I found this :) hope it helps
Answer:
With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.
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:
For this problem, we have that:
95% confidence level
So 
, z is the value of Z that has a pvalue of 
, so 
.
With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is
We need a sample size of at least n, in which n is found M = 0.04.
With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.
1/40 would be your answer because 0.025 is would be 25/1000. Simplify that then you would get 1/40.
The surveyor selects the twentieth student who enters the building. The student do not enter the building to be surveyed but for other reasons and the researcher take advantage of that and selects them. This is a form of continence sampling.
The researcher selects every twentieth student and not any other. Therefore, the sampling is systematic in nature.
Finally, students selected to participate in the survey are free to choose if to participate in he survey or not. Therefore, it is voluntary kind of sampling.
Therefore, the applicable sampling methods are convenience, systematic, and voluntary.
1) 17, 29, 32, 9, 30, 14, 8, 39, 11, 32, 23<br>
Minimum :<br>
Maximum:<br>
Q1:<br>
Q2:<br>
Q3:
Lisa [10]
Answer:
Minimum: 8
Maximum: 39
Q1: 11
Q2: 23
Q3: 32
Step-by-step explanation:
Set the numbers into numerical order
8 9 11 14 17 23 29 30 32 32 39
Then separate into quarters
[8 9 11] [14 17 23] [29 30 32] 32 39
