Question

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

Answer 1

Answer:

We are 90% confident that the true mean lies between 8..6563 and     9.9437  

Step-by-step explanation:

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: brainly.com/question/2141785