Question 1194260: A survey was given to a random sample of 650 voters about their preference for a presidential candidate. those surveyed, 481 respondents preferred Candidate A. Determine a 95% confidence interval for the proportion of people who prefer Candidate A, rounding values to the nearest thousandth.
Answer by math_tutor2020(3817)   (Show Source):
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p = population proportion of people who prefer candidate A.
x = number of people who prefer candidate A
n = sample size
x = 481
n = 650
phat = best estimate of p
phat = sample proportion of people who prefer candidate A
phat = x/n
phat = 481/650
phat = 0.74
74% of the sample prefers candidate A.
At 95% confidence, the z critical value is roughly z = 1.960 which you use a reference table to determine.
E = margin of error
E = z*sqrt(phat*(1-phat)/n)
E = 1.960*sqrt(0.74*(1-0.74)/650)
E = 0.033721
This is approximate
L = lower boundary of the confidence interval
L = phat - E
L = 0.74 - 0.033721
L = 0.706279
L = 0.706
U = upper boundary of the confidence interval
U = phat + E
U = 0.74 + 0.033721
U = 0.773721
U = 0.774
The 95% confidence interval is roughly (0.706, 0.774) which is of the format (L, U)
This is equivalent to writing 0.706 < p < 0.774 which is in the format L < p < U
We are 95% confident the population proportion (p) of people who prefer candidate A is between 0.706 and 0.774
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