Question

[6-34] The manufacturer of a new fiberglass tire claims that its average life will be at least 40,000 miles. To verify this claim, sample of 15 tires are tested. Sample mean was 41.1 (in 1000 miles), and sample standard deviation was 2.50. Assuming the measurements were random sample from a Normal distribution, calculate 95% one-sided confidence interval for true mean tire life. Choose lower-bound or upper-bound, whichever is appropriate. (Round to 2 decimals.) Your Answer: Answer

Answer 1

The critical value is t^* = 1.761

The 95% lower-bound confidence interval is

$\bar x$ - t^* * s/$\sqrt n$

= 41.1 - 1.761 * 2.5/$\sqrt 15$

= 41.1 - 1.14

= 39.96