Biostatistics Assignment Solutions and Examples

This page provides solutions and examples for Biostatistics assignments, including calculating confidence intervals, critical values, effect size, and more. It also includes a comparison of risk-taking behavior between boys and girls using side-by-side box plots.

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Biostatistics Assignment

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ANS: Confidence interval for mean is
x
−
± t0 . 95
s
√ n , where x
−
=2998 g is the mean and s=800 g is
the standard deviation of the sample, n=49 is the sample size, and tα is the value of the statistic
for given confidence level ( tα=±2. 01 for 95 % CI , two tailed ). The confidence interval is
[2998−2. 01∗800
√ 49 , 2998+ 2. 01∗800
√ 49 ]
= [2768 . 29 , 3227 . 71 ] for 95% confidence level.
From the sampling attributes it is assessed that there is a 95% chance that average birth weight of
SIDS cases will be within the limits of confidence interval.
ANS:
Sample mean =
7 . 40+ 7. 36+7 . 45
3 =7. 40 , population standard deviation is σ = 0.070. As
titer measurement follows normal distribution, 95% confidence interval is calculated as
x
−
± z0. 95
σ
√ n

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Hence, the 95% confidence interval is
[7 . 40−1. 96∗0. 07
√ 3 , 7 . 40+1. 96∗0. 07
√ 3 ]=[7 . 32 , 7. 48 ]
where
z0. 95=1 . 96 .
ANS: a (False), b (False), c (True)
ANS: Critical value of t-statistics for α = 0.05 are as follows:
Right-tail tcrit =1.725 (degrees of freedom (DOF) = n -1 = 20), left-tail tcrit = - 1.725 (degrees of
freedom (DOF) = n -1 = 20). Therefore, P ( t stat≤−1. 725 ) = 0.05 and P ( tstat≥−1. 725 ) = 0.05.
Two-tail critical value of t-statistics is t crit =± 2.086 (DOF = 20), which implies
P ( −2. 086≤tstat≤2. 086 ) = 0.05.

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ANS: Both z-procedure and t-procedure follows Gaussian distribution. But, in absence of the
knowledge of population standard deviation, sample standard deviation is used for estimating the
population parameters. Sample standard deviation is known to follow t-distribution instead of z-
procedure, and for estimating population parameter t-procedure is followed.
ANS: Sample size: n = 26 boys, sample mean = 63.8 inches, and sample standard deviation = 3.1
inches.
Population standard deviation is not known, and sample S.D is known to follow t-distribution.
Degrees of freedom = 26 – 1 =25. Hence,
tcrit =2.086 for two tail implies that the 95%
confidence interval is estimated as
x
−
± t0 . 95
s
√ n .
The calculated 95% confidence interval for the mean height of the population is as follows: -
[63 . 8−2 . 086∗ 3. 1
√ 26 , 63. 8+2 . 086∗ 3. 1
√ 26 ]=[62 . 53 , 65. 07 ]

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Average population height is expected to be within 62.53 inches and 65.07 inches.
ANS: a. Paired Samples
b. Independent Samples
c. Single Sample
ANS: The null hypothesis: H0: μd=0 and the alternate hypothesis HA: μd≠0
The sample size is
n= ( Z1−α
2
+ Z1−β
Effect Size )
2
, where
Z1−α
2 = 1.96 for α = 0.05 (two-sided) and Z1−β =
1.282 for 90% power (Williams, 2017).

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The effect size =
|Δμ|
σ = 0 .25
0 .67 =0 .37 , and
n= ( Z1−α
2
+ Z1−β
Effect Size )
2
=( 1 .96 +1. 282
0 . 37 )
2
=76 .78
Therefore, sample size of 77 has 90% power at α = 0.05 (two-sided).
ANS: Back-to-back steam and Side-by-side box plots can compare two independent groups of
quantitative data.
ANS: A measure of index of risk-taking behavior in boys and girls has been explored utilizing
group differences with side-by-side box plots.

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6 0 8 0 1 0 0 1 2 0 1 4 0
Index_Boys Index_Girls
Figure 1: Side-by-side Box Plot for index of risk-taking behavior
The index for boys is left skewed, whereas index for girls is slightly right skewed. Median of
girls is higher than that of the boys. From the box plots an outlier index value of 125 is clearly
visible for the boys. Risk taking attitude is visibly greater for the girls compared to that of the
boys (Norman, & Streiner, 2008).
References
Norman, G. R., & Streiner, D. L. (2008). Biostatistics: The Bare Essentials. PMPH-USA.
Williams, B. (2017). Biostatistics : Concepts and Applications for Biologists. CRC Press.
https://doi.org/10.1201/9781315150314