Question

Reliance on solid biomass fuel for cooking and heating exposes many children from developing countries to high levels of indoor air pollution. An article presented information on various pulmonary characteristics in samples of children whose households in India used either biomass fuel or liquefied petroleum gas (LPG). For the 753 children in biomass households, the sample mean peak expiratory flow (a person's maximum speed of expiration) was 3.9 L/s, and the sample standard deviation was 1.2. For the 751 children whose households used liquefied petroleum gas, the sample mean PEF was 4.29 and the sample standard deviation was 1.85.

(b)

Carry out a test of hypotheses at significance level 0.01 to decide whether true average PEF is lower for children in biomass households than it is for children in LPG households (the article included a P-value for this test). (Round your test statistic to two decimal places and your P-value to four decimal places.)

State the relevant hypotheses.

H₀: μ_biomass − μ_LPG = 0
H_a: μ_biomass − μ_LPG ≥ 0

H₀: μ_biomass − μ_LPG = 0
H_a: μ_biomass − μ_LPG ≠ 0    

H₀: μ_biomass − μ_LPG = 0
H_a: μ_biomass − μ_LPG ≤ 0

H₀: μ_biomass − μ_LPG = 0
H_a: μ_biomass − μ_LPG > 0

H₀: μ_biomass − μ_LPG = 0
H_a: μ_biomass − μ_LPG < 0

Calculate the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.)

z=???

P-value = 0.000 Correct

Answer 1

Ans:

Test statistic:

z=(3.9-4.29)/SQRT((1.2^2/753)+(1.85^2/751))

z=-4.85

p-value=P(z<-4.849)=0.000