Question

1. A sample consists of 75 TV sets purchased several years ago. The replacement times of those TV sets have a mean of 8.2 years. Assume σ= 1.1 years. Find the 95% confidence interval.

Answer 1

8.2+/-0.25

= ( 7.95, 8.45) years

the 95% confidence interval (a,b) = (7.95, 8.45) years

Explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 8.2 years

Standard deviation r = 1.1 years

Number of samples n = 75

Confidence interval = 95%

z value(at 95% confidence) = 1.96

Substituting the values we have;

8.2+/-1.96(1.1/√75)

8.2+/-1.96(0.127017059221)

8.2+/-0.248953436074

8.2+/-0.25

= ( 7.95, 8.45)

Therefore the 95% confidence interval (a,b) = (7.95, 8.45) years