Question

The quality-control manager at a compact fluorescent light bulb (CFL) factory needs to determine whether the mean life of a large shipment of CFL's is equal to 7,500 hours. The population standard deviation is 1,000 hours. A random sample of 64 CFL's indicates a sample mean life of 7,250 hours. Level of significance is 0.05.

1. What is the p-value? Round to 4 decimal places i.e 0.0432

2. What is the critical value?

Round answer to 4 decimal places i.e. 1.2568

Answer 1

Puye Solution : - The null and alternative hypothesid gol. - Hoil a 7500 Hailu & 7500 Heze, =64X - 7250, 5= 1000 The test Statistic z 2. X-M on za-2 is. . 7250 - 7500 1000 / √64 & Sign in Ha => Two tailed test 1. P-value = P(Z > 2.00) + P(Z <-2.00) + 0.0228 -2 table = 0.0228 -0.0456 .. p-value = 0.0456

2 The caitical value = = 0.05 2 Using the Caitiral value appagach. Caitical value is Z = +/-1.6450

Puye Solution : - The null and alternative hypothesid gol. - Hoil a 7500 Hailu & 7500 Heze, =64X - 7250, 5= 1000 The test Statistic z 2. X-M on za-2 is. . 7250 - 7500 1000 / √64 & Sign in Ha => Two tailed test 1. P-value = P(Z > 2.00) + P(Z <-2.00) + 0.0228 -2 table = 0.0228 -0.0456 .. p-value = 0.0456

2. The Caitical value = Significance level = 0.05 Using caitical value approach, Za = +/-1.6450