SI
Assignment 2 .
Haris Kalim .
27839
On
% = 52 - 040
.D
5 = 2 .
807
21 = 95%
1 -
0 .
95 = 0 .
05
dF = 30 -
1 : 29
T -
tax - 1))
52 040 .
- 1 .
69(
Ft17g we are
6) 51 that confident that
17g 95 %
·
mean
of small at least
the true mean
bags is
51
Ag
·
.
2)ithepackagclaim hut each,a
conclude that
B is
highe sample ,
the advatised
,
we mean
i
is
e
highefor consumers
weightfavorable than
weight .
d)
-
07
Yes we may a t-table test as
,
6 is unknown an n > 5.
,
⑬ Eo
a)
% : =0
zu
:
6) (1 = 95 % 1-0 9570
. .
05
df 20-1 19.
-
, =
2551 =
2 .
093 (1937)
Jo rental
ne
= 301 3 .
5 1 000
Earl
=:
=
F
S =
22 .
35
1-6 09
:
5
025 0
%
-
df = 20 - 1 = 19
71 2 093 .
(3)
81 4 .
60 .
5
S D
3 pop
.
Sp
: sample
Exl Hotel Guest Bill Anderson
a) No H = 600
Hall > 600
b) We can conclude that there is not
evidence that mean
enough to
suggest
guest bills are
increasing
) We would conclude that the is
amount of evience that
significant
indicates that mean
guest
bills one
increasing more $600 .
Ho 14
Ea M :
HaM > 14
b) If He cannot be rejected ,
then we
conclude that there is not evidence
enough
tosuggestthat
the new busine is
a
2) If Ho is rejected it indicates that there
,
is sufficient evidence that the new
will increase sale.
plan
6) Ho H-1
Ha(1 > 1
6 Type I error would be
reject
to the Ho
that I litre containe contains
stating
> 1 fat. ,
leading to False
advatising a
unfit to consume for consumes on
daily basis
2) A Type It evol would be to not
est
reg
the Ho when it is
actually
fake .
Anduson 9 - 11 p465
-20
3/558
:
E
00017
017 c 05 so
1) since p is 0 we con
. .
conclude that there is sufficient eviden
a)to ejeHoisusLaR h
, v
a)
T
b) 0009D
C) at x = 0 .
01
,
0 .
9292 > 0 .
01 we
can conclude there isn't sufficient
evidence
eject
to Ho
d) at 0 . 01 the critical approach give
us the value of 1 17 x-2 331 .
so
do
we not
ejectIts
a) samples
.
D =
145-15 => 7003
.
350
b)
as
at 05 conclude that
C) a = 0 we can
.
then is sufficient evidum to
rejectHo as 0 .
0008 20.05
2) Whing CVA
get-1
we 645 that is
.
1-2 .
003
,
here
vejet Ho
En 15 ,
16 , 20 p 466 Anderson
15a) +o = ( = 1856
Ha =
M < 1056
Thunjecting Ho will support seance a
b)
10525
prawe 0003YY =
C) at 2 = 0 . 05
,
we can conclude that
0 . 0344 c 0 . 05 so there is
sufficient evidence to
rejectHo
D)
Using critical
approach ,
we
get-1 -
645
which -1 . 825 ,
hunce can
leject to
16) a) Ho = 5173
Ha T
, 3173
b) 059
5565-3173
: =
2 .
=
M0793
c) at 0 .
05 20 .
9793
, so we conclude
that we don't have sufficient evidum to
Reject Ho
Ho = 838
Ha < 745
838
b)
> 100082
d) at a = 0 . 01 > 0 .
0082
,
so we can
conclude that then is sufficient evidence
Ho
reject
to
Andersen p472 E28 , 29 31 30 , ,
Ho 9
28)
Ha 9
1)
500066
c) at a : 01
,
0 .
0062 < 0 ·
01
,
home we
on
reject Ho
29)a) Ho = 21 62 ·
Ha = 21 62 .
b) = 0 5
p
.
c) No
d)
30)a) Ho = 6 .
4
Ha * 6 4 .
D) X = 6 4 .
05
31.Ho Y :3
: a = 0 .
2-
2
.
2 = 0 . 986470 .
5199
ot enoughea
to
suggest Ma
3)
