Using the z-distribution, as we are working with a proportion, it is found that 1016 constituents are required.
<h3>What is a confidence interval of proportions?</h3>
A confidence interval of proportions is given by:

In which:
is the sample proportion.
The margin of error is given by:

In this problem, we have a 95% confidence level, hence
, z is the value of Z that has a p-value of
, so the critical value is z = 1.96.
The estimate is of
, while the margin of error is of M = 0.03, hence solving for n we find the minimum sample size.






Rounding up, 1016 constituents are required.
More can be learned about the z-distribution at brainly.com/question/25890103
Answer:
The answer is b.
Step-by-step explanation:
x-4=-2
+4. +4
x= 2
Answer: 99% of confidence interval for the population proportion of employed individuals who work at home at-least once per week
//0.20113,0.20887[/tex]
Step-by-step explanation:
<u>step 1:-</u>
Given sample size n=200
of the 200 employed individuals surveyed 41 responded that they did work at home at least once per week
Population proportion of employed individuals who work at home at least once per week P = 
Q=1-P= 1-0.205 = 0.705
<u>step 2:-</u>
Now 
=0.0015
<u>step 3:-</u>
<u>Confidence intervals</u>
<u>using formula</u>


=0.20113,0.20887[/tex]
<u>conclusion:</u>-
99% of confidence interval for the population proportion of employed individuals who work at home at-least once per week
//0.20113,0.20887[/tex]