Comparison of Tar Yield in Side Stream and Main Stream Smokers

This document compares the average tar yield of side stream smokers and main stream smokers. It includes the null and alternate hypotheses, assumptions, test statistics, p-value, and conclusion. The Mann-Whitney U test is also performed as a non-parametric alternative.

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Assignment 3

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Answer 1
(a) Average heart rate for N = 88 patients who survived cardiac arrest after being admitted
to hospital for critical illness was calculated as x
¿
= 104.97 (SD = 29.44), which varied
between minimum heart rate of 25 and maximum heart rate of 217.
The confidence interval for population mean with sample SD is
x
¿
± tα
2
∗ s
√ n
Here significance level is α=2 % and degrees of freedom N – 1 = 87.
Using calculator and t-table t ( 87 ) =2 .37 for α=0 . 02
Hence,
x
¿
± tα
2
∗ s
√ n = [ 104 . 97−2. 37∗29 . 44
√ 88 , 104 . 97+2 .37∗29 . 44
√ 88 ] =[ 97 .53 , 112. 41 ]
Therefore, with 98% confidence it can be said that population average heart rate will be
somewhere between 97.53 beats/min and 112.41 beats/min for all critically ill patients
who survived cardiac arrest.

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Figure 1: Confidence interval for t-distribution with 87 degrees of freedom
(b) The assumptions for validity of confidence interval or hypothesis test for the population
mean are randomization, independence condition (10% rule), and sample size condition.
Randomization: Data for all critically ill patients was sampled randomly with a good
sampling methodology.
Independence: Population size for all critically ill patients surviving cardiac arrest after
being admitted to hospital is large enough for the sample size of N = 88 to be less than
10% .
Sample Size: Sample size of N = 88 was large enough (n > 30) to apply Central Limit
Theorem. The distribution of critically ill patients surviving cardiac arrest was plotted in
a histogram and a Q-Q plot to identify the possible outliers and shape of the distribution.
Histogram in Figure 2 indicates almost normally shaped histogram with few outliers.
This pattern was confirmed by Q-Q plot in Figure 3 with presence of at least three
outliers. Shapiro Wilk test indicated that the distribution was not at all significant (W =
0.96, p < 0.05).
Figure 2: Histogram of Heart rate for critically ill patients surviving cardiac arrest

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Figure 3: Q-Q plot of Heart rate for critically ill patients surviving cardiac arrest
(c) Null hypothesis: H0: μ ≤ 90: Average heart rate of adult critically ill patients who
survived cardiac arrest is not more than 90 beats per min.
Alternate hypothesis: HA: μ > 90 (Right Tail): Average heart rate of adult critically ill
patients who survived cardiac arrest is significantly than 90 beats per min.
(d) The t-statistic at n – 1 = 87 degrees of freedom is calculated using the formula
t ( 87 ) = x
¿
−μ
s
√ n where x
¿
=104 . 97 , s = 29.44, n = 88, μ = 90.
Therefore,
t ( 87 ) = x
¿
−μ
s
√ n
=104 . 97−90
29 . 44
√ 88
=4 . 77
(e) The p-value is calculated as the probability P ( T >tcal =4 .77 )=0 . 000<0 . 01 at n – 1 = 87
degrees of freedom (using t-table).
At 1% level of significance, the null hypothesis was rejected to signify that average heart
rate of adults was significantly greater than 90 beats per min for critically ill patients who
were brought to hospital and survived cardiac arrest.
Figure 4: Critical region for t-test with p-value

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(f) Significance level: 1%
Test Statistic: The t-stat = t ( 87 ) =4 .769 with p < 0.01
SPSS output for one sample t-test at 1% level of significance was done for two tailed
alternate hypothesis. For one tail the p-value will be = 2* 0.000 = 0.000. Hence, the
conclusion from hand evaluated test statistic remains unaltered that the null hypothesis
has to be rejected at 1% level.
Figure 5: T-test SPSS output for heartbeat of critically ill patients who survived cardiac arrest.

