BB113 BB113 Statistics and Its Its Application Application
Tutorial Tutorial 7
Answer Tutorial Tutorial 7 - Fundamental of Hypothesis Testing – One Sample Tests Tests
One-Sample Tests
1.
If you use use a 0.05 level level of signifi significance cance in a (two-t (two-tail) ail) hypothes hypothesis is test, test, what will will you decide decide if if Z STAT = – 0.7! STAT =
1.
"nswe#
$ecision #ule% &e'ect H &e'ect H 0 if Z if Z STAT – 1. o# Z o# Z STAT * + 1.. STAT * $ecision% ince Z ince Z STAT = – 0.7 is in etween the two c#itical values, do not #e'ect H #e'ect H 0. STAT =
.
If you use use a 0.05 level level of signifi significance cance in a (two-t (two-tail) ail) hypothes hypothesis is test, test, what will will you decide decide if if Z STAT = +.1! STAT =
.
"nswe#
$ecision #ule% &e'ect H &e'ect H 0 if Z if Z STAT – 1. o# Z o# Z STAT * + 1.. STAT * $ecision% ince Z ince Z STAT = + .1 is g#eate# than the uppe# c#itical value of + 1., #e'ect H #e'ect H 0. STAT =
/.
If you use a 0.10 level level of signif significa icance nce in a (two-t (two-tail ail)) hypothes hypothesis is test, test, what is you# you# decisio decision n #ule fo# #e'ecting a null hypothesis that the pop ulation ean is 500 if you use the Z the Z test!
/.
"nsw "nswe# e#%% $ecis $ecisio ion n #ule #ule%% &e'e &e'ect ct H H 0 if Z if Z STAT – 1.5 o# Z o# Z STAT * + 1.5. STAT STAT *
.
If you use a 0.01 level level of signif significa icance nce in a (two-t (two-tail ail)) hypothes hypothesis is test, test, what is you# you# decisio decision n #ule fo# #e'ecting H #e'ecting H 0 % = = 1.5 if you use the Z the Z test!
.
"nswe#% $eci ecision #ul #ulee% &e &e'ect H ct H 0 if Z if Z STAT – .52 o# Z o# Z STAT * + .52. STAT STAT *
5.
3hat is the p-value p-value if, in a two-tail hypothesis test, Z test, Z STAT = +.00! STAT =
5.
"nswe#% p-value p-value = (1 - .77) = 0.05
.
3hat is the p-value p-value if, in a two-tail hypothesis test, Z test, Z STAT = -1./2! STAT =
.
"nswe#% p-value p-value = 0.17
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BB113 BB113 Statistics and Its Its Application Application
Tutorial Tutorial 7
7. 4he uality-cont#ol anage# at a light ul facto#y needs to dete#ine whethe# the ean life of a la#ge shipent of light uls is eual to /75 hou#s. 4he population standa#d deviation is 100 hou#s. " #ando saple of light uls indicates a saple ean life of /50 hou#s. (a) "t the 0.05 level of significance, significance, is the#e evidence that the ean life is diffe#ent diffe#ent f#o /75 hou#s! () 6oput 6oputee the p the p-value -value and inte#p#et its eaning. (c) 6onst# 6onst#uct uct a 5 confidenc confidencee inte#v inte#val al estia estiate te of the popula populatio tion n ean ean life life of the light uls. (d) 6opa#e the #esults of (a) and (c). 3hat conclusions do you #each! 7.
"nswe# H 0% = /75. 4he ean life of a la#ge la#ge shipent of light uls is eual to /75 hou#s.
(a)
H 1% /75. 4he ean life of a la#ge la#ge shipent of light light uls diffe#s diffe#s f#o /75 hou#s. $ecision #ule% &e'ect H 0 if |Z STAT STAT | * 1. 4est statistic% Z STAT
=
X - s 8
n
=
/50 - /75 100 8
= -
$ecision% $ecision% ince |Z STAT 1., #e'ect #e'ect H 0 . 4he#e is is enough evidence evidence to conclude conclude that that the STAT 9 * 1., ean life of a la#ge shipent of light uls diffe#s f#o /75 hou#s. () p-valu p-valuee = 0.05. If the population population ean life of a la#ge shipent shipent of light uls is indeed indeed eual to /75 hou#s, the p#oaility of otaining a test statistic that is o#e than units away f#o 0 is 0.05. (c)
X
�Z a 8
s
n
=
/50 � 1.)
100
/5.5005 � �/7.))5
(d) :ou a#e 5 confide confident nt that the populati population on ean life of a la#ge la#ge shipent shipent of light uls uls is soewhe#e etween /5.5005 and /7.5 hou#s. ince the 5 confidence inte#val does not contain the hypothesi;ed value of /75, you will #e'ect H 0 . 4he conclusions a#e the sae.
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BB113 BB113 Statistics and Its Its Application Application
2.
(a)
Tutorial Tutorial 7
If, in a saple of n = 1 selected f#o a no#al population,
X =
5 and S = = 1, what
is the value of t STAT if you a#e testing the null hypothesis H hypothesis H 0 % = = 50! STAT if ()
In <#o <#ole le 2(a), 2(a), how how any any deg# deg#ees ees of f#eedo f#eedo does the t test have!
(c) (c)
In <#ol <#ole ess 2(a) 2(a) and 2() 2(),, what what a#e the the c#it c#itic ical al valu values es of t if the level of significance, , is 0.05 and the alte#native hypothesis, H hypothesis, H 1 , is 50!
(d) (d)
In <#ole <#oles s 2(), 2(), 2(c) 2(c) and and 2(d) 2(d),, what what is you# you# stati statist stic ical al decisi decision on if the the alte alte#n #nat ativ ivee hypothesis, H hypothesis, H 1 , is 50!
