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Unformatted text preview: Reneegd'i Review 35“ Problem No. Vale" Name: Problem No. Md Name: Dist.: Areas/Scores & Table Reading Template (9.09) Dist: Areas/Scores & Table Reading Template (9.09)
1a. Draw (and label) above the number line below, the la. Draw (and label) above the number line below, the
distribution under consideration. Putting numbers below the distribution under consideration. Putting numbers below the
line, and areas above shaded regions (either know or use line, and areas above shaded regions (either know or use
unknown symbols) [only use “2” on a N(O,l)], represent ALL unknown symbols) [only use “2” on a N(0,l)], represent ALL
the problem information in this picture.) the problem information in this picture.)
1b. If you do NOT have a Table for the distribution in la, 1b. If you do NOT have a Table for the distribution in 1a,
convert the distribution picture in la to one in 1b for which convert the distribution picture in la to one in lb for which
you have a Table. [Example: 1a N(u,62) => 1b N(0,1) using you have a Table. [Examplez 1a N(u,62) => 1b N( ,1) using
z=(xu)/G ] Usma 251—“ __> Z ___ 98  Z=(XH)/6 ] U$lﬂi "(4’) 8 : ﬁig e —’—. 0’ (is(4.))[gszsrs \0
1a i; 2a. Symbol for unknown in 1a: , is an area 0 . 2a. Symbol for unknown in 1a: A , is anr score? 2b.Bound on unknown: " < g “é 2b.Bound on unknown: 0 < A 5 l
3. Draw TP in 3 below. 4. Put 1b (la ifno 1b) in terms ofthe 3. Draw TP in 3 below. 4. Put 1b (la ifno lb) in terms ofthe
TP. Look up table values representing them as TP and get TP. Look up table values representing them as TP and get
your answer in terms of the unknown. your answer in terms of the unknown. . . By .
3 Table Plcture (TP). I 4 3 Table P1cture(TP):  4 1 l
l
l
l o ,z 4 (cont) 0 2 12; 0 o 2375
am Tana (twang WWW?! t ‘
closed‘ 2 mine} Na“) / N 0’ )
H2, «‘2‘: ‘d'l‘ + {J C: t E :: — Mu Usm3>H abwa “2:.(P8ivbj/8 4 (c033,): 23%.
—h'+\ ‘5 P84"é ___> f.“ 6 ’thl) f A—ch 'Za g = —6 ~ n.z.8
= " I 728 4. (Cont) Final answer in terms of unknown in 2a: 4. (cont) Final answer in terms of unknown in 2a: 'Pg:“\'1.28 A: 058‘?“ 5. Check if ﬁnal answer is consistent with bound in 2b. V 5. Check if ﬁnal answer is consistent with bound in 2b. \/ ’Reywa 92H 45 ﬁrm: 7% PQNQ“*i\e o“ Problem N0. M3 Name: Problem No. Name: Dist.: Areas/Scores & Table Reading Template (9.09) Dist.: Areas/Scores & Table Reading Template (909)
1a. Draw (and label) above the number line below, the la. Draw (and label) above the number line below, the
distribution under consideration. Putting numbers below the distribution under consideration. Putting numbers below the
line, and areas above shaded regions (either know or use line, and areas above shaded regions (either know or use
unknown symbols) [only use “2” on a N(0,1)], represent ALL unknown symbols) [only use “2” on a N(0, 1)], represent ALL
the problem information in this picture.) the problem information in this picture.)
