Answer: a. 1.981 < μ < 2.18
b. Yes.
Step-by-step explanation:
A. For this sample, we will use t-distribution because we're estimating the standard deviation, i.e., we are calculating the standard deviation, and the sample is small, n = 12.
First, we calculate mean of the sample:


2.08
Now, we estimate standard deviation:


s = 0.1564
For t-score, we need to determine degree of freedom and
:
df = 12 - 1
df = 11
= 1 - 0.95
α = 0.05
0.025
Then, t-score is
= 2.201
The interval will be
± 
2.08 ± 
2.08 ± 0.099
The 95% two-sided CI on the mean is 1.981 < μ < 2.18.
B. We are 95% confident that the true population mean for this clinic is between 1.981 and 2.18. Since the mean number performed by all clinics has been 1.95, and this mean is less than the interval, there is evidence that this particular clinic performs more scans than the overall system average.
The two bars ( | | ) symbolize absolute value
Absolute value is essentially just making the number given positive.
If the given number is already positive then the absolute value is the same as the original number
|-13| is 13
The absolute value of -13 is 13
Hope this helped!
~Just a girl in love with Shawn Mendes
The answer is 18 I’m pretty certain so D
Answer:
Standard deviation of given data = 3.16227
Step-by-step explanation:
<u><em>Step(i)</em></u>:-
Given sample size 'n' = 5
Given data 4, 6,8,10,12

Mean of the sample x⁻ = 8
Standard deviation of the sample

<u><em>Step(ii)</em></u>:-
Given data
x : 4 6 8 10 12
x-x⁻ : 4 - 8 6-8 8-8 10-8 12-8
(x-x⁻) : -4 -2 0 2 4
(x-x⁻)² : 16 4 0 4 16

S.D = √10 = 3.16227
<u><em> Final answer</em></u>:-
The standard deviation = 3.16227