Answer:
The 90% confidence interval for the mean score of all takers of this test is between 59.92 and 64.08. The lower end is 59.92, and the upper end is 64.08.
Step-by-step explanation:
We have that to find our 
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of 
.
So it is z with a pvalue of 
, so 
Now, find the margin of error M as such
In which 
is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 62 - 2.08 = 59.92
The upper end of the interval is the sample mean added to M. So it is 62 + 2.08 = 64.08.
The 90% confidence interval for the mean score of all takers of this test is between 59.92 and 64.08. The lower end is 59.92, and the upper end is 64.08.
Answer:
The point estimate for this problem is 0.48.
Step-by-step explanation:
We are given that a University wanted to find out the percentage of students who felt comfortable reporting cheating by their fellow students.
A survey of 2,800 students was conducted and the students were asked if they felt comfortable reporting cheating by their fellow students. The results were 1,344 answered "Yes" and 1,456 answered "no".
<em>Let </em>
<em> = proportion of students who felt comfortable reporting cheating by their fellow students</em>
<u></u>
<u>Now, point estimate (</u><u>) is calculated as;</u>
where, X = number of students who answered yes = 1,344
n = number of students surveyed = 2,800
So, Point estimate () = 
= <u>0.48 or 48%</u>
Answer:
y=20x
Step-by-step explanation:
Since y represents the total cost y is the total. 20x because it is 20 dollars an hour and x after 20 because we do not know how many hours they will plan for lessons.
Answer:
I think it would be Column 2
Step-by-step explanation:
Answer:
19,25
Step-by-step explanation:
chia hi