2.
"nswe# (a)
t STAT
=
X – S n
5 – 50 =
= .00
1 1
() d!f! = d!f! = n – 1 = 1 – 1 = 15 (c) >o# a two-tailed two-tailed test with with a 0.05 level of significance, the c#itical values a#e .1/15. (d) (d) inc incee t STAT = .00 is etween the c#itical ounds of .1/15, do not #e'ect H #e'ect H 0. STAT =
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BB113 BB113 Statistics and Its Its Application Application
.
Tutorial Tutorial 7
?ne ?ne of the a'o# a'o# easu# easu#es es of the the ual ualit ity y of se#vi se#vice ce p#ovid p#ovided ed y any any o#ga o#gani ni;at ;atio ion n is the speed with which it #esponds to custoe# coplaints. " la#ge faily-held depa#tent sto#e selling fu#nitu#e and floo#ing, including ca#pet, had unde#gone a a'o# e@pansion in the past seve#al yea#s. In pa#ticula#, the floo#ing depa#tent had e@panded f#o installation c#ews to an instal installat lation ion supe#v supe#viso iso##, a easu# easu#e# e#,, and 15 instal installat lation ion c#ews. c#ews. 4he sto#e sto#e had the usiness o'ective of ip#oving its #esponse to coplaints. 4he va#iale of inte#est was defined as the nue# of days etween when the coplaint was ade and when it was #esolved. $ata we#e collected f#o 50 coplaints that we#e ade in the past yea#. 4he data, sto#ed in Furniture, a#e as follows% 5
5
/5
1/7 /1
11
1 1
1 1 110
1
1/
10
15
1/ 21
7
7
110
1 1
/5 /5
/1 /1
5
15 /
2
5
1
1
1/
5
5
/0
/
0
/
//
2
7
7
4he installation supe#viso# clais that the ean nue# of days etween the #eceipt of a coplaint and the #esolution of the coplaint is 0 days. "t the 0.05 level of significance, is the#e evidence that the clai is not t#ue (i.e., that the ean nue# of days is diffe#ent f#o 0)!
"nswe# PHStat output:
Data Null Hypothesis m= "e#el of Signifi$an$e Sample Si'e Sample (ean Sample Standard De#iation
! !%!& &! )*%!) )+%,!&7*
Inte#ediate 6alculations tanda#d B##o# of the Cean $eg#ees of >#eedo t Te Test Statisti$
5./2/ *%..&.,+
Two-Tail Test "ower /riti$al 0alue 1pper /riti$al 0alue p-0alue
-%!!,&7&+,, %!!,&7&+,, !%!!!*!*
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BB113 BB113 Statistics and Its Its Application Application
Tutorial Tutorial 7
2e3e$t the null hypothesis H 0 % = 0
H 1 %
0
$ecision #ule% &e'ect H 0 if t STAT STAT * .00 4est statistic%
t STAT =
/.0 - 0
X - S 8
d!f! = d!f! =
=
n
1.)18
50
= /.2252
$ecision% ince t STAT #e'ect H 0 . 4he#e is enough evidence to conclude STAT * .00, #e'ect that the ean nue# of days is diffe#ent f#o 0.
10. In a #ecent yea#, yea#, the >ede#al 6ounicati 6ounications ons 6oission 6oission #epo#ted #epo#ted that the ean wait fo# #epai#s fo# De#i;on custoe#s was /.5 hou#s. In an effo#t to ip#ove this se#vice, suppose that a new #epai# se#vice p#ocess was developed. 4his new p#ocess, used fo# a saple of 100 #epai#s, #esulted in a saple ean of /.5 hou#s and a saple standa#d deviation of 11.7 hou#s. Is the#e evidence that the population ean aount is less than /.5 hou#s! (Ese a 0.05 level of significance.) 10. "nswe# H0 % = /.5
H1 % < /.5
$ecision #ule% &e'ect H 0 if t STAT STAT -1.0 4est statistic% t STAT
=
X - S8 n
=
/.5 /.5 - /.5 /.5 11.7 8
100
= - 1.70)
$ecision% ince t STAT -1.0 , #e'ect H 0 . 4he#e is enough evidence to conclude that that the STAT population ean aount is diffe#ent f#o /.5 hou#s.
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BB113 BB113 Statistics and Its Its Application Application
Tutorial Tutorial 7
11. (a) If, in in a #ando #ando saple saple of of 00 ites, ites, 22 a#e a#e defective defective,, what is the the saple saple p#opo#ti p#opo#tion on of defective ites! () In 11 (a), (a), if the null hypothesis is that 0 of the ites in the population a#e defective, what is the value of Z of Z STAT STAT ! (c) In (a) and (), (), suppose suppose you a#e a#e testing testing the null null hypothesi hypothesiss H 0 % $ = = 0.0 against the twotail alte#native alte#native hypothesis hypothesis H H 1 % $ 0.0 and you choose the level of significance significance = 0.05. 3hat is you# statistical decision! 11. "nswe# (a) p = p =
()
X n
Z STAT
22 =
= 0.
00
p -
=
1 -
=
0. - 0.0
0.0 0.2 00
n
o# (c) H 0% H 1%
Z STAT =
=
= 1.00
22
X - n n 1 -
=
-
00 0.0
00 0 .0 0.20
= 1.00
0.0
0.0
$ecision #ule% If Z If Z – 1. o# Z o# Z * * 1., #e'ect H #e'ect H 0. 4est statistic%
Z STAT
=
p -
1 -
n
=
0. - 0.0
0.0 0.2
= 1.00
00
$ecision% ince Z ince Z = = 1.00 is etween the c#itical ounds of 1., do not #e'ect H #e'ect H 0.
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