1 / 3100%
a a a a 1
Homeworka 13
Questiona 40:a Entera thea dataa intoa youra calculatora ora computera p-value
Questiona 41:a a alphaa =a 0.05
Decision:a wea rejecta thea nulla hypothesis
Conclusion:a therea isa noa sufficienta evidencea toa affirma thea samplea dataa set.a
Questiona 42:a alphaa =a 0.01
Decision:a wea rejecta thea nulla hypothesisa
Conclusion:a therea isa noa substantiala evidencea toa affirma thea samplea dataa set.a
Questiona 43:a Namea onea assumptiona thata musta bea true
Thea populationa ofa eacha ofa thea samplesa area normallya distributed
Questiona 44:a whata isa thea othera assumptiona thata musta bea true
Thea samplesa area independenta froma onea anothera
Questiona 45:a Statea thea nulla anda alternativea hypotheses
Nulla hypothesis:a σ
1
a =a σ
2
Alternativea hypothesis:a σ
1
a <a σ
2
Questiona 46:a Whata isa s1a ina thisa problem?
S2a =a 2.01
Questiona 47:a Whata isa s2a ina thisa problem?
S2a =a 4.11
Questiona 51:a Isa thea claima accurate
Thea claima isa nota accurate.a Thisa isa becausea ata 10%a levela ofa significancea thea nulla hypothesisa
isa nota rejecteda anda thea dataa doa nota showa thata thea variationsa ina drivea timesa fora thea firsta
workera isa lessa thata thea variationsa ina drivea timesa fora thea seconda workera
Questiona 52:a Statea thea nulla anda alternativea hypotheses
Nulla hypothesis:a 𝜎
1
2
= 𝜎
2
2
a
Alternativea hypothesis:a 𝜎
1
2
< 𝜎
2
2
Questiona 53:a Whata isa thea Fa statistic
F-a statistica =a 2.8675
Questiona 59:a Statea SS
between
,a SS
within
,a anda Fa statistic
a a a a 2
Fa statistica =a 0.265
SS
between
a
=
a
26
SS
within
a
=
a
441
Questiona 62:a degreesa ofa freedoma a denominator:a dfa (denom)
Dfa (denom)a =15
Questiona 64:a Usinga aa significancea levela ofa 10%,a testa thea hypothesisa thata thea threea
formulasa producea thea samea meana weighta gain
Nulla hypothesis:a µ
L
a =a µ
T
a =a µ
Ja
Alternativea hypothesis:a µ
L
a a µ
T
a a µ
J
Degreea ofa freedoma (df)a =a 2
p-valuea =a 0.5304;a undera 5%a levela ofa significance.a Therefore,a doa nota rejecta thea nulla
hypothesis.a
Questiona 68:a Assumea thata alla distributionsa area normal,a thea foura populationa standarda
deviationsa area approximatelya thea same,a anda thea dataa werea collecteda independentlya anda
randomly.a Usea aa levela ofa significancea ofa 0.05
p-valuea =a 07435;a undera 5%a significancea level.a Therefore,a wea accepta thea nulla hypothesis.a
Questiona 70:a Assumea thata alla distributionsa area normal,a thea foura populationa standarda
deviationsa area approximatelya thea same,a anda thea dataa werea collecteda independentlya anda
randomly.a Usea aa levela ofa significancea ofa 0.05
p-valuea =a 0.6571;a undera 5%a significancea level.a Therefore,a wea accepta thea nulla hypothesis
Questiona 73:a
Thea dataa froma thea charta appeara toa bea normallya distributed.a
Nulla hypothesis:a μ
1
a =a μ
2
a =a μ
3
Alternativea hypothesis:a μ
1
a a μ
2
a a μ
3
P-valuea =a 0.0004;a undera 1%a levela ofa significance.a Therefore,a wea rejecta thea nulla hypothesisa
asa thea evidencea isa sufficienta toa affirma thata thea numbera ofa eggsa laida ina 14a daysa isa equala ina
alla samples.a
Questiona 77:a Whicha twoa magazinea typesa doa youa thinka havea thea samea variancea ina
length
Homea decoratinga anda newsa magazines
a a a a 3
Questiona 79:a Isa thea variancea fora thea amounta ofa money,a ina dollars,a thata shoppersa spenda
ona Saturdaysa ata thea malla thea samea asa thea variancea fora thea amounta ofa moneya thata
shoppersa spenda ona Sundaysa ata thea mall?
Nulla hypothesis:a 𝜎
1
= 𝜎
2
a
Standarda deviationa (Saturday)=a 42.67
Standarda deviationa (Sunday)a =a 36.80
P-a valuea =a 0.00367;a undera 5%a levela ofa significance.a Therefore,a wea rejecta thea nulla
hypothesis.a
Questiona 86:a
Nulla hypothesis:a μa a 129
Alternativea hypothesis:a μa <a 129
Thea t-testa =a 1.209a witha aa p-valuea ofa 0.8792a undera aa 5%a significancea level.a Therefore,a
therea isa sufficienta evidencea nota toa rejecta thea nulla hypothesis.a
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