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 Without performing any calculations for the hypothesis test, what will be a range of values of the p-value based on the 90% confidence interval calculated in question 5 to test the hypothesis in question 8 at α = 0.10? Briefly explain. The options are (a) p-value is less than 0.05 (b) p-value is less than 0.10 (c) p-value is greather than 0.05 (d) p-value is greater than 0.10 10. Is there enough evidence to reject H0 at the α = 0.1 level of significance? What does this mean in context of this project? 11. What is the critical value of this hypothesis test at the α = 0.1 level of significance?

Answer 1

5.

a.
TRADITIONAL METHOD
given that,
standard deviation, σ =0.7
sample mean, x =1.873
population size (n)=38
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 0.7/ sqrt ( 38) )
= 0.11
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
value of z table is 1.645
margin of error = 1.645 * 0.11
= 0.19
III.
CI = x ± margin of error
confidence interval = [ 1.873 ± 0.19 ]
= [ 1.69,2.06 ]
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DIRECT METHOD
given that,
standard deviation, σ =0.7
sample mean, x =1.873
population size (n)=38
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
value of z table is 1.645
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 1.873 ± Z a/2 ( 0.7/ Sqrt ( 38) ) ]
= [ 1.873 - 1.645 * (0.11) , 1.873 + 1.645 * (0.11) ]
= [ 1.69,2.06 ]
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interpretations:
1. we are 90% sure that the interval [1.69 , 2.06 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean

b.
Given that,
population mean(u)=1.8
standard deviation, sigma =0.7
sample mean, x =1.873
number (n)=38
null, Ho: μ=1.8
alternate, H1: μ!=1.8
level of significance, alpha = 0.1
from standard normal table, two tailed z alpha/2 =1.645
since our test is two-tailed
reject Ho, if zo < -1.645 OR if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 1.873-1.8/(0.7/sqrt(38)
zo = 0.643
| zo | = 0.643
critical value
the value of |z alpha| at los 10% is 1.645
we got |zo| =0.643 & | z alpha | = 1.645
make decision
hence value of |zo | < | z alpha | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 0.643 ) = 0.52
hence value of p0.1 < 0.52, here we do not reject Ho
ANSWERS
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null, Ho: μ=1.8
alternate, H1: μ!=1.8
test statistic: 0.643
critical value: -1.645 , 1.645
decision: do not reject Ho
p-value: 0.52
we do not have enough evidence to support the claim
option:d
p value is greater than 0.1