Answer:
0.3830,0.6170
Step-by-step explanation:
Given that a process for manufacturing an electronic component yields items of which 1% are defective.
n =100 and p = 0.01
Here X no of defectives is binomial since independence and two outcomes.
Approximation to normal would be
X is N(
X is N(1,0.995)
a) the probability that the process continues given the sampling plan described
=
(with continuity correction)
=
b) the probability that the process continues even if the process has gone bad (i.e., if the frequency of defective components has shifted to 5.0% defective)
1-0.3830
=0.6170
Answer:
m24,m249,m416,kar98k
Step-by-step explanation:
過文はまやたらあわまはやたはたやはたやたなまやたはたはまはたやあはたはたはたはたはた
Answer:
.12
Step-by-step explanation:
set up two ratios the same way
=
cross multiply and solve for y
4.8(y) = .09(6.4)
4.8y = .576
y = .12
Answer:
1.
The null hypothesis
H0: Ud=0
The alternate hypothesis
H1:Ud≠0
2. Test statistic = -3.04
3. P value = 0.039
4. Reject h0
Step-by-step explanation:
X = right arm
Y = left arm
d = difference between both arms
X. Y. d(x-y). d²
145. 173. -28. 784
142. 163. -21. 441
116. 182. -66. 4356
133. 148. -15. 225
134. 149. -15. 225
Total
d = -145
d² = 6031
We have sample space n = 5
d' = -145/5
= -29
Sd = √1/n-1(Σd²-(Σd)²/n
Sd = √1/4(6031-(-145)²/5
= √1/4(6931-4205)
= √456.5
= 21.366
To get t
=d'/(sd/√n)
= -29/(21.366/√5)
= -3.035
In summary
This is a two tailed test
1.
The null hypothesis
H0: Ud=0
The alternate hypothesis
H1:Ud≠0
2. Test statistic = -3.04
3. P value = 0.039
4. We have pvalue to be 0.039 which is less than the level of significance 0.05
0.039<0.05, so we take decision to reject h0 which is the null hypothesis
We have enough evidence to support claim that there is a measurement difference between left and right arms