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Answer 2
(a) Variable of interest was the nominal variable called “Survival” which has two categories,
namely survived and died.
(b) Null hypothesis: p
^¿=0 .55
¿ : Proportion of critically ill patients surviving cardiac arrest
was equal to 0.55 or 55%
Alternate hypothesis: p
^¿≠0 .55
¿ : (two tailed): Proportion of critically ill patients surviving
cardiac arrest was significantly no more 0.55 or 55%
(c) The observations of survival were taken randomly from population of critically ill
patients.
Population of critically ill patients surviving cardiac arrest would be more than 10 times
of favourable number of cases here (N = 88).
Here, n∗p=88∗0 .55=48 . 4 >10 and n∗( 1− p ) =88∗0 . 45=39 .6> 10
Hence, assumptions for the z-test for proportions are satisfied.
(d) The test statistic is calculated as
Z =p− p
^¿
√ p
^¿ ¿¿
¿ ¿ ¿ ¿
¿¿ ¿
Here p = Proportion of critically ill patients surviving cardiac arrest in the sample =
88/146 = 0.60.
p
^¿=
¿ Proportion of critically ill patients surviving cardiac arrest according to the
researcher = 0.55

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Therefore,
Z =p− p
^¿
√ p
^¿ ¿¿
¿ ¿ ¿ ¿
¿¿ ¿
(e) The p-value is calculated as the probability
P ( Z > zcal =0 . 96 ) =1−P ( Z≤0. 96 ) =1-0 . 8315=0 .1685 .
P value, greater than 0.01 signifies that the null hypothesis failed to get rejected. This
indicates that proportion of survival for critically ill patients surviving cardiac arrest was
statistically equal to 55%, and the researcher’s postulation was not correct.
Figure 6: Region for p-value in the normal curve
(f) The true proportion for critically ill patients surviving cardiac arrest was p
^¿=
¿ 0.6
The margin of error is
Z α
2
∗ √ p
^¿¿ ¿
¿ ¿ ¿ ¿
where α=0 . 01 and n = 88
Now, at α=0 . 01 , Z = 2.575 is the critical value.
Now, according to the researcher, margin of error for critically ill patients surviving
cardiac arrest after being admitted to hospital will be within 0.02. In conservative method
p
^¿=0 .5
¿
Therefore,
Z α
2
∗ √ p
^¿¿ ¿
¿ ¿ ¿ ¿
=> 0 .5∗0 .5
n ≤( 0 .02
2. 575 ) 2
=> n
0 .25 > ( 2 . 575
0 . 02 )
2
=> n>0 . 25∗16576 . 56 =>n> 4144 . 14

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Hence, minimum sample size required is 4145 critically ill patients to obtain margin of
error within 2%.
(g) The margin of error is
Z α
2
∗ √ p
^¿¿ ¿
¿ ¿ ¿ ¿
where α=0 . 01 and n = 88
Now, at α=0 . 01 , Z = 2.575 is the critical value.
Now, according to the researcher, margin of error for critically ill patients surviving
cardiac arrest after being admitted to hospital will be within 0.02.
Therefore,
Z α
2
∗ √ p
^¿¿ ¿
¿ ¿ ¿ ¿
=> 0 .6∗0 . 4
n ≤( 0 . 02
2 .575 )
2
=> n
0 .24 > ( 2. 575
0 .02 )
2
=> n> 0. 24∗16576 .56 => n>3978 . 37
Hence, minimum sample size required is 3979 critically ill patients to obtain margin of
error within 2%.
The impact of calculation sample size with true proportion from the sample will reduce
the required number of samples to 3979 from sample size of conservative method (N =
4145).

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Answer 3
(a) Null hypothesis: H0: ( μs=μm ) : Average tar yield of side stream smokers was equal to
that of the mainstream smokers.
Alternate hypothesis: H0: ( μs≥μm ) : Average tar yield of side stream smokers was
significantly greater that of the mainstream smokers.
(b) Three assumptions of independent sample t-test are: Independent observations, normality
of the dependent variable, and homogeneity between the two samples.

