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.
Given:
Population proportion,
= 57% = 0.57
Population standard deviation, σ = 3.5% = 0.035
Sample size, n = 40
Confidence level = 95%
The standard error is
The confidence interval is
where
= sample proportion
z* = 1.96 at the 95% confidence lvvel
The sample proportion lies in the interval
(0.57-1.96*0.0783, 0.57+1.96*0.0783) = (0.4165, 0.7235)
Answer: Between 0.417 and 72.4), or between (41% and 72%)
We have that
smaller figure
ha=8.7
ra=1.6
larger figure
hb=10.44
rb=1.92
ha/hb=8.7/10.44----> 0.83
ra/rb=1.6/1.92-------> 0.83
ratio smaller figure to the larger figure
1.6/1.92-------> divided by 1.6 both members[1.6/1.6]/[1.92/1.6]-------> 1/1.2
the ratio ha/hb is equal to the ratio ra/rb
so
the smaller figure and the larger figure are similar
and
the ratio smaller figure to the larger figure is equal to----> 1/1.2
the answer is
the option yes 1/1.2
