Question

A consumer magazine has contacted a simple random sample of 58 owners of a certain model of automobile and asked each owner how many defects had to be corrected within the first 2 months of ownership. The average number of defects was 4.9 with a sample standard deviation of 2.1 defects. Construct the 90% confidence interval for the population mean.

Answer 1

Answer:

(4.439 ; 5.361)

Explanation:

Given :

Confidence level = 90%

α = (1 - 0.9) / 2 = 0.05

Sample size (n) = 58

Degree of freedom (df) = n - 1 = 58 - 1 = 57

Mean number of defect (m) = 4.9

Standard deviation (s) = 2.1

m ± t0.05, 57 * (s /√n)

t0.05, 57 = 1.672029 (t value calculator)

4.9 ± 1.672 * (2.1 /√58)

4.9 ± 0.4610431

4.9 - 0.4610431 ; 4.9 + 0.4610431

(4.4389569 ; 5.3610431)

(4.439 ; 5.361)