Answer:
(a) 0.0833 or 8.33%
(b) 0.40 or 40%
Step-by-step explanation:
Parts line one (n1) = 1,000 parts
Defects line one (d1) = 100 parts
Parts line two (n2) = 2,000 parts
Defects line two (d2) = 150 parts
Total number of parts (n) = 3,000 parts
a. Probability of a randomly selected part being defective:
The probability is 0.0833 or 8.33%
b. Probability of a part being produced by line one, given that it is defective:
The probability is 0.40 or 40%.
Answer:
The P-value is 0.0353.
Step-by-step explanation:
We are given the six measurements of a sprinter's reaction time show below;
X = 0.152, 0.154, 0.166, 0.147, 0.161, and 0.159 seconds.
Let = <u><em>mean sprinter's reaction time</em></u>
So, Null Hypothesis, :
0.150 seconds {means that the mean sprinter's reaction time is at most 0.150 seconds}
Alternate Hypothesis, :
> 0.150 seconds {means that the mean sprinter's reaction time is more than 0.150 seconds}
The test statistics that will be used here is One-sample t-test statistics because we don't know about the population standard deviation;
T.S. = ~
where, = sample mean =
=
= 0.1565 seconds
s = sample standard deviation = = 0.0068 seconds
n = sample of measurements = 6
So, <u><em>the test statistics</em></u> = ~
= 2.341
The value of t-test statistics is 2.341.
<em>Now, the P-value of the test statistics is given by the following formula;</em>
P-value = P( > 2.341) = 0.0353.
{Interpolating between the critical values at 5% and 2.5% significance level}