Operational Statics Measures Interview Preparation Guide Download PDF
Statics Measures of association Interview Question and Answers will guide us now how to get preparation for the jobs of Statics Measures of association with the help of this Statics Measures of association Interview Question with Answers guide
29 Statics Measures Questions and Answers:
Table of Contents:
1 :: List some measure used in the study of static measures?
The measures used in this study are eight:
☛ Number of sentences
☛ Number of atomic conditions per decision
☛ Total number of decisions
☛ Number of equalities
☛ Number of inequalities
☛ Nesting degree
☛ McCabe's cyclomatic complexity
☛ Branch coverage
☛ Number of sentences
☛ Number of atomic conditions per decision
☛ Total number of decisions
☛ Number of equalities
☛ Number of inequalities
☛ Nesting degree
☛ McCabe's cyclomatic complexity
☛ Branch coverage
2 :: What are the kinds of measures?
There are 2 kinds of measures:
Dynamic Measures:
Which requires the execution of the program.
Static Measures:
Which does not require the execution.
Dynamic Measures:
Which requires the execution of the program.
Static Measures:
Which does not require the execution.
3 :: What does cyclomatic complexity apply to?
Cyclomatic complexity may also be applied to individual functions, modules, methods or classes within a program, and is formally defined as follows:
v(G) = E − N + 2P
where E is the number of edges of the graph, N is the number of nodes of the graph and P is the number of connected components.
v(G) = E − N + 2P
where E is the number of edges of the graph, N is the number of nodes of the graph and P is the number of connected components.
4 :: How is cyclomatic complexity computed?
Cyclomatic complexity is computed using the control flow graph of the program, the nodes of the graph correspond to indivisible groups of sentences of a program and a directed edge connects two nodes if the second sentence might be executed immediately after the first sentence.
5 :: What is a cyclomatic complexity?
Cyclomatic complexity is a complexity measure of code related to the number of ways there exists to traverse a piece of code. This measure determines the minimum number of test cases needed to test all the paths using linearly independent circuits.
6 :: What is statement coverage?
Statement coverage is defined as the percentage of statements that are executed.
7 :: What is a branch coverage?
branch coverage is the percentage of branches exercised in a program. This coverage measure is used in most of the related papers in the literature.
8 :: What is a nesting degree?
The nesting degree is the maximum number of conditional statements that are nested one inside another.
9 :: What is number of inequalities?
The number of (in)equalities is the number of times that the operator is found in atomic conditions of a program.
10 :: What are quantitative models?
Quantitative models are frequently used in different engineering disciplines for predicting situations, due dates, required cost and so on. These quantitative models are based on some kinds of measure performed on project data or items.
11 :: Should you use a one-tailed, or a two-tailed hypothesis when doing a chi square test?
1. Two-tailed.
2. It doesn't matter.
3. One-tailed.
4. SPSS will include the right one in the output.
Answer: It doesn't matter.
2. It doesn't matter.
3. One-tailed.
4. SPSS will include the right one in the output.
Answer: It doesn't matter.
12 :: One serious complication associated with the analysis of more than three levels (4 x 5) is?
1. The sample size would have to be so large that chi square analysis would not be powerful enough to interpret the data.
2. It can be difficult to interpret accurately all of the relationships within a large contingency table.
3. That type of analysis would not meet the criteria for a chi square test.
4. They would have to be analyzed by hand as SPSS has no option for tables larger than 3x3.
Answer: it can be difficult to interpret accurately all of the relationships within a large contingency table.
2. It can be difficult to interpret accurately all of the relationships within a large contingency table.
3. That type of analysis would not meet the criteria for a chi square test.
4. They would have to be analyzed by hand as SPSS has no option for tables larger than 3x3.
Answer: it can be difficult to interpret accurately all of the relationships within a large contingency table.
13 :: Data for a chi square test should be assumed to have no less than one participant per cell. If there is less than one participant per cell, it is sometimes useful to combine cells together into one category?