1b. If you do NOT have a Table for the distribution in 1a, 1b. If you do NOT have a Table for the distribution in 1a,
convert the distribution picture in 1a to one in lb for which convert the distribution picture in la to one in 1b for which
you have a Table. [Example: 1a N(u,o2) => 1b N(0,1) using you have a Table. [Example: 1a N(u,oz) => lb N(0,1) using
Z=(Xu)/G ] Z=(Xu)/G ] 1 a 1b 1 a 1b ’2. 1L... l
l

l
i
i
I A .67 \
2a. Symbol for unknown in 1a: A , is an area or score? 2a. Symbol for unknown in 1a: P] , is an area or
2b.Bound on unknown: 0 :A < ‘ S 2b.Bound on unknown: 0 < < 00
3. Draw TP in 3 below. 4. Put 1b (1a ifno lb) in terms ofthe 3. Draw TP in 3 below. 4. Put 1b (1a ifno 1b) in terms ofthe
TP. Look up table values representing them as TP and get TP. Look up table values representing them as TP and get
your answer in terms of the unknown. your answer in terms of the unknown. 3 Table Picture (TP): 3 Table Picture (TP): ,ol§< A<0§ 4. (cont) Final answer in terms of unknown in 2a: 4. (cont) Final answer in terms of unknown in 2a: .01'5" A <05 \33'42 < P1 4 “355% 5. Check if ﬁnal answer is consistent with bound in 2b. \/ 5. Check if ﬁnal answer is consistent with bound in 2b. Problem No. SREWetiB SS» i "I it 7 Parametric Hypothesis Testing Template (9.23.09) 1.a Identify the relevant given summary statistics and
other information in the problem and summarize in
terms of the usual statistical symbols: W225 Pg.hocmo.i gafZQ SzthO as: .‘0 lb.Identify and deﬁne the unknown parameter(s) and if
several parameters, the possible parameter combination M: Lm known Papu\5hcm vmeom lc.In terms of this parameter(s) (or parameter
combination), what question are you being asked to
answer when you conclude your hypothesis test? 13 M< (Deiecmmei b3 wicnj 9mm 3M?“ “04 Ht) 1d.Identify the = value in terms of the parameter(s) or
parameter combination. Zlo 1e.Answer the question identiﬁed above if you believe
the parameter or parameter combination are <, =, or >
the equal value identified above, i.e. if you believe the
parameter or parameter combination is <, =, > than the
equal value identiﬁed above, you would answer the
above identiﬁed question respectively either YES or NO. < equal value ANS: ‘29;
= equal value ANS: NO > equal value, ANS: N0 1f.Set up the null and alternative hypotheses in terms of
the parameter(s) (or parameter combination). H0 always
includes the “=” value and any other values for which ((#3) you would take the same action as the — value. Ho: ‘5 = 9 H03 g z 3
H]: E 2L H11 k 24 ; NAME: 2.a What statistic would estimate the parameter (or parameter
combination)? In the case of multiple parameters, give the
statistic typically used to test these hypotheses. v.
2b. Find a distribution fact which is directly related to this statistic. Note that it will also typically contain the parameter
or parameter combination. Write DF # and the DP. including Sim4 distribution \/ e M
D.F.#Z ;i.e., 3N6 2c.From this distribution fact, give the test statistic and known
distribution under the null hypothesis. Identify this test
statistic with a brief identiﬁer (like 20b5, tobs, xobs, F obs, etc.)
which reminds you of the distribution under the null
hypothesis. (In some cases, the test statistic and distribution
under the null hypothesis are already given in the distribution
fact above.) We typically derive the test statistic from the «:99 above distribution fact by substituting the value of the
unknow_n_parameter (or parameter combinatiop) under the
null hypothesis to get the test statistic taking care NOT to substitute for the estimate of the unknown parameter (or
parameter combination. (In some cases, a distribution fact will actually be the test statistic in the next step.) E obs: (‘7 '26) /( \D/V’z’g) 3646”? ~ i: 24
3.Look at the test statistic:(20)k7 '— 26) and
(It)le < 26 . Decide what extreme values of the test siatistic would tend to indicate that H] is true (general direction). We do this in two steps when possible. under H0 3a.Identify the estimate of the unknown parameter (or
parameter combination) in the hypotheses (given typically in
2a above), and indicate what extreme values of this estimate
would indicate the alternative hypothesis is true (general ' direction)
Reject HO if: \/ \$ MUCH less Jam“ % 3b.Use this information as it applies to the test statistic, and
indicate what extreme values of the test statistic would
indicate the alternative hypothesis is true (general direction) Reject Ho if: t .Obs ‘3 Very neg‘JhUt Problem No. (cont) RQN \QSJJ 31".” l 1499 7 3c.Draw a number line, label it under the right side {3 .obs,
and shade and identify where you would Reject Ho m t
o Jcbhs 4.Use the results of #2 and #3 along with the value of the
probability of a type I error to ﬁnd the cutoff point(s) for the
acceptance and rejection region(s). Note oc=Pr{Reject HolHO true}.lN THE SPACE 4ac belowza. To do this, we recopy
the previous number line including the shaded parts of the
