Assume that a sample is used to estimate a population proportion .

Question

Assume that a sample is used to estimate a population proportion p. Find the 90% confidence interval for a sample of size 107 with 58 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

The answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Answer 1

The sample proportion is the number of successes divided by the sample size:

p = 58/107 = 0.5413

The standard error of the sample proportion is given by:

SE = sqrt(p * (1 - p) / n) = sqrt(0.5413 * (1 - 0.5413) / 107) = 0.0343

The 90% confidence interval is given by:

p - Z_critical * SE <= p <= p+ Z_critical * SE

where Z_critical is the critical value for the normal distribution such that 90% of the area under the curve is to the left of Z_critical. For a 90% confidence interval, Z_critical = 1.645 (to 3 decimal places).

0.5413 - 1.645 * 0.0343 <= p <= 0.5413 + 1.645 * 0.0343

So, the 90% confidence interval for p is approximately:

0.475 <= p <= 0.608