b. ans
p
0.XX
n
200
X
-X.XX
P XXXXX = 0.XXXX
C ans: Here p XXXXX XX X.XXX XXX X=0.01 so XX XXX XXXXXX XXXX XXXXXXXXXX
We may conclude XXXX XXX true proportion is less XXXX 62%
9
a ans:
B XXX :
XX
xbar
73.40
s
XX.XXX
t
3.XXX
P XXXXX= 0.0035
C XXX: Here p XXXXX XX 0.0035 and α=0.01 so we XXX reject XXXX hypothesis
XX may conclude that XXX XXXX mean is XXXXXXX then XX
10 a XXX:
a XXX:
:
r
-0.951
10
a
0.01
XXXX
-8.662
X value
0.XXXX
X XXXXXXXXXX XXXX XX X = -X.62133X + XX.41659
XXX estimate of XX we XXX = -X.XXXXX×40 + 91.XXXXX =66.XXXX