* True
* False
Answer: TRUE
* False
Answer: TRUE
14 :: Refer back to the example in question 14 which look at submission of essays and time planning. In terms of essay submission, number of early, late and on time students are counted. The number of students who planned their time was also counted, leading to two levels of time planning or not. How would this analysis be described?
1. 3x2
2. 3x3
3. 2x2
4. 3x1
Answer: 3x2
2. 3x3
3. 2x2
4. 3x1
Answer: 3x2
15 :: For a 2x2 chi square test, which of the following equations would be used to calculate the degrees of freedom?
1. (r-1) x (c-1) x (n-1)
2. (r-c) x (n-1)
3. (r-1) x (c-1)
4. (r+1) x (c+1)
Answer: r-1) x (c-1)
2. (r-c) x (n-1)
3. (r-1) x (c-1)
4. (r+1) x (c+1)
Answer: r-1) x (c-1)
16 :: What does the Fishers Exact Probability test show?
1. Fisher's Exact Probability Test shows the F statistic associated with the chi square value when the null is assumed to be true.
2. The Fisher's Exact Probability Test shows the probability of reaching the assumption of 25% of cells with an expected frequency of less than 5.
3. The Fisher's Exact Probability Test shows the percentage of variation which one variable accounts for in the other.
4. The Fisher's Exact Probability Test shows the probability of obtaining the chi square value when the null is assumed to be true.
Answer: The Fisher's Exact Probability Test shows the probability of obtaining the chi square value when the null is assumed to be true.
2. The Fisher's Exact Probability Test shows the probability of reaching the assumption of 25% of cells with an expected frequency of less than 5.
3. The Fisher's Exact Probability Test shows the percentage of variation which one variable accounts for in the other.
4. The Fisher's Exact Probability Test shows the probability of obtaining the chi square value when the null is assumed to be true.
Answer: The Fisher's Exact Probability Test shows the probability of obtaining the chi square value when the null is assumed to be true.
17 :: You conduct a study exploring whether or not students planned their time and whether or not they submitted their assignment on time, your SPSS output shows a value for Cramers V of 0.42. How would you interpret this?
1. 8% of the variation in frequency counts of essay submission timing (on time or late) can be explained by time planning.
2. 42% of the variation in frequency counts of essay submission timing (on time or late) can be explained by time planning.
3. 64.8% of the variation in frequency counts of essay submission timing (on time or late) can be explained by time planning.
4. 4.2% of the variation in frequency counts of essay submission timing (on time or late) can be explained by time planning.
Answer: 8% of the variation in frequency counts of essay submission timing (on time or late) can be explained by time planning.
2. 42% of the variation in frequency counts of essay submission timing (on time or late) can be explained by time planning.
3. 64.8% of the variation in frequency counts of essay submission timing (on time or late) can be explained by time planning.
4. 4.2% of the variation in frequency counts of essay submission timing (on time or late) can be explained by time planning.
Answer: 8% of the variation in frequency counts of essay submission timing (on time or late) can be explained by time planning.
18 :: What is Cramers V used for?
1. Cramer’s V is used instead of X2 in analyses which are bigger than 2x2.
2. Cramer's V is used when assumptions for conducting chi square are violated.
3. Cramer's V is a measure of effect used for tests of association.
4. Cramer's V is a way of reporting the ratio between the observed and expected scores.
Answer: Cramer's V is a measure of effect used for tests of association.
2. Cramer's V is used when assumptions for conducting chi square are violated.
3. Cramer's V is a measure of effect used for tests of association.
4. Cramer's V is a way of reporting the ratio between the observed and expected scores.
Answer: Cramer's V is a measure of effect used for tests of association.
19 :: When reporting your results, what elements should you include from the SPSS output?
1. The number of participants, the X2 value, and the probability level
2. The Pearson's X2, degrees of freedom and the probability level.
3. The number of participants, the degrees of freedom, X2, and the probability level
4. The X2 and the probability level.
Answer: The Pearson's X2, degrees of freedom and the probability level.
2. The Pearson's X2, