2. A survey of first-time home buyers found that the sample mean annual income was $46,300. Assume that the survey used a sample of 28 first-time home buyers and that the sample standard deviation was $1,100. Compute and explain a 90% confidence interval estimate of the population mean.
Accepted Solution
A:
Answer:the confidence interval is from 46641.96 to 45958.04Step-by-step explanation:Using this formula X ± Z (s/√n)WhereX = 46300 --------------------------MeanS = 1100----------------------------- Standard Deviationn = 28 ----------------------------------Number of observationZ = 1.645 ------------------------------The chosen Z-value from the confidence table belowConfidence Interval Z80%. 1.28285% 1.44090%. 1.64595%. 1.96099%. 2.57699.5%. 2.80799.9%. 3.291Substituting these values in the formulaConfidence Interval (CI) = 46300 ± 1.645(1100/√28)CI = 46300 ± 1.645(1100/5.2915)CI = 46300 ± 1.645(207.8806)CI = 46300 ± 341.9636CI = 46300 + 341.9636. ~. 46300 - 341.9636CI = 46641.96. ~. 45958.04In other words the confidence interval is from 46641.96 to 45958.04