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In terms of the present study, tar yield of smokers for eight brands of cigarettes are
independent of each other. Tar yield of Side stream and main stream smokers is
supposed to be normally distributed. This assumption is specially required when sample
size is less than 30. Standard deviations of tar yield of both types of smokers should be
identical.
(c) Average tar yield for side stream smokers
x
¿
s=15. 8+16 . 9+21 . 6+18 .8+29. 3+20 . 7+18 . 9+25
8 =20. 875
Average tar yield for main stream smokers
x
¿
m=18 .5+17 +17 .2+19 . 4+ 15. 6+16 . 4 +13 .3+10 . 2
8 =15 . 95
Standard deviation for tar yield for side stream smokers
ss= √ ( 15 . 8−20. 875 ) 2 + ( 16 .9−20 . 875 ) 2+. ..+ ( 25−20 . 875 ) 2
8−1 =4 . 44
Standard deviation for tar yield for main stream smokers
sm= √ ( 18. 5−15. 95 ) 2 + ( 17−15 . 95 ) 2+.. .+ ( 10 .2−15. 95 ) 2
8−1 =2. 96
For unequal variances, the test statistics is
t= x
¿
s−x
¿
m
√ ss
2
n + sm
2
n
=20 . 875−15 . 95
√ 4 . 442
8 + 2. 962
8
=2. 61
(d) The p-value is calculated for right tailed test as P ( T >t=2 .61 ) =0 . 0102> 0. 01 at 14
degrees of freedom.
(e) Therefore, the null hypothesis will fail to get rejected at 1% level of significance. But, at
5% level of significance, the null hypothesis will get rejected. Therefore, it can be
concluded that average tar yield of side stream smokers was not significantly greater than
main stream smokers at 1% level of significance. But, at 5% level average tar yield of
side stream smokers was greater that of the main stream smokers.
(f) Mann-Whitney U Test (non-parametric) has been performed instead of t-test.
Null hypothesis: Population of smokers of side stream and main stream are equal versus
Alternate hypothesis: Population of smokers of side stream and main stream are not
equal
Significance level = 5%

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Test Statistic Calculation: Mann-Whitney U test statistic is smaller of the following two
values (MacFarland, and Yates, 2016, pp. 103-132).
U1=n1 n2 + n1 ( n1+1 )
2 −R1 And U2=n1 n2 + n2 ( n2 +1 )
2 −R2
Population 1 Sample
Sample Size = 8
Sum of Ranks = 89
Population 2 Sample
Sample Size = 8
Sum of Ranks = 47
Table 1: Rank of two streams of smokers
Stream Value Rank
Mainstream 10.2 1
Mainstream 13.3 2
Mainstream 15.6 3
Sidestream 15.8 4
Mainstream 16.4 5
Sidestream 16.9 6
Mainstream 17 7
Mainstream 17.2 8
Mainstream 18.5 9
Sidestream 18.8 10
Sidestream 18.9 11
Mainstream 19.4 12
Sidestream 20.7 13
Sidestream 21.6 14
Sidestream 25 15
Sidestream 29.3 16
Intermediate Calculations
Total Sample Size n = 16
T1 Test Statistic = 89
T1 Mean = 68
U1=n1 n2 + n1 ( n1+1 )
2 −R1=8∗8+ 8∗9
2 −89=11

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U2=n1 n2 + n2 ( n2 +1 )
2 −R2=8∗8+ 8∗9
2 −47=53
Hence, test statistic U = 11
Z Test Statistic =
U−n1 n2 /2
√ n1 n2 ( n1+n2+1 )
12
=
2.205
Two-Tail Test
Lower Critical Value = -1.96
Upper Critical Value = 1.96
The p-Value = P (Z > 2.205) = 0.0274
Decision: The null hypothesis is rejected at 5% level of significance, signifying that tar
yield in both the populations are not same at 5% level of significance.
(g) At 5% level of significance, there is no difference between the decision taken in t-test
and Mann-Whitney U test. Sample size was considerably less than 30 (where CLT can
be applied to infer normality), and therefore, non-parametric test result should be
considered for a better decision.

