A recent study of 28 city residents showed that the mean of the time they had lived at their present address was 9.3 years. The standard deviation of the population was 2 years. Find the 90% confidence interval of the true mean? Assume that the variable is
approximately normally distributed. Show all your steps

The mean of your estimate plus and minus the range of that estimate constitutes a confidence interval. Within a specific level of confidence, this is the range of values you anticipate your estimate to fall within if you repeat the test. Confidence is another word for probability.

mean (x) = 9.3

standard deviation (s) = 2

sample size (n) = 28

T- Distribution:

degree of freedom = n - 1 = 28 - 1 = 27

90% Confidence Interval ==> α = 0.1

critical value, t = 1.703 ( t-table)

standard error, SE = s/√n = 2 /√28 = 0.3780

margin of error, E = t × SE= 1.703 × 0.3780 = 0.6437

Confidence Interval: x ± E = 9.3 ± 0.6437

lower limit: x - E = 9.3 - 0.6437 = 8.6563

upper limit: x + E = 9.3 + 0.6437 = 9.9437

Confidence Interval: 8.6563 < µ < 9.9437

Z - Distribution:

Formula: Confidence Interval = x ± z (s/√n)

Sample Mean = s/√n = 2/√28 = 2/5.29 = 0.3781

z = 1.645 (from z - table)

From formula, z (s/√n): 0.3781 x 1.645 = 0.620 (rounded)

x ± 0.62

lower limit: x - 0.62 = 9.3 - 0.62 = 8.68

upper limit: x + 0.62 = 9.3 + 0.62 = 9.92

Confidence Interval: 8.68 < µ < 9.92

Learn more about Confidence Intervals here: https://brainly.com/question/2141785