Question

/books/9781323584736/cfi/6/3781/4/2/16/10@0:100 Case Study Fire-Safe Cigarettes In 2011, 90,000 fires in the United States were started by a lighted tobacco product The dollar value of the property lost in these fires is staggering. In that same year fires caused by dropped or discarded cigarettes resulted in 540 deaths. Twenty-five percent of the victims of cigarette induced fires were not the smokers, and about 40% of cigarette-induced fires started in the bedroom. (Source. www.firesafecigarettes.org) Certainly it makes sense to develop a cigarette Nat extinguishes itself when left unattended (as would happen if someone falls asleep while smoking). The state of New York decided to require cigarettes sold in its stores to be designed so they self- extinguished when left unattended. Two types of cigarettes were tested to determine their propensity to self-extinguish. For each type (A and B), 40 cigarettes were lit and allowed to burn unattended until the cigarette either extinguished or did not. Cigarette brand A was designed to have ultra thin concentric paper bands affixed to the traditional cigarette paper. These bands are referred to as "speed bumps and cause extinguishing of the cigarette by restricting the flow of oxygen to the burning ember. Cigarette brand B was a traditionally designed model. The results of the experiment are in the following table, where E represents an extinguished cigarette and F represents a full-burn cigarette BRAND A

Answer 1

Solution

Part (a)

Comparison of self-extinguishing property

Let X = number of sample cigarettes which extinguished themselves (E) for Brand A

and Y = number of sample cigarettes which extinguished themselves (E) for Brand B

Then, X ~ B(n₁, p₁), and Y ~ B(n₂, p₂) where

n₁ and n₂ are sample sizes and

p₁ = probability that Brand A cigarette extinguishes itself and

p₂ = probability that Brand B cigarette extinguishes itself,

which are also equal to the corresponding proportion in the respective the population.

Claim : Proportion of cigarettes extinguishing themselves is more for Brand A than for Brand B [i.e., p₁ > p₂]

Hypotheses:

Null H₀ : p₁ = p₂   Vs H_A : p₁ > p₂

Test Statistic:

Z = (p₁cap – p₂cap)/√[pcap(1 - pcap){(1/n₁) + (1/n₂)} where

p₁cap and p₂cap are sample proportions,

n₁, n₂ are sample sizes and

pcap = {(n₁ x p₁cap) + (n₂ x p₂cap)}/(n₁ + n₂).

Zcal = 6.094

Calculations:

Given
n1 =	40
n2 =	40
x =	27
y =	1
p1cap =	0.675
p2cap =	0.025
pcap =	0.35
Zcal =	6.094494
α =	0.05
Zcrit	1.644854
p-value	5.49E-10

Distribution, Critical Value and p-value:

Under H₀, distribution of Z can be approximated by Standard Normal Distribution

So, given a level of significance of α%, Critical Value = upper α% of N(0, 1), and

p-value = P(Z > Zcal)

Assuming a 5% significance level, Critical Value and p-value are found to be as given in the above table. [Using Excel Function: Statistical NORMSINV and NORMSDIST]

Decision:

Since Zcal > Zcrit, or equivalently p-value < α, H₀ is rejected.

Conclusion :

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

i.e., ‘speed bumps’ would reduce the incidence of fires due to unattended cigarette ends. Answer

Part (b)

Comparison of Nicotine Content

Let X = Nicotine Content in Brand A

      Y = Nicotine Content in Brand B

Then, we assume: X ~ N(µ₁, σ₁²) and Y ~ N(µ₂, σ₂²), where σ₁² = σ₂² = σ², say and σ² is unknown.

Claim: Nicotine content is the same in both brands. i.e., µ1 = µ2

Hypotheses:

Null: H₀: µ1 = µ2 Vs Alternative: H_A: µ1 ≠ µ2

Test Statistic:

t = (Xbar - Ybar)/{s√(2/n)} where

s² = (s₁² + s₂²)/2;

Xbar and Ybar are sample averages

s₁,s₂ are sample standard deviations based on n observations each on X and Y.

Calculations

Summary of Excel calculations is given below:

n =	15
Xbar =	1.206667
Ybar =	1.234
s1 =	0.131511
s2 =	0.078813
s^2 =	0.011753
s =	0.108413
|tcal| =	0.690467
α =	0.05
tcrit =	2.048407
p-value =	0.495585

So, |tcal| = 0.690

Distribution, Critical Value and p-value:

Under H₀, t ~ t_{2n - 2}. Hence, for level of significance α%, Critical Value = upper (α/2)% point of t_{2n - 2} and p-value = P(t_{2n - 2} > | tcal |).

Using Excel Functions: Statistical TINV and TDIST, the above are found to be as shown in the above table.

Decision:

Since | tcal | < tcrit, or equivalently, p-value > α, H₀ is accepted.

Conclusion:

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

i.e., the nicotine content is the same in both brands on an average. Answer

DONE