Question

a light bulb manufacturer guarantees that the mean life of a certain type of light bulb is at least 757 hours. a random sample of 20 light bulbs has a mean life of 728 hours. assume the population is normally distributed and the population standard deviation is 61 hours. at \\( \alpha = 0.02 \\), do you have enough evidence to reject the manufacturers claim? complete parts (a) through (e).
(a) identify the null hypothesis and alternative hypothesis.
a. \\( h _ { 0 } : \mu \geq 757 \\) (claim) \\( h _ { a } : \mu < 757 \\)
b. \\( h _ { 0 } : \mu > 757 \\) \\( h _ { a } : \mu \leq 757 \\) (claim)
c. \\( h _ { 0 } : \mu \leq 728 \\) \\( h _ { a } : \mu > 728 \\) (claim)
d. \\( h _ { 0 } : \mu < 728 \\) (claim) \\( h _ { a } : \mu \geq 728 \\)
e. \\( h _ { 0 } : \mu \
eq 757 \\) (claim) \\( h _ { a } : \mu = 757 \\)
f. \\( h _ { 0 } : \mu = 728 \\) \\( h _ { a } : \mu \
eq 728 \\) (claim)
(b) identify the critical value(s). use technology.
\\( z _ { 0 } = \\)
(use a comma to separate answers as needed. round to two decimal places as needed.)

Explanation:

Step1: Determine the type of test

This is a left - tailed test because the alternative hypothesis \(H_{a}:\mu < 757\).

Step2: Find the critical value

For a left - tailed test with \(\alpha = 0.02\), we use the standard normal distribution \(Z\).
The critical value \(z_{0}\) is the value such that \(P(Z<z_{0})=\alpha = 0.02\).
Using a standard normal table or technology (e.g., in Excel: NORM.S.INV(0.02)), we get \(z_{0}=- 2.05\)

Answer:

\(-2.05\)