Assessment Task Tutorial

This tutorial covers various assessment tasks including population parameter and sample statistic, descriptive statistics and inferential statistics, nominal scale and ordinal scale, primary data source and secondary data source, joint probability table, conditional probability, probability of having salary, percentage of graduate receive tax break, number of graduates, and SAT mean scores.

Assessment Task Tutorial
Questions

Question 1:
(a) Population Parameter and Sample Statistic
A parameter defines characteristics of the entire population like average of data, standard
deviations and more. While sample statistics describes characteristics of a small sample of data,
which helps in drawing more accurate conclusion about biased or unbiased estimates.
(b) Descriptive Statistics and Inferential Statistics
Both terms are considered as two main categories of statistics, where descriptive statistics
help in describing a sample by using tools like central tendency, dispersion, skewness and
kurtosis etc. While inferential statistics drawn data from a sample, to draw conclusion and
generalise the same to a population. For this purpose, it utilises tools like hypothesis test,
regression analysis, confidence interval, ANOVA and more.
(c) Nominal Scale and Ordinal Scale
Both measurement scales are used for analysing the information in a detailed manner,
where nominal scale utilises for categorising the data into specific form like gender, ethnicity,
colour and more. While ordinary scale analyses information by arranging data into a particular
sequence, such as on the basis of ranks.
(d) Primary Data Source and Secondary Data Source
Primary data source includes methods like questionnaire, online and offline survey, etc. for
gathering and analysing original information. While secondary data source includes census
report, statistical survey, business data and more etc., for obtaining result.
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Question 2
a. A Joint Probability Table
Given contingency table
Uses Social Media and
Other Websites to
Voice Opinions About
Television Programs
Doesn’t Use Social
Media and Other
Websites to Voice
Opinions About
Television Programs
Total
Female 395 291 686
Male 323 355 678
Total 718 646 1364
P (male) = 678/ 1364 = 0.49
P (female) = 686/ 1364 = 0.50
P (uses social media) = 718/1364 = 0.53
P (do not use social media) = 646/1364 = 0.47
P (female use social media) = 395/1364 = 0.29
P (male use social media) = 323/1364 = 0.24
P (female do not use social media) = 291/1364 = 0.21
P (male do not use social media) = 355/1364 = 0.26
So, Joint Probability table
Uses Social Media and
Other Websites to
Voice Opinions About
Television Programs
Doesn’t Use Social
Media and Other
Websites to Voice
Opinions About
Television Programs
Total
Female 0.29 0.21 0.50
Male 0.24 0.26 0.50
Total 0.53 0.47 1
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b) Probability of respondent who is female
From above
P (female use social media) = 0.29
c) Conditional probability
P (social media │female) = 395/718 = 0.55
d) To check if F and A are independent
Two events are independent when P (A∩B) = P(A). P(B)
here, F denote the event that the respondent is female
and, A denote the event that the respondent uses social media
then,
P (F∩A) = P(F). P(A)
0.29 ≠ 0.50 x 0.50
Hence, F and A are not independent
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Question 3
(a) Probability of having salary at least 30,400
P (X ≥ 30,400) = X – Mean
std. dev
= 30,400 – 20,000
8000
= 1.3
(b) Probability of having salary exactly 30,400
P (X = 30,400) = 0
(c) Percentage of graduate receive tax break
P (X < 15600) = 15600 – 20000
8000
= -0.55
so, 55% of graduate having salary less than 15,600 will receive tax break
(d) Number of graduates current year
Number of students having salaries at least 32,240 = 189
Then total number of graduates =
P (X ≥ 32,240) = X – Mean
std. dev
= 32,240 – 20,000
8000
= 1.53
P (X ≥ 32,240) = Number of students having at least 32240
total number of graduates
1.53 = 189
n
n = 124
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Question 3
SAT Mean Scores
Critical reading 502
Mathematics 515
Writing 494
Standard deviation is 100
(a) z = X – Mean
std. dev/√n
= ± 10
100/ √90
≈ ±0.94
P = (-10 < z < 10) = P (-0.94 < z < 0.94)
= 1 – 2P (z < -0.95) using z table
= 1 – 2 x 0.1711 ≈ 0.66
(b) z = X – Mean
std. dev/√n
= ± 10
100/ √90
≈ ±0.94
P = (-10 < z < 10) = P (-0.94 < z < 0.94)
= 1 – 2P (z < -0.95) using z table
= 1 – 2 x 0.1711 ≈ 0.66
(c) (a) z = X – Mean
std. dev/√n
= ± 10
100/ √100
≈ ±1
P = (-10 < z < 10) = P (-1 < z < 1)
= 1 – 2P (z < -1) using z table
= 1 – 2 x 0.1587 ≈ 0.68
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