Question

(1) Let µ be the mileage of a certain brand of tire. A sample of n = 22 tires is taken at random, resulting in the sample mean x = 29, 132 and sample variance s²= 2, 236. Assuming that the distribution is normal, find a 99 percent confidence interval for µ.

(2)We need to estimate the average of a normal population and from measurements on similar populations we estimate that the sample mean is s² = 9. Find the sample size n such that we are 90 percent confident that the estimate of x is within ±1 unit of the true mean µ.

Answer 1

1)

t critical value at 0.01 level with 21 df = 2.831

99% confidence interval for is

- t * S / sqrt(n) < < + t * S / sqrt(n)

29132 - 2.831 * sqrt( 2232) / sqrt( 22) < < 29132 + 2.831 * sqrt( 2232) / sqrt( 22)

29103.48 < < 29160.52

99% CI is ( 29103.48 , 29160.52 )

2)

Sample size = Z² * S² / E²

= 1.6449² * 9 / 1²

= 24.35

Sample size = 25 (Rounded up to nearest integer)