Can someone please give me the answers to this? ... please ...
1 answer:
You might be interested in
The expected frequency of east campus and passed is C. 42 students
<h3>How to calculate the value?</h3>The table for expected frequency is ,
East Campus West Campus Total
Passed (84*100)/22=42 (84*100)/200 =42 84
Failed (116*100)/200=58 (116*100)/22=58 116
Total 100 100 200
Passed = 84×100/200
= 42
Therefore, the expected frequency of East Campus and Passed is 42 students.
Learn more about frequency on:
#SPJ4
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.