Answer:
We cannot say that the mean wake time are different before and after the treatment, with 98% certainty. So the zopiclone doesn't appear to be effective.
Step-by-step explanation:
The goal of this analysis is to determine if the mean wake time before the treatment is statistically significant. The question informed us the mean wake time before and after the treatment, the number of subjects and the standard deviation of the sample after treatment. So using the formula, we can calculate the confidence interval as following:
![IC[\mu ; 98\%] = \overline{y} \pm t_{0.99,n-1}\sqrt{\frac{Var(y)}{n}}](https://tex.z-dn.net/?f=IC%5B%5Cmu%20%3B%2098%5C%25%5D%20%3D%20%5Coverline%7By%7D%20%5Cpm%20t_%7B0.99%2Cn-1%7D%5Csqrt%7B%5Cfrac%7BVar%28y%29%7D%7Bn%7D%7D)
Knowing that 
:
![IC[\mu ; 98\%] = 98.9 \pm 2.602\frac{42.3}{4} \Rightarrow 98.9 \pm 27.516](https://tex.z-dn.net/?f=IC%5B%5Cmu%20%3B%2098%5C%25%5D%20%3D%2098.9%20%5Cpm%202.602%5Cfrac%7B42.3%7D%7B4%7D%20%5CRightarrow%2098.9%20%5Cpm%2027.516)
![IC[\mu ; 98\%] = [71.387 ; 126,416]](https://tex.z-dn.net/?f=IC%5B%5Cmu%20%3B%2098%5C%25%5D%20%3D%20%5B71.387%20%3B%20126%2C416%5D)
Note that ![102.8 \in [71.384 ; 126.416]](https://tex.z-dn.net/?f=102.8%20%5Cin%20%5B71.384%20%3B%20126.416%5D)
so we cannot say, with 98% confidence, that the mean wake time before treatment is different than the mean wake time after treatment. So the zopiclone doesn't appear to be effective.
Answer:
0.2941 = 29.41% probability that it was manufactured during the first shift.
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
In which
P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Defective
Event B: Manufactured during the first shift.
Probability of a defective item:
1% of 50%(first shift)
2% of 30%(second shift)
3% of 20%(third shift).
So
Probability of a defective item being produced on the first shift:
1% of 50%. So
What is the probability that it was manufactured during the first shift?
0.2941 = 29.41% probability that it was manufactured during the first shift.
A quadratic in vertex form can be represented as
a represents reflection over the x-axis, and a vertical stretch or compress
- is reflection and a fraction (1/2) represents a compression.
-h represents a shift of that many units to the right (-2 shifts to the
right two units)
k represents a shift up or down (-2 is shifting down 2 units_
Reflected over the x-axis, Vertically compressed by a factor of 1/2, shifted 2 units to the right, and shifted 2 units down
Answer:
This is 0.14 to the nearest hundredth
Step-by-step explanation:
Firstly we list the parameters;
Drive to school = 40
Take the bus = 50
Walk = 10
Sophomore = 30
Junior = 35
Senior = 35
Total number of students in sample is 100
Let W be the event that a student walked to school
So P(w) = 10/100 = 0.1
Let S be the event that a student is a senior
P(S) = 35/100 = 0.35
The probability we want to calculate can be said to be;
Probability that a student walked to school given that he is a senior
This can be represented and calculated as follows;
P( w| s) = P( w n s) / P(s)
w n s is the probability that a student walked to school and he is a senior
We need to know the number of seniors who walked to school
From the table, this is 5/100 = 0.05
So the Conditional probability is as follows;
P(W | S ) = 0.05/0.35 = 0.1429
To the nearest hundredth, that is 0.14
Answer:
18,500. Years. 103%
