Question

1. You are interested in whether the average lifetime of Duracell AAA batteries is less than the average lifetime of Energizer AAA batteries. You lay out your hypotheses as follows: Null Hypothesis: μ₁ ≥ μ₂, Alternative Hypothesis: μ₁ < μ₂ (Duracell = Group 1, Energizer = Group 2). You perform an experiment on 20 randomly selected Duracell AAA batteries and 40 randomly selected Energizer AAA batteries and test them using a battery drainer. The Duracells had an average lifetime of 29.19 hours (SD = 1.887 hours) while the Energizer batteries had an average lifetime of 29.16 hours (SD = 2.309 hours). Assuming the population standard deviations to be equal, what is the test statistic and p-value of this test?

Test Statistic: -0.05, P-Value: 0.4801

Test Statistic: -0.05, P-Value: 0.5199

Test Statistic: 0.05, P-Value: 0.4801

Test Statistic: 0.05, P-Value: 1.0398

Test Statistic: 0.05, P-Value: 0.5199

2. The owner of a local golf course wants to examine the difference between the average ages of males and females that play on the golf course. Specifically, he wants to test if the average age of males is different from the average age of females. If the owner conducts a hypothesis test for two independent samples and calculates a p-value of 0.9205, what is the appropriate conclusion? Label males as group 1 and females as group 2.

We did not find enough evidence to say the average age of males is larger than the average age of females.

The average age of males is equal to the average age of females.

We did not find enough evidence to say a significant difference exists between the average age of males and females.

We did not find enough evidence to say the average age of males is less than the average age of females.

The average age of males is significantly different from the average age of females.

Answer 1

n₁= 20 x̅₁= 29.19 sd₁= 1.887

n₂= 40 x̅₂= 29.16 sd₁= 2.309

to test

H₀: μ₁ ≤ μ₂

H₁: μ_{1 >} μ₂

_{test statics t =} (x̅₁− x̅₂)/s_p√(1∕n₁)+(1∕n₂)

s_p = √((n₁-1)sd₁² + (n₁-1)sd2²∕(n₁+n₂-2))

= √19(1.887²)+40(2.309²)∕58

= √67.654+213.259/58

= √4.8432

= 2.2007

t = (29.19-29.16)/2.2007√(1/20)+(1/40)

t = 0.05

p- value = P( t > 0.05) = 0.4801

2.

p-value > 0.05 Thus we don't have enough evidence to support our claim.

We did not find enough evidence to say the average age of males is less than the average age of females.