Sukumar R. answered 10/03/22
Experienced Statistics Tutor
Population proportion
Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 128 with 24% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percent’s) accurate to three decimal places.
- Confidence interval = ()
- Express the same answer as a tri-linear inequality using decimals (not percent’s) accurate to three decimal places. < p <
- Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places. p = ±
Step-1:
Sample Size = n = 128
Sample proportion of success = p̂ = 0.24
Sample proportion of no success (failures) = q̂ = 1 - p̂ = 1 – 0.24 = 0.76
Check n * p̂ * q̂ = 128 *0.24*0.76 = 24.3472 >= 10
Hence, the distribution can be approximated as a normal distribution.
Sample point estimate of success = 0.24*128 = 30.72
Step-2
Confidence Interval = 99%
For a normal distribution, Zc for 99% confidence interval:
Zc = ± 2.58
Step-3
Margin of Error (MoE) = Zc * SQRT(p̂ * q̂ / n) = ± 2.58 * SQRT(0.24 * 0.76 / 128) = ±0.097
Step-4
Confidence Interval Low Limit = CIL = (Sample Proportion – MoE) = 0.24–0.097 = 0.143
Confidence Interval High Limit = CIH = (Sample Proportion + MoE) = 0.24+0.097 = 0.337
Population mean Low Limit = 0.143*128 = 18.304
Population mean High Limit = 0.337*128 = 43.136
Sample point estimate of success = 0.24*128 = 30.72
Answers:
- Confidence interval = (0.337-0.143) = (0.194)
- 0.143< p <0.337 (99% sure that true population proportion will lie between these two proportions of 0.143 and 0.337)
- 18.304 < 30.72 < 43.136 (99% sure that true population mean will lie between these two values of 18.304 and 43.136)