Using the standard z table, a 95% confidence interval for p is (0.5327,0.6173).
In the given question,
A poll is taken in which 302 out of 525 randomly selected voters indicated their preference for a certain candidate.
We have to find a 95% confidence interval for p.
From the question, x=302, n=525
So estimation point,
P=x/n
P=302/525
P=0.575
Now the z value of 95% confidence interval is 1.960 using the standard z table.
Margin of Error (E)=z×√{P(1−P)}/n
Now putting the value
E=1.960×√{0.575(1−0.575)}/525
E=1.960×√(0.575×0.425)/525
E=1.960×√0.244/525
E=1.960×√0.000465
E=1.960×0.0216
E=0.0423
At 95% confidence interval for p is
P−E ≤ p ≤ P+E
Now putting the value
0.575−0.0423≤ p ≤0.575+0.0423
0.5327≤ p ≤0.6173
Hence, a 95% confidence interval for p is (0.5327,0.6173).
To learn more about confidence interval link is here
brainly.com/question/24131141
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