What is the margin of error for a yes/no survey of 200 people with a proportion of 65% who answered \"yes\"?. A. ±3.37%. B. ±5.0
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Using the z-distribution, a sample of 142,282 should be taken, which is not practical as it is too large of a sample.
<h3>What is a z-distribution confidence interval?</h3>The confidence interval is:
The margin of error is:
In which:
is the sample mean.
- z is the critical value.
- n is the sample size.
is the standard deviation for the population.
Assuming an uniform distribution, the standard deviation 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 sample size is found solving for n when the margin of error is of M = 0.006, hence:
n = 142,282.
A sample of 142,282 should be taken, which is not practical as it is too large of a sample.
More can be learned about the z-distribution at brainly.com/question/25890103
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Hoi!
To solve this, first plug in the values for x and y.
x = 2, so anywhere you see x, put 2 in its place.
y = 3, so anywhere you see y, put 3 in its place.
4 × 2 = 8
× 3 = 
= 
is your answer.
To solve this, first plug in the values for x and y.
x = 2, so anywhere you see x, put 2 in its place.
y = 3, so anywhere you see y, put 3 in its place.
4 × 2 = 8