Answer:
Kindly check explanation
Step-by-step explanation:
Given the following :
Group 1:
μ1 = 59.7
s1 = 2.8
n1 = sample size = 12
Group 2:
μ2 = 64.7
s2 = 8.3
n2 = sample size = 15
α = 0.1
Assume normal distribution and equ sample variance
A.)
Null and alternative hypothesis
Null : μ1 = μ2
Alternative : μ1 < μ2
B.)
USing the t test
Test statistic :
t = (m1 - m2) / S(√1/n1 + 1/n2)
S = √(((n1 - 1)s²1 + (n2 - 1)s²2) / (n1 + n2 - 2))
S = √(((12 - 1)2.8^2 + (15 - 1)8.3^2) / (12 + 15 - 2))
S = 6.4829005
t = (59.7 - 64.7) / 6.4829005(√1/12 + 1/15)
t = - 5 / 2.5108165
tstat = −1.991384
Decision rule :
If tstat < - tα, (n1+n2-2) ; reject the Null
tstat < t0.1,25
From t table :
-t0.1, 25 = - 1.3163
tstat = - 1.9913
-1.9913 < - 1.3163 ; Hence reject the Null
Answer:
Step-by-step explanation:
i'll do a few of them
1) one inch plus a half inch plus an eighth inch plus a sixteenth inch
1 + 1/2 + 1/8 + 1/16 read the tick marks
1 + 8/16 + 2/16 + 1/16 find a common denominators
1 and 11/16 add the sixteenth numerator
2)
3)
4) 6 inch plus an eighth inch plus a sixteenth inch
6 + 1/8 + 1/16 read the tick marks
6 + 2/16 + 1/16 find a common denominators
6 and 3/16 add the sixteenth numerator
5) nine inch plus a quarter inch plus an eighth inch
9 + 1/4 + 1/8 read the tick marks
9 + 2/8 + 1/8 find a common denominators
9 and 3/8 add the sixteenth numerator
3 1/3 times 2 1/2 =3 2/6 times 2 3/6 =6 15/6=8 3/6
8 3/6 I THINK
The answer is B
hopefully this helps