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Answer 4
(a) The side-by-side boxplot representing distribution of Circulation time depending on
patients’ survival status has been presented in the following figure.
Figure 7: Side-by-side Boxplot representing distribution of Circulation time depending on patients’ survival status
(b) Geometrically, average circulation time for critically ill patients surviving cardiac arrest
was noted to be below 200 seconds. Therefore, in 50% cases circulation time for
critically ill patients surviving cardiac arrest was below 200 seconds. In lowermost 25%
cases circulation time for critically ill patients surviving cardiac arrest was noted to be
approximately below 140 seconds. The topmost 25% cases circulation time for critically
ill patients surviving cardiac arrest was noted to be approximately above 240 seconds.
From the above side-by-side box plot 6 outliers were noted for circulation time for

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survived patients. Circulation time for mid-50% critically ill patients surviving cardiac
arrest was noted to have a spread of approximately 100 seconds (Potter et al., 2010, pp.
823-832).
Geometrically, average circulation time for critically ill patients died after cardiac arrest
was noted to be below 250 seconds. Therefore, in 50% cases circulation time for
critically ill patients surviving cardiac arrest was approximately below 250 seconds. In
lowermost 25% cases circulation time for critically ill patients died after cardiac arrest
was noted to be approximately below 170 seconds. The topmost 25% cases circulation
time for critically ill patients died after cardiac arrest was noted to be approximately
above 350 seconds. From the above side-by-side box plot no outliers were noted for
circulation time for died patients. Circulation time for mid-50% critically ill patients died
after cardiac arrest was noted to have a spread of approximately 180 seconds.
(c) Null hypothesis: H 0: ( μs=μd ) :Average times to pump blood for patients surviving and
dying after being critically ill due to cardiac arrest was equal.
Alternate hypothesis: HA : ( μs < μd ) :Average time to pump blood for patients surviving
takes less time than dying patients after being critically ill due to cardiac arrest.
(d) Three assumptions of independent sample t-test are: Independent observations, normality
of the dependent variable, and homogeneity between the two samples.
Independent observations: Blood circulation time for different patients is assumed to be
independent on each other.
Normality: The dependent variable, blood circulation time is noted to be distributed in a
non-normal shape. The Shapiro-Wilk (W (146) = 0.92, p < 0.05) test provided the proof
of the distribution being not normal. But, sample size was more than 30, and therefore
using CLT the distribution was assumed to be normal (Ghasemi, and Zahediasl, 2012,
p.486).
Homogeneity: Using Levene’s test, the equality between the standard deviations was
tested. Standard deviations of circulation time of surviving and dying patients after being
critically ill due to cardiac arrest was identified to be statistically equal (F = 3.87, p =
0.051) at 5% level (Kim, 2015, p.540).

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(e) Test statistic for independent sample t-test is
t = x
¿
s−x
¿
d
s∗
√ 1
n1
+ 1
n2 where “s” is the pooled
SD.
x
¿
s=206 .6 , ss=93 .84 , x
¿
d =255 .31 , sd=111 . 39 ,n1=88 , n2=58 (SPSS output)
Pooled SD is
s= √ 87∗93 . 842 +57∗111. 392
144 =101. 15
So,
t = x
¿
s−x
¿
d
s∗
√ 1
n1
+ 1
n2
=206. 6−255 .31
101 . 15∗ √( 1
88 + 1
58 ) =−2. 847
(f) The p-value for this left tailed t-test is calculated as the probability,
P ( T <t=−2 . 847 ) =0. 0025 < 0.05 for 144 degrees of freedom.
Therefore, the null hypothesis is rejected at 5% level of significance, implying that
average time to pump blood for patients surviving takes significantly less time than
dying patients after being critically ill due to cardiac arrest.
(g) SPSS output for hypothesis testing for a both tailed t-test is provided below.
Figure 8: The SPSS t-Test output for comparing Circulation time between survived and died

