Question

Have you ever wondered what it means to click the “offset carbon emissions” button when you book a flight or train trip? It adds a small cost to your ticket, but how does this reduce emissions? The money is typically used to fund projects that reduce carbon emissions. One such project type is the introduction of more efficient cooking stoves into communities. Much of the world uses inefficient charcoal or wood stoves that result in excessive indoor air pollution, deforestation, and carbon emissions. Switching millions of families to more efficient stoves can result in a significant reduction in carbon emissions. You may read more about such a project here. In order for a project to claim carbon credits, they must provide accurate estimates of how much carbon that project is saving. An important parameter for cook-stove projects is the reduction in fuel that results from switching to the more efficient stove. Statisticians are needed to design the experiments; it is expensive to do these tests, so figuring out how big the sample size should be in order to get sufficiently accurate estimates, or to detect significant differences between the stove types, is important. The EXCEL file, stove.xlsx, for this lab contains data from a pilot study using 19 randomly selected cooks. The numbers refer to the weight of firewood (in kg) to cook a regular meal. Each row in the spreadsheet corresponds to the same cook cooking the same meal. Use this data to answer the following questions. You may assume the conditions to carry out a hypothesis test are satisfied. You can assume (based on many similar studies) that the population standard deviation of reduction of firewood used is 0.7kg. Try to store as many decimal places as possible in intermediate steps. Old Stove Improved Stove 3.9 1.8 3.8 2.65 3.65 1.5 3.2 2.2 2.6 1.25 2.4 1.65 2.3 1.4 2.25 1.7 2.2 2.15 2.1 1.8 2 1.4 2 1.05 1.9 0.8 1.9 1.75 1.8 0.55 1.55 0.9 1.4 1.3 1.4 1.1 1.15 0.75 5. Find the 90% CI for the true mean reduction in firewood used. You may assume the population standard deviation of reduction in firewood used is 0.7. What is the margin of error (round to 4 decimal places)? 6. For a project to qualify for carbon credits, the required precision for estimates of the amount of wood saved per new stove adopted is 90/10, i.e. the 90% confidence interval must have a margin of error no greater than 10% of the value of the estimate. Will the data from the pilot study enable the project to qualify for carbon credits? 7. What is the minimum sample size required to meet the 90/10 precision requirement? 1 8. We want to know if the weight of wood used with the improved stove is significantly less than the weight of wood used with the old stove. State the null and alternative hypotheses for such a test.

Answer 1

Solution

Let

X = weight of firewood (in kg) to cook a regular meal with old stove

Y = weight of firewood (in kg) to cook a regular meal with improved stove

Then, reduction in firewood used with improved stove = d = X – Y

d ~ N(µ, σ²), where σ is given to 0.7…………………………………………………… (1)

Part (5.1)

100(1 - α) % Confidence Interval for μ, when σ is known: Xbar ± (Z_{α /2})σ/√n ……………… (2)

where

Xbar = sample mean,

Z_{α /2} = upper (α /2)% point of N(0, 1),

σ = population standard deviation and

n = sample size.

So, 90% Confidence Interval for the true mean reduction in firewood used with improved stove

= [0.5755, 1.1038] Answer 1

Details of calculations

α =	0.1
n =	19
σ	0.7
Xbar =	0.8396
Zα/2 =	1.644854
95% CI for μ				0.8396		±		0.26414871
Lower Bound =				0.57545
Upper Bound =				1.10375

Part (5.ii)

Margin of error = 0.2641   Answer 2

Part (6)

10% of the estimate = 0.8396 x 0.1 = 0.08396

So, margin of error for the CI > 10% of the estimate.

Since for a project to qualify for carbon credits, the required precision for estimates of the amount of wood saved per new stove adopted is 90/10, i.e. the 90% confidence interval must have a margin of error no greater than 10% of the value of the estimate, the data from the pilot study does not enable the project to qualify for carbon credits. Answer 3

Part (7)

For margin of error for the CI < 10% of the estimate, we must have:

(Z_{α /2})σ/√n < 0.08396

i.e., (1.6449 x 0.7)/√n < 0.08396

or, √n > (1.6449 x 0.7)/0.08396 = 13.71

=> n > 188.07

Thus, minimum sample size required = 189 Answer 4

Part (8)

Test

Claim:

Reduction in firewood used with improved stove is significant

Hypotheses:

Null: H₀: µ_d = 0 Vs Alternative: H_A: µ_d > 0

Test Statistic:

t = (√n) (dbar)/σ where

dbar is the sample average

σ is the population standard deviation

n is the sample size.

Calculations

Summary of Excel calculations is given below:

i	xi	yi	di
1	3.90	1.80	2.1
2	3.80	2.65	1.15
3	3.65	1.50	2.15
4	3.20	2.20	1
5	2.60	1.25	1.35
6	2.40	1.65	0.75
7	2.30	1.40	0.9
8	2.25	1.70	0.55
9	2.20	2.15	0.05
10	2.10	1.80	0.3
11	2.00	1.40	0.6
12	2.00	1.05	0.95
13	1.90	0.80	1.1
14	1.90	1.75	0.15
15	1.80	0.55	1.25
16	1.55	0.90	0.65
17	1.40	1.30	0.1
18	1.40	1.10	0.3
19	1.15	0.75	0.4

n	19
dbar	0.831578947
σd	0.7
√n	4.358898944
Zcal	5.17824085
α	0.05
Zcrit	1.644853627

Distribution, Significance level (α), Critical Value and p-value:

Under H₀, Z ~ N(0, 1). Hence, for level of significance α%,

Critical Value = upper α% point of N(0, 1)

α = 5% [i.e., 0.05 (given/assumed)]

Using Excel Function: Statistical TINV, the above are found to be as given in the above table.

Decision:

Since Zcal > Zcrit, H₀ is rejected.

Conclusion:

There is sufficient evidence to suggest that the claim is valid.

i.e., improved stove leads to significant reduction in consumption of fire wood. Answer 5

DONE