Question

AS11ER2 AD3E62E60465... 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 754 children in biomass households, the sample mean peak expiratory flow (a person's maadimum speed of expiration) was 3.8 US, and the sample standard deviation was 1.2. For the 751 children whose households used quefied petroleum gas, the sample mean PE was 4.47 and the sample standard deviation was 1.92. (a) Calculate a confidence interval at the 95% confidence level for the population mean PEF for children in biomass households and then do likewise for children in LPG households. (Round your answers to two decimal places.) biomass households LPG households What is the simultaneous confidence level for the two intervals? (Round your answer to two decimal places.) (6) Carry out a test of hypotheses a 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. Ho biomass -heo- biomase MLPGO M Wo biomass Pus biomass PuG SO Morsomas Puso biomassa Ho: Mbiomass Hups = 0 "biomass LPGO biomass biomass IPGO LPG 20 calculate the test statistic and p-value. (Round your test statistic to two decimal places and your p-value to three decimal places.)

Answer 1

For Biomass :

x̅1 = 3.8, s1 = 1.2, n1 = 754

For LPG :

x̅2 = 4.47, s2 = 1.92, n2 = 751

a) 95% confidence interval for Biomass Households:

At α = 0.05 two tailed critical value, z_c = ABS(NORM.S.INV(0.05/2)) = 1.960

Lower Bound = x̅1 - z_c*s1/√n1 = 3.8 - 1.96 * 1.2/√754 = 3.71

Upper Bound = x̅1 + z_c*s1/√n1 = 3.8 + 1.96 * 1.2/√754 = 3.89

3.71 < µ < 3.89  

95% confidence interval for LPG Households:

At α = 0.05 two tailed critical value, z_c = ABS(NORM.S.INV(0.05/2)) = 1.960

Lower Bound = x̅2 - z_c*s2/√n2 = 4.47 - 1.96 * 1.92/√751 = 4.33

Upper Bound = x̅2 + z_c*s2/√n2 = 4.47 + 1.96 * 1.92/√751 = 4.61

4.33 < µ < 4.61

Simultaneous confidence level = 0.95*0.95 = 90.25%

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b) Null and alternative hypothesis:

Ho : µ_biomass - µ_LPG = 0

Ha : µ_biomass - µ_LPG < 0

Test statistic:  

z = (x̅1 - x̅2)/√(σ1²/n1 + σ2²/n2 ) = (3.8 - 4.47)/√(1.2²/754 + 1.92²/751) = -8.11

p-value :  

p-value = NORM.S.DIST(-8.11, 1) = 0.000

Conclusion:

p-value < α, Reject the null hypothesis

Reject Ho. The data suggest that the true average PEF is lower for children in biomass households than it is for children in LPG households.

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c) x̅ = 2.8, σ = 0.5, n = 754

95% Confidence interval for FEV1:  

At α = 0.05 two tailed critical value, z_c = ABS(NORM.S.INV(0.05/2)) = 1.960

Lower Bound = x̅ - z_c*σ/√n = 2.8 - 1.96 * 0.5/√754 = 2.76

Upper Bound = x̅ + z_c*σ/√n = 2.8 + 1.96 * 0.5/√754 = 2.84

2.76 < µ < 2.84  

No, the two confidence interval in a) were from two independent random sample and so the confidence interval could be multiplied

However, two variables collected on the same group of 754 children does not constitute independent random samples.

Hence, we cannot say that the two confidence intervals for the populations have the same simultaneous confidence level in a).