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(h) The t-test statistic in SPSS output is equal to hand calculated t-value, but, the p-value
needs to be differently interpreted. In SPSS the test is two tailed, and in actual the
alternate hypothesis is left-tailed. So, the p-value for one tail is the double of the value of
SPSS output = 2*0.005 =0.01 < 0.05. Hence, the decision does not change.
Answer 5
(a) Mean age of patients who enter hospital with cardiac arrest μ=50 years. Standard
deviation of age of patients’ σ =15 years.
Let age of patients X ~ N ( 50 ,152 )
Now, probability of patients who are 45 years or younger is P ( X≤45 )
So,
P ( X≤45 ) =P ( X −μ
σ ≤45−μ
σ )=P ( Z≤45−50
15 ) =P ( Z≤−0. 33 )
P ( Z≤−0. 33 ) =P ( Z≥0 . 33 ) (From symmetry of normal curve)
P ( Z≥0 . 33 ) =1−P ( Z≤0 . 33 ) =1−0 .629=0 . 371
Hence, percentage of patients aged 45 years or younger = 0.371*100 = 37.1%.
(b) Any sample with sample size equal or more than 30, chosen from a normal population
will have the characteristics of the normal population. For random sampling the sample
mean is known to be an unbiased estimator of population mean. Therefore the shape of
the distribution will be normal in nature with mean age of 50 years as the central
location. Standard deviation of the sampling distribution will be
s= σ
√ n =15
√ 50 =2. 12
.

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Figure 9: Shape of the population and sampling distribution
Therefore within 50± 2. 12 or within 47.88 years to 52.12 years, 68.2% population was
aged. Similarly, 95.44% population was aged between 50± 2∗2 .12 or within 45.76 years
and 54.24 years. Again, 99.73% population was aged between 50± 3∗2 .12 or within
43.64 years and 56.36 years.
Figure 10: 68%-95%-99.7% Rule for Normal Curve
(c) Population Mean μ=50 years, sampling SD = 2.12 years, sample size n = 50
Probability of a mean age of 45 years or younger = P ( X≤45 )
Hence,
P ( X≤45 ) =P ( Z≤45−50
2 . 12 ) =P ( Z≤−2 . 36 ) =1−P ( Z≤2 .36 ) =1−0. 991=0 .009
Therefore, probability of a mean age of 45 years or younger is 0.009.
Hence, the probability of a mean age of 45 years or younger = 0.009 signifies that the
null hypothesis assuming that the mean of population ( μ=50 ) and mean of the sample (
μ=45 ) statistically equal, is rejected at 1% level of significance.

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So, statistically, the claim of the second researcher was true that sample mean of 45 that
is chosen from a population with mean μ=50 years, SD = 15 years is unusual.
But, mean of sampling distribution for large samples is considered equal to the
population mean (using CLT), irrespective of sample means. Therefore, mean of one
sample can be different from population mean. From this point of view, the claim of the
first researcher seems to be correct, but due to the fact that sample mean is an unbiased
estimator of population mean. Sample mean cannot be systematically smaller or larger
than population mean. Therefore claim of the second researcher was true that sample
mean of 45 for a single sample is unusual.
References
Kim, T.K., 2015. T test as a parametric statistic. Korean journal of anesthesiology, 68(6),
p.540.
Ghasemi, A. and Zahediasl, S., 2012. Normality tests for statistical analysis: a guide for non-
statisticians. International journal of endocrinology and metabolism, 10(2), p.486.
Potter, K., Kniss, J., Riesenfeld, R. and Johnson, C.R., 2010, June. Visualizing summary
statistics and uncertainty. In Computer Graphics Forum (Vol. 29, No. 3, pp. 823-832).
Oxford, UK: Blackwell Publishing Ltd.
MacFarland, T.W. and Yates, J.M., 2016. Mann–whitney u test. In Introduction to
nonparametric statistics for the biological sciences using R (pp. 103-132). Springer, Cham.
Appendices
SPSS output of Mann Whitney U test (3f)

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