Question

A phone manufacturer wants to compete in the touch screen phone market. Management understands that the leading product has a less than desirable battery life. They aim to compete with a new touch screen phone that is guaranteed to have a battery life more than two hours longer than the leading product. A recent sample of 123 units of the leading product provides a mean battery life of 5 hours and 50 minutes with a standard deviation of 41 minutes. A similar analysis of 101 units of the new product results in a mean battery life of 8 hours and 30 minutes and a standard deviation of 54 minutes. It is not reasonable to assume that the population variances of the two products are equal.

All times are converted into minutes. Let new products and leading products represent population 1 and population 2, respectively. (You may find it useful to reference the appropriate table: z table or t table)

a. Set up the hypotheses to test if the new product has a battery life more than two hours longer than the leading product.

• H₀: μ₁ − μ₂ = 120; H_A: μ₁ − μ₂ ≠ 120

• H₀: μ₁ − μ₂ ≥ 120; H_A: μ₁ − μ₂ < 120

• H₀: μ₁ − μ₂ ≤ 120; H_A: μ₁ − μ₂ > 120

b-1. Calculate the value of the test statistic. (Round all intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.)

Test statistic:

Answer 1

a) Correct answer: option (D) H₀: μ₁ − μ₂ ≤ 120; H_A: μ₁ − μ₂ > 120

b) Given New product: mean battery life of 8 hours and 30 minutes means = 8*60+30 = 510

leading product : mean battery life of 5 hours and 50 minutes means = 5*60 +50 =350