Answer:
The required sample size for the new study is 801.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of 
, and a confidence level of 
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of 
.
The margin of error is:
25% of all adults had used the Internet for such a purpose
This means that 
95% confidence level
So 
, z is the value of Z that has a pvalue of 
, so 
.
What is the required sample size for the new study?
This is n for which M = 0.03. So
Rounding up:
The required sample size for the new study is 801.
123 rounded to the nearest hundred is 100 i believe
Answer:
6
Step-by-step explanation:
just separate the ramge and domain
Answer:
Step-by-step explanation:
Remark
My guess is that what is confusing you is not what you have to do, but why it is disguised as g(n)
What you are doing in effect is setting up a table. You are also not certain where the table starts. And that is a problem. I will start it at zero, but it might be 1.
zero
n = 0
g(0) = 34 - 5*0
g(0) = 34
One
n = 1
g(1) = 34 - 5*1
g(1) = 34 - 5
g(1) = 29
Two
g(2) = 34 - 5*2
g(2) = 34 - 10
g(2) = 24
Three
g(3) = 34 - 5*3
g(3) = 34 - 15
g(3) = 19
Four
g(4) = 34 - 5*4
g(4) = 34 - 20
g(4) = 19
Answer
0 1 2 3 4
34 29 24 19 14
