Biostatistics Assignment: Analysis of Statistical Problems

This biostatistics assignment provides detailed solutions to ten questions covering key concepts in biostatistics. The assignment addresses topics such as the comparison of prospective and retrospective studies, quantitative and categorical scales of measurement, sampling frames in telephone surveys, the probability of random digits, construction and interpretation of stem-and-leaf plots, measures of central location, calculation of mean and standard deviation, probability calculations in a lottery game, probability of events, and the definition of random variables. The solutions are comprehensive, providing clear explanations and calculations to help students understand the concepts and solve similar problems. This assignment is a valuable resource for students studying biostatistics, offering insights into various statistical methods and their applications. The assignment covers topics from study design and data analysis to probability and statistical measures, providing a holistic understanding of biostatistics.

Running head: BIOSTATISTICS ASSIGNMENT
BIOSTATISTIC ASSIGNMENT
Name of Student
Name of University
Author Note

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1. Prospective studies on nutrition often require subjects to keep detailed daily dietary logs.
In contrast, retrospective studies often rely on recall. Which method (dietary logs or
retrospective recall) do you believe is more likely to achieve accurate results? Explain your
response.
Answer: A prospective study is one that follows its subjects through the course of their
experiences with respect to the interest of the study and observes the reaction, in this case their
nutrition using dietary logs. This allows them to control for external factors that may interfere
with the study and induce bias. Comparatively retrospective study counts on the subject to recall
about the experiences and this has the risk of reporting bias and other errors due to factors that
are outside the control of the study. The former thus has fewer sources of unaccounted for error
and bias than the latter and is therefore preferred.
2. We often have a choice of whether to record a given variable on either a quantitative or a
categorical scale. How does one measure age quantitatively? Provide an example by which
age can be measured categorically.
Ans: Age measured in quantitative scale mere assumes the value of the number of years since
birth in a continuous scale. Say a person is born on the 1st of January 1995 then the age of that
person on 1st June would be 0.5 years as it is exactly half a year since his birth. Now measuring
the age categorically, one can divide age in m groups, say 4 groups, namely 0 to 18 years as A,
19 to 34 years as B, 35 to 54 years as C and greater than 54 years as D. Then the individual
would be assigned the value which corresponds to the interval his age falls in.

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3. Telephone surveys may use a telephone directory to identify individuals for study.
Speculate on the type of household that would be under covered by using this sampling
frame.
Answer: Telephonic surveys rely on the requirement that their population units that is
households are listed in the telephone directory that is they own a telephone. Hence a house
which may not have a telephone would not show up in the sampling frame. Households which
fall in the younger or unemployed or economically and socially deprived section of society are
usually expected to not have fixed phones at home. Thus it is this group which is expected to be
under represented in the sampling frame.
4. Could the number “0000” appear in a table of random digits? If so, how likely is this?
Answer: In a table with values ranging from 0000 to 10000 any number has an equal chance
to occur. The probability that any number between 0000 and 10000 would occur in 1/10000 and
so the number 0000 has as good chance of appearing as any other number in the range of
numbers in the table.
5. Body weights of 18 diabetics expressed as a percentage of ideal (defined as body weight
divided by ideal body weight x 100) are listed: {107, 119, 99, 114, 120, 104, 88, 114, 124, 116,
101, 121, 152, 100, 125, 114, 95, 117}. Construct a stem-and-leaf plot of these data and
interpret your findings.
Answer:

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9 5 9
10 0 1 7
11 4 4 4 6 7 9
12 0 1 4 5
13
14
15 2
The stem and leaf plot shows that the key is 11/4 = 114. The median is given by the average
of the 8th and 9th number as there are a total of 16 observations. The median is then (114+116)/2
= 115. The data is positively skewed.
6. Name three measures of central location.
Answer: Three measures of central location are mean, median and mode.
7. To assess the air quality in a surgical suite, the presence of colony-forming spores per
cubic meter of air is measured on three successive days. The results are as follows: {12, 24,
30}. Calculate the mean and standard deviation for these data.
Answer: The mean is computed as sum of the numbers divided by total number of numbers=
(12+24+30)/3 = 66/3 = 22.
The standard deviation is the square root of the mean squared deviation from the mean and is
given by the formula
√ ∑
i=1
3
(xi−x )2
3
= √ {(12−22)2 +(24−22)2 +(30−22)2 }
3

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= √ {(−10)2 +(2)2 +( 8)2 }
3
=
√ {100+ 4+ 64 }
3
= √ {168 }
3 =√56 = 7.483
8. In a lottery game, a person must select 5 numbers from a total of 40. Tracy has chosen 7,
8, 9, 10, 11. Jaime has chosen 39, 17, 37, 5, 28. Who has a greater chance of winning?
Answer: The probability for selecting the first set of numbers is 1/ ( 40
4 ) that is 0.0000109. This is
because the given arrangement of 4 numbers is not independently drawn and can be selected in
only 1 way out of all the ways that 4 numbers can be drawn from 40 numbers without repetition.
The chance for selecting the second selection is 1
40 × 1
39 × 1
38 × 1
37 that is 0.00000045. This is
because the numbers are drawn independently without replacement from a total of 40 numbers.
Then the chance that the first combination is selected is higher than the second one. Therefore
Tracy is more likely to win.
9. In a box, there are 8 orange, 7 blue, and 6 red balls. One ball is selected randomly. What
is the probability that it is neither orange nor red?
Answer: The case where neither an orange nor red ball is selected implies that the ball selected
must be blue. Then the chance that out of 8 orange, 7 blue and 6 red balls that is a total of
8+7+6= 21 balls, a blue ball is selected is 7/21 which is 1/3.

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10. A ___Random Variable___ is a numerical quantity that takes on different values
depending on chance. There are two types of random variables. ___Discrete_______ form a
countable set of possible values. ___Continuous_______ form an unbroken continuum of
possible values.