axis, drawing and labeling the distribution of the test statistic
under H0 above the number line. b. We then shade the areas
under the curve above the shaded parts of the axes and label
these areas above the curve in terms of 0F, typically as or: or
oc=/2. c. We label below the axes the border(s) between the
shaded and unshaded regions in terms of X cm where “X” is
the symbol we used when we wrote X obs above. Sometimes
we will write —X cm and +X cm , and at other times we may use
X cm, Lower and X critauppef when we have multiple critical
values. Complete 4d4i below to ﬁnd the critical values. 4d. If you do NOT have a Table for the distribution in la,
convert the distribution picture in 4a—c to one in 4d for which you have a Table. [Examplez 4a—c N(].L,62) => 4d N(0,l) using
Z=(Xu)/6 1 4ac 4d 0 4e. Symbol for unknown in 4a—c: teat, is an area or 4f.Bound on unknown: "" 0‘3 < m< O terd 4g. Draw TP in 4g below. 4h. Put 4d (4a—c ifno 4d) in terms
of the TP in 4h below. Look up table values representing
them as TP and get your answer in terms of the unknown. 4g Table Picture (TP):
+é3 CL 4h i 4h (cont? { l
i
i
i
l
5
ago. 'tc’l ' 4h (cont) 4h. (cont) Final answer in terms of unknown in 4e: tank = .38 4i. Check if ﬁnal answer is consistent with bound in 4f. / 5.Clearly state your decision rules. It may be advantageous to
NOT include the critical boundary points in the Accept Ho
region and to list the Reject Ho region as “otherwise” AcceptHo +abS>‘—L3‘8 ; Reject Ho otherwise 6a. Calculate below your test statistic given in 2c using
information in la. bobs 5 "—26:20 1'3 6b. Using the decision rules in #5, indicate whether you will
Accept H0 or Reject Ho: 6c. Interpret your results. (If you have transformed the
problem. typically we interpret the results in terms of the
original parameters.) Usﬁ'g CL i0% lend a? 5396*}:an
isided hypothesis ‘i'esi’lme. Condqu
p 4 2(0 Problem No. Rgvtm E: I 3 #80. Parametric Hypothesis Testing Template (9.23.09) 1.a Identify the relevant given summary statistics and
other information in the problem and summarize in terms of the usual statistical symbols: (is 25 POE “OHMGA "125520 cl: «)0 ($33K) 1b.Identify and deﬁne the unknown parameter(s) and if
several parameters, the possible parameter combination Ms Lm known papuiainm mean 1c.ln terms of this parameter(s) (or parameter
combination), what question are you being asked to answer when you conclude your hypothesis test? Is p424? 1d.ldentify the = value in terms of the parameter(s) or
parameter combination. 21a 1e.Answer the question identiﬁed above if you believe
the parameter or parameter combination are <, =, or >
the equal value identiﬁed above, i.e. if you believe the
parameter or parameter combination is <, =, > than the
equal value identiﬁed above, you would answer the
above identiﬁed question respectively either YES or NO. < equal value ANS: YES * equal value ANS: No > equal value, ANS: N 0 1f.Set up the null and alternative hypotheses in terms of
the parameter(s) (or parameter combination). H0 always
includes the “=” value and any other values for which
you would take the same action as the “=” value. > ,
H0: M = 9 H01 F g i
H12 '5 HI; I l ; NAME: 2.a What statistic would estimate the parameter (or parameter
combination)? In the case of multiple parameters, give the
statistic typically used to test these hypotheses. '3‘ __1__________ 2b. Find a distribution fact which is directly related to this
statistic. Note that it will also typically contain the parameter
or parameter combination. Write DF # and the DR including distribution "'
11‘
D.F.# \ ;i.e., cr/Vﬁ 449‘) 2c.From this distribution fact, give the test statistic and known
distribution under the null hypothesis. Identify this test
statistic with a brief identiﬁer (like 20b5, tabs, xobs, Fobs, etc.)
which reminds you of the distribution under the null
hypothesis. (In some cases, the test statistic and distribution
under the null hypothesis are already given in the distribution
fact above.) We typically derive the test statistic from the “:15 above distribution fact by substituting the value of the
unknown parameter for parameter combination) under the null hypothesis to get the test statistic taking care NOT to
substitute for the estimate of the unknown parameter (or parameter combination. (In some cases, a distribution fact will
actually be the test statistic in the next step.) 2 ...=L7~26>/<w/ws~>s<m>/z
N Nto, I)
3.Look at the test statistic:(20) ($745) [2 and (1t)le ‘5 < 210 . Decide what extreme values of
the test s atistic would tend to indicate that H1 is true (general direction). We do this in two steps when possible. under H0 3a.ldentify the estimate of the unknown parameter (or
parameter combination) in the hypotheses (given typically in
2a above), and indicate what extreme values of this estimate
would indicate the alternative hypothesis is true (general
