Question

4. Find the margin of error E. In a random sample of 151 college students, 84 had part-time jobs. Find the margin of error E for the 95% confidence interval used to estimate the population proportion. Round your answer to four decimal places. Answer 0.0792

5. Find the minimum sample size required to estimate the population proportion p: Margin of error: 0.10; confidence level: 95%; from a prior study, is known to be 66%.

6. Find the minimum sample size required to estimate the population proportion p: In a certain population, body weights are normally distributed with a mean of 152 pounds and a standard deviation of 26 pounds. How many people must be surveyed if we want to estimate the percentage p who weigh more than 180 pounds? Assume that we want 97.5% confidence that the error is no more than 3.9 percentage points; no prior results on estimating p are available.

Answer #1

4)

Sample proportion = 84 / 151 = 0.5563

= 1 - 0.95 = 0.05

Margin of error = Z/2 * sqrt [ ( 1 - ) / n ]

= 1.96 * sqrt [ 0.5563 ( 1 - 0.5563) / 151 ]

= **0.0792**

5)

Sample size = Z2/2 * p ( 1 - p) / E2

= 1.962 * 0.66 ( 1 - 0.66) / 0.102

= 86.2

**Sample size = 87** (Rounded up to nearest
integer)

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