Final Exam Solutions: Applied Statistical Methods and Analysis

This document provides a comprehensive solution to an applied statistical methods assignment. It begins by differentiating between descriptive and inferential statistics, followed by the analysis of a dataset of teacher satisfaction scores. The solution includes the construction of both ungrouped and grouped frequency distributions, along with a histogram representing the grouped data. The assignment then proceeds to calculate the mean, median, mode, and standard deviation of the data. Further, the solution addresses hypothesis testing, comparing the mean test scores of two groups with different sample sizes and evaluating the significance of the results. The document concludes with solutions to questions involving correlation and regression analysis, including the prediction of a value using a regression equation.

APPLIED STATISTICAL METHODS
STUDENT ID:
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Question 1
Difference between descriptive statistics and inferential statistics
 Descriptive statistics includes study of the population whereas inferential statistics
includes the study of sample analysis and observation from which the various conclusions
would be drawn about population.
 Descriptive statistics generally analyse and organise the sample data in meaningful ways
whereas inferential statistics compares, test and predicts the population parameters.
 Descriptive statistics merely explains research but does not make any
conclusions/prediction while inferential statistics provides possible conclusions and
prediction about the event.
 Descriptive statistics represent results in form charts, graphs and tabular forms whereas
inferential statistics represent results in the form of probability.
Question 2
Ungrouped frequency distribution of scores
2

Grouped frequency distribution of score
Question 3
Histogram of the grouped frequency for the scores
4-7 8-11 12-15 16-19 20-23 24-27 28-31
0
2
4
6
8
10
12
14
16
18
20 Histogram
Scores
Frequency
Question 4
Option B (25)
Question 5
Mean=∑ f ( y )
n = 3+5+4 +3+2+3
6 = 20
6 =3.33
Median
3

Here, the number of observations is even and thus,
Median= ( n
2 )th value+( n
2 +1 )th value
2
Scores in ascending data
Median= 1
2 {( 6
2 )th value+( 6
2 +1 )th value }= 1
2 { ( 3+3 ) }=3
Mode=Maximum frequency has ovserved for 3∧thus ,
Mode = 3
Question 6
Standard deviation
Standard deviation= √ ∑ (f ( y )¿−mean)2
n−1 = √ 5.3334
6−1 =1.03 ¿
4

Question 7
Group 1
Number of participants = 15
Mean = 75
Variance = 120
Standard deviation = sqrt (variance) = sqrt (120) = 10.95
Group 2
Number of participants = 15
Mean = 86
Variance = 100
Standard deviation = sqrt (variance) = sqrt (100) = 10
(a) Null and alternative hypothesis
Null hypothesis H0 :μgroup 1−μgroup2 =0
Alternative hypothesis Ha : μgroup 1−μgroup 2 ≠ 0
The standard deviation of difference Sd= sqrt (s1^2 /n1 + s2^2/n2)
= sqrt(10.95^2 /15 + 10^2 /15) = 3.830
Now,
t value= x 1− x 2
Sd =75−86
3.830 =−2.872
Degree of freedom = n1+n2-2 = 15+15-2 =28
The p value for t value and degree of freedom (two tailed) = 0.00769
Significance level = 0.05
5

It can be said based on the above that p value is lower than significance level and thus,
sufficient evidence is present to reject the null hypothesis and to accept the alternative
hypothesis. Therefore, it can be said that the mean test scores of the two groups is statistically
different.
(b) When there are 7 participants in each group.
The standard deviation of difference Sd= sqrt (s1^2 /n1 + s2^2/n2)
= sqrt (10.95^2 /7 + 10^2 /7) = 5.606
Now,
t value= x 1− x 2
Sd =75−86
5.606 =−1.962
Degree of freedom = n1+n2-2 = 7+7-2 = 12
The p value for t value and degree of freedom (two tailed) = 0.0597
Significance level = 0.05
It can be said based on the above that p value is higher than significance level and thus,
insufficient evidence is present to reject the null hypothesis. Therefore, it can be said that the
mean test scores of the two groups is not statistically different.
Question 8
Answer: Option a) -4.3
Question 9
Answer: Option a) -0.81
Question 10
Option C (0.50)
6

Question 11
The regression equation can be formed based on the given information.
y=4.4+2 x
x=score=12
Predicted value Y =4.4+(2∗12)=28.4
Question 12
Option A If two variables share 25% of their variation, the correlation between the two variables is:
7