direction) Reject Ho if: 7 i5 "Mix [(35 iban 3b.Use this information as it applies to the test statistic, and
indicate what extreme values of the test statistic would
indicate the alternative hypothesis is true (general direction) Reject HO if: 2 .ob, very heap:th Problem No. (cont) ReVWW l 3 a: 3c.Draw a number line, label it under the right side 2 .Obs,
and shade and identify where you would Reject Ho all“ t
e labs 4.Use the results of #2 and #3 along with the value of the
probability of a type I error to ﬁnd the cutoff point(s) for the acceptance and rejection region(s). Note 0t=Pr{Reject H0H0 true}.IN THE SPACE 4a—c belowza. To do this, we recopy
the previous number line including the shaded parts of the
axis, drawing and labeling the distribution of the test statistic
under H0 above the number line. b. We then shade the areas
under the curve above the shaded parts of the axes and label
these areas above the curve in terms of 0F, typically as 0t: or
oc=/2. c. We label below the axes the border(s) between the
shaded and unshaded regions in terms of X W where “X” is
the symbol we used when we wrote X obs above. Sometimes
we will write —X m and +X cm , and at other times we may use
X cm, Lower and X Cmgupper when we have multiple critical
values. Complete 4d4i below to ﬁnd the critical values. 4d. ’If you do NOT have a Table for the distribution in la,
convert the distribution picture in 4ac to one in 4d for which you have a Table. [Example: 4a—c N(u,c52) => 4d N(0, 1) using
z=(xu)/o ] 4ac 4d l
l , NOW i l .l0 l
‘  Zr} 0
4e. Symbol for unknown in 4a—c: h, is an area or@ 4f.Bound on unknown: 9 6‘) < Zu¢< 0 4g. Draw TP in 4g below. 4h. Put 4d (4ac if no 4d) in terms
of the TP in 4h below. Look up table values representing
them as TP and get your answer in terms of the unknown. 4g Table Picture (TP): 4h (cont) 4h. (cont) Final answer in terms of unknown in 4e:
2654* s ' \~ 4i. Check if ﬁnal answer is consistent with bound in 4f. 1 5.C1early state your decision rules. It may be advantageous to
NOT include the critical boundary points in the Accept Ho
region and to list the Reject Ho region as “otherwise” AcceptHo 2657‘llzg ; Reject Ho otherwise 6a. Calculate below your test statistic given in 2c using
information in 1a. ﬁbs 4; 5 '3 6b. Using the decision rules in #5, indicate whether you will
Accept H0 or Reject Ho: new» 6c. Interpret your results. (If you have transformed the
problem, typically we interpret the results in terms of the
original parameters.) Using a. “3% level ciggniftmnca) [W‘ujd t“, 9&9.st legawe Problem No. VQVfw 3% l; 4* Power (or [3) Calculation Template (9.23 .09) 1. Since we generally end up with ﬁnding an area under
a curve as the answer, call that area A, and write for your
problem below: A = Pr{Accept H0 or Reject Ho (pick
one)  HI true} A: E ‘l Acmé“ol\’\i+mg} (6'37 2. Next substitute in for the phrase to the left of the
the decision rule region (see HT. Step 5) and substitute
to the right of the “l” ”6 = a”, where “9” is the unknown
parameter and “a” is the value of the unknown parameter
at which we are making this evaluation. = Pr‘i mywtlﬁl P5 22} 3. Unless you know the distribution under “9 = a”, you
will next substitute for the observed statistic in the
expression to the left of the “I” using its deﬁnition (see H.T. Step 20). = M 't‘aZ">”""8 W22} 4. Next solve for the statistic in the expression to the left
of the “”. FA ? > 2e~<zxt283l a: 22} PA 5; > 23m \ W22}
5 .State the relevant distribution fact: UsingD.F.il : Which in this case, substituting the known values and
with “6 = a”, the DF for our problem becomes: 131‘“ V ’U N ( 2’2) “’72; =63? 1: 22) s M22521) 6.a. Below, draw a number line, label the variable
(to the left of the “I” at the end of step 4 above)
under the right side of the number line. Name: 6b. Draw (and label) above the number line below the
distribution of this statistic from the end of Step 5 above. 6c. Mark on the number line below & shade above the
region identiﬁed to the left of the “I” at the end of HT.
Step 4 above. Label this shaded area in terms of A. 6d. If you do NOT have a Table for the distribution in 1a,
convert the distribution picture in 6a—c to one in 6d for which you have a Table. [Examplez 6a—c N(u,62) => 6d N(0,1) using
z=(xu)/o] 233.“22 é f a 6a—c Z 6d
. 
“(13%): We")
I . // A I /’A
I 4” . //
22 Q .12
6e. Symbol for unknown in 6a—c: A , is aor score?
0 < A < 05 6g. Draw TP in 6g below. 6h. Put 6d (6a—c if no 6d) in terms
of the TP in 6h below. Look up table values representing
them as TP and get your answer in terms of the unknown. 6f.Bound on unknown: 6g Table Picture (TP): f 6h 6h (cont) 6h. (cont) Final answer in terms of unknown in 6e: 6i. Check if ﬁnal answer is consistent with bound in 6f. Y ...
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This document was uploaded on 11/01/2011 for the course MANAGEMENT 385 at Rutgers.
 Fall '10
 Szatrowski

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