Question

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.

Answer 1

the confidence interval is from 46641.96 to 45958.04

Step-by-step explanation:

Using this formula X ± Z (s/√n)

Where

X = 46300 --------------------------Mean

S = 1100----------------------------- Standard Deviation

n = 28 ----------------------------------Number of observation

Z = 1.645 ------------------------------The chosen Z-value from the confidence table below

Confidence Interval Z

80%. 1.282

85% 1.440

90%. 1.645

95%. 1.960

99%. 2.576

99.5%. 2.807

99.9%. 3.291

Substituting these values in the formula

Confidence Interval (CI) = 46300 ± 1.645(1100/√28)

CI = 46300 ± 1.645(1100/5.2915)

CI = 46300 ± 1.645(207.8806)

CI = 46300 ± 341.9636

CI = 46300 + 341.9636. ~. 46300 - 341.9636

CI = 46641.96. ~. 45958.04

In other words the confidence interval is from 46641.96 to 45958.04