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 756 children in biomass households, the sample mean peak expiratory flow (a person's maximum speed of expiration) was 3.4 L/s, and the sample standard deviation was 1.2. For the 752 children whose households used liquefied petroleum gas, the sample mean PEF was 4.28 and the sample standard deviation was 1.19. Required:
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. What is the simultaneous confidence level for the two
intervals?
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 cited article included a P-value for this test).

c. FEVI, the forced expiratory volume in 1 second, is another measure of pulmonary function. The cited article reported that for the biomass households the sample mean FEY, was 2.3 L/S and the sample standard deviation was .5 L/s. If this information is used to compute a 95% Cl for population mean FEVI, would the simultaneous confidence level for this interval and the first interval calculated in (a) be the same as the simultaneous confidence level deter-mined there? Explain.

A) 95% confidence interval for the population mean PEF for children in biomass households = (3.314, 3.486)

95% confidence interval for the population mean PEF for children in LPG households

= (4.195, 4.365)

Simultaneous confidence interval for both = (3.314, 4.365)

B) The result of the hypothesis test is significant, hence, the true average PEF is lower for children in biomass households than it is for children in LPG households.

C) 95% confidence interval for the population mean FEY for children in biomass households = (2.264, 2.336)

Simultaneous confidence interval for both = (2.264, 4.365)

This simultaneous interval cannot be the same as that calculated in (a) above because the sample mean obtained for children in biomass households here (using FEY) is much lower than that obtained using PEF in (a).

Step-by-step explanation:

A) Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the z-distribution. This is because although, there is no information provided for the population standard deviation, the sample sizes are large enough for the sample properties to approximate the population properties.

Finding the critical value from the z-tables,

z-critical value for 95% confidence level = 1.960 (from the z-tables)

For the children in the biomass households

Sample mean = 3.40

Standard error of the mean = σₓ = (σ/√N)

σ = standard deviation of the sample = 1.20

N = sample size = 756

σₓ = (1.20/√756) = 0.04364

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 3.40 ± (1.960 × 0.04364)

CI = 3.40 ± 0.08554

95% CI = (3.31446, 3.48554)

95% Confidence interval = (3.314, 3.486)

For the children in the LPG households

Sample mean = 4.28

Standard error of the mean = σₓ = (σ/√N)

σ = standard deviation of the sample = 1.19

N = sample size = 752

σₓ = (1.19/√752) = 0.043395

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 4.28 ± (1.960 × 0.043395)

CI = 4.28 ± 0.085054

95% CI = (4.1949, 4.3651)

95% Confidence interval = (4.195, 4.365)

Simultaneous confidence interval for both = (3.214, 4.375)

B) The null hypothesis usually goes against the claim we are trying to test and would be that the true average PEF for children in biomass households is not lower than that of children in LPG households.

The alternative hypothesis confirms the claim we are testing and is that the true average PEF is lower for children in biomass households than it is for children in LPG households.

Mathematically, if the true average PEF for children in biomass households is μ₁, the true average PEF for children in LPG households is μ₂ and the difference is μ = μ₁ - μ₂

The null hypothesis is

H₀: μ ≥ 0 or μ₁ ≥ μ₂

The alternative hypothesis is

Hₐ: μ < 0 or μ₁ < μ₂

Test statistic for 2 sample mean data is given as

Test statistic = (μ₂ - μ₁)/σ

σ = √[(s₂²/n₂) + (s₁²/n₁)]

μ₁ = 3.40

n₁ = 756

s₁ = 1.20

μ₂ = 4.28

n₂ = 752

s₂ = 1.19

σ = √[(1.20²/756) + (1.19²/752)] = 0.061546

z = (3.40 - 4.28) ÷ 0.061546 = -14.30

checking the tables for the p-value of this z-statistic

Significance level = 0.01

The hypothesis test uses a one-tailed condition because we're testing in only one direction.

p-value (for z = -14.30, at 0.01 significance level, with a one tailed condition) = 0.000000001

The interpretation of p-values is that

When the p-value > significance level, we fail to reject the null hypothesis and when the p-value < significance level, we reject the null hypothesis and accept the alternative hypothesis.

Significance level = 0.01

p-value = 0.000000001

0.000000001 < 0.01

Hence,

p-value < significance level

This means that we reject the null hypothesis, accept the alternative hypothesis & say that true average PEF is lower for children in biomass households than it is for children in LPG households.

C) For FEY for biomass households,

Sample mean = 2.3 L/s

Standard error of the mean = σₓ = (σ/√N)

σ = standard deviation = 0.5

N = sample size = 756

σₓ = (0.5/√756) = 0.018185

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 2.30 ± (1.960 × 0.018185)

CI = 2.30 ± 0.03564

95% CI = (2.264, 2.336)

Simultaneous confidence interval for both = (2.264, 4.365)

This simultaneous interval cannot be the same as that calculated in (a) above because the sample mean obtained for children in biomass households here (using FEY) is much lower than that obtained using PEF in (a).

Hope this Helps!!!

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step-by-step explanation: