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Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal

places.
95% confidence; the sample size is 10,000, of which 40% are successes
0.0096
0.0126
0.0072
0.0110
Ik The answer is 0.0096 but don’t know how to get that. Would someone show steps?

User Jmattheis
by
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2 Answers

15 votes
15 votes

Final answer:

To calculate the margin of error for a population proportion with a 95% confidence level and a sample size of 10,000 with 40% successes, use the formula E = Z * sqrt[ (p * (1 - p)) / n ], where Z is 1.96. The calculated margin of error E is 0.009604, which rounds to 0.0096.

Step-by-step explanation:

To find the margin of error E for a population proportion p with a 95% confidence level, given a sample size of 10,000 with 40% successes, you use the formula for the confidence interval of a population proportion:

E = Z * √[ (p * (1 - p)) / n ]

Where:

E is the margin of error

Z is the Z-score corresponding to the confidence level

p is the sample proportion (0.40 in this case)

n is the sample size (10,000)

For a 95% confidence level, the Z-score is approximately 1.96. Plugging the values into the formula:

E = 1.96 * √[ (0.40 * (1 - 0.40)) / 10,000 ]

E = 1.96 * √(0.24 / 10,000)

E = 1.96 * √(0.000024)

E = 1.96 * 0.0049

E = 0.009604, which when rounded to four decimal places is 0.0096.

User Grkvlt
by
3.0k points
26 votes
26 votes
Just put 0.0096 in the ansewr bar
User Barrowc
by
2.9k points
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