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CS706 Assignment 2 Solution and Discussion

zareen Cyberian's Gold last edited by
Assignment No. 02
SEMESTER Fall 2019

CS706 – Software Quality Assurance

Due Date: December 05, 2019

Instructions to Solve Assignments

The purpose of assignments is to provide you with application-oriented exercise to strength the concepts. It is expected that students will solve the assignments themselves. Following rules will apply during the evaluation of assignment.

• Cheating from any source will result in zero marks in the assignment.
• In case of cheating from any other student, both students will be awarded “F” grade in the course.
• In case of question No. 03 direct copy and paste from the research paper will be awarded zero marks.
• No assignment after due date will be accepted.

Question No. 1 [Marks 21]
You have studied different types of cohesion e.g. procedural cohesion, temporal cohesion, communicational cohesion etc. In the table below you are given Different code snippets, some details are also given against each code piece. You are required to identify and mention the type of cohesion each of these code snippets represents:

Question No. 2 [Marks 09]
Given below is pseudo code of a “BankAccount” Module. You are required to identify the issue(s) with this module (from quality perspective). Justify you answer with solid reasons.
```
Class BankAccount {
Public String AccountNo;
Public String AccountType
Public Float AccountBalance;
Private OpenCurrentAccount() {
}
Private UpdateCurrentAccount() {
}
Private CloseCurrentAccount() {
}
Private OpenSavingAccount() {
}
Private UpdateSavingAccount() {
}
CloseSavingAccount() {
}
} // Comment: End of Class
```
Description: Module contains three public/global data members to store different contents of information about a bank account. Module also has different private/local functions to perform different types of operations. Furthermore, module can handle two types of accounts which are current accounts and saving accounts.

Question No. 3 [Marks 20]
Read the paper titled “Continuous Code Quality: Are We (Really) Doing That?” attached with this assignment and answer the following questions:
1. Provide a brief summary of the paper (less than or equal to 250 words)
2. Provide critical review (i.e. strengths and weaknesses of the paper)
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@zareen said in EDU431 Assignment 3 Solution and Discussion:

Differentiate between Classical Test Theory and Item response Theory of item analysis

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.CPP File Code
using namespace std; #include <stdlib.h> #include <iostream> struct StudentDetail{ string name; string vuid; }; //class Node* head; class Node{ struct StudentDetail newStd; class Node* next; class Node* prev; void Set(string name,string vuid) { newStd.name=name; newStd.vuid=vuid; } Node Get() { } void setNext(string name,string vuid){ if(next==NULL) { } else { class Node* newNode= new Node(); newStd.name=name; newStd.vuid=vuid; newNode->next= newNode; } } string getNext() { next; } void setPrev(string name,string vuid){ if(next==NULL) { } else { class Node* newNode= new Node(); newStd.name=name; newStd.vuid=vuid; newNode->next= newNode; } } string getPrev() { prev; } }; class DoublyLinkedList{ public: struct StudentDetail newStd; class DoublyLinkedList* headPtr; class DoublyLinkedList* curPtr; class DoublyLinkedList* nextPtr; int size; //dfdf class DoublyLinkedList* headDlinkList=NULL; void addAtBegining(string vuid, string name) { class DoublyLinkedList* dNode= new DoublyLinkedList(); dNode->newStd.vuid=vuid; dNode->newStd.name=name; dNode->nextPtr=headDlinkList; headDlinkList= dNode; dNode->curPtr=dNode; } void addAtEnd(string vuid, string name) { class DoublyLinkedList* dNode= new DoublyLinkedList(); dNode->newStd.vuid=vuid; dNode->newStd.name=name; dNode->nextPtr=headDlinkList; headDlinkList= dNode; dNode->curPtr=dNode; } void delNode() { class DoublyLinkedList* temp1=curPtr; class DoublyLinkedList* temp2= temp1; temp1->nextPtr= temp2->nextPtr; free(temp2); } void print() { class DoublyLinkedList* temp= headDlinkList; while(temp!=NULL) { cout<<temp->newStd.vuid<<" "<<temp->newStd.name<<endl; temp= temp->nextPtr; } } }; int main() { string vuid,name; cout<<"Add your vuID and Name at First Position "<<endl; cout<<"_ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ __ _ _"<<endl; DoublyLinkedList dlist1; cin>>vuid; cin>>name; dlist1.addAtBegining(vuid,name); dlist1.print(); cout<<"Insertion At Beginning in doubly Link List "<<endl; cout<<"_ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ __ _ _"<<endl; cin>>vuid; cin>>name; dlist1.addAtBegining(vuid,name); dlist1.print(); cout<<"Insertion At End in doubly Link List "<<endl; cout<<"_ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ __ _ _"<<endl; cin>>vuid; cin>>name; dlist1.addAtEnd(vuid,name); dlist1.print(); cout<<"Deletion of Current Node (Last Node) "<<endl; cout<<"_ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ __ _ _"<<endl; dlist1.delNode(); dlist1.print(); }
Download .cpp File
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Solution:
Question no. 1:
Test the hypothesis that the mean of a normal population with known variance 70 is 31, if a sample of size 13 gave x ̅ = 34. Let the alternative hypothesis be H1: µ > 31, and let α = 0.10.
Solution:
Formulation of Hypothesis:
Ho: µ = 31
H1: µ > 31
Level of Significance:
α = 0.10
Test Statistics:
z=(x ̅-μ)/(σ⁄√n)
Calculation:
z=(34-31)/(√70⁄√13)

z = 1.29.
Critical Region:
Reject Ho, if
Z > Zα
1.29 > Z0.10
1.29 > 1.28
We reject Ho.
Conclusion:
Since 1.29 > 1.28 fall in the critical region, so we reject Ho.
Questions no. 2:
Explain in detail why it is not a good statistical procedure to perform several t-test on pairs of means , when several means are to be compared. Suggest the alternative statistical procedure and also give its assumptions.
Solution:
Whenever we compare more than two population means, we apply the two-sample t-test to all possible pairwise comparisons of means. For example, if we wish to compare 4 population means, there will be 6 pairs and to test the hypothesis that all four population means are equal, would require six two-sample t-test. This type of multiple two-sample t-test has two disadvantages. First, the procedure is difficult and time consuming and secondly, the level of significance increases as the number of t-test increases. Thus, a series of two-sample t-test is not a good procedure. ANOVA is a technique that measure the variations between the means.
Assumptions of ANOVA:
1: Experimental errors are normally distributed.
2: Equal variance between the treatments.
3: Samples are independent.

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