5‘2: 3 :2.) 85356 “a: Q3 .1: o 031 W: (I 76 “RR R Comma“ 3 glVKE gas is “a "1 WI: ‘11, RETECT THE H
‘ Fame» T T$ST W; A researcher wants to conduct a before/after study on 15 subjects to determine if a treatment results in any
difference in scores. The null hypothesis is that the average difference is zero while the altemative hypothesis is
that the average difference is not zero. Scores are obtained on the subjects both before and after the treatment.
After subtracting the after scores from the before scores, the average difference is computed to be ‘~2.13 with a sample standard deviation of 1.21. The degrees of freedom for this test are 1“
.r'l 2 = »
8 Hozbm vs 3:0160 ‘1 "‘3
g, 15 ‘' Au: mu
. 14 is suaSz<15=b4(:,~—i:q
/ 13 is. A researcher wants to conduct a before/after study on 15 subjects to determine if a treatment results in any
difference in scores. The null hypothesis is that the average difference is zero while the alternative hypothesis is
that the average difference is not zero. Scores are obtained on the subjects both before and after the treatment.
After subtracting the after scores from the before scores, the average difference is computed to be —2.13 with a
sample standard deviation of 2.51. The observed t value for this test is: " ‘329 HEM A. 21.3w
r", 1.76 .— ‘b
J\ " \ ~1J3 '0
1:; ‘2.87 t = : ..___.__________ 2 . 3.187
.S\ f
3.04 All 1"; X I I: l H. . A researcher wants to conduct a before/after study on 15 subjects to determine if a treatment results in any
difference in scores. The null hypothesis is that the average difference is zero while the alternative hypothesis is
that the average difference is not zero. Stores‘a‘re obtained on the subjects both before and after the treatment.
After subtracting the after scores from the before scores, the average difference is computed to be —2.13 with a
sample standard deviation of 1.21. Using an alpha of .05, the table t value for this test is: (‘1, 2.131
2145 :3 'r H <  =
f 2252 u c ‘ W 1‘ toxs‘m °“ "Foam ”is g. 1.761 ...
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