A point moves in the plane in the following sequence.A point moves down 2 units, reflects over the x-axis, moves right 4 units,
1 answer:
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Answer:
The sample does not contradicts the manufacturer's claim.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 200 mg
Sample mean, = 211.5 mg
Sample size, n = 20
Alpha, α = 0.05
Sample standard deviation, s = 18.5 mg
a) First, we design the null and the alternate hypothesis for a two tailed test
We use Two-tailed t test to perform this hypothesis.
b) Formula:
Putting all the values, we have
c) Now,
Since, the calculated t-statistic does not lie in the range of the acceptance region(-2.0930,2.0930), we reject the null hypothesis.
Thus, the sample does not contradicts the manufacturer's claim.
d) P-value = 0.011934
Since the p-value is less than the significance level, we reject the null hypothesis and accept the alternate hypothesis.
Yes, both approach the critical value and the p-value approach gave the same results.
Answer:
The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.
Step-by-step explanation:
Using Bayes' Theorem
P(A|B) =
where
P(B|A) is probability of event B given event A
P(B|a) is probability of event B not given event A
and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.
For this problem,
Let P(B1) = Probability of machine B1 = 0.3
P(B2) = Probability of machine B2 = 0.2
P(B3) = Probability of machine B3 = 0.5
Let P(D) = Probability of a defective product
P(N) = Probability of a Non-defective product
P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003
P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006
P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010
Likewise,
P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997
P(N|B2) be probability of a non-defective product produced by machine 2 = 1 - P(D|B2) = 1 - 0.006 = 0.994
P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990
For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)
P(B1|N) = =
= 0.1138
Hence the probability that a non-defective product is produced by machine B1 is 11.38%.