the top question is the second answer and the second question is also the second answer
Answer:
95% confidence interval: (0.325 ,0.383)
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 231
Sample mean = 0.354 ppm
Sample standard deviation = 0.231 ppm
95% confidence interval:
Putting the values, we get,
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.
You want to do base × height to get your answer
3×2 = 6
So the area of the poster is 6 sq in.
If this helps pls mark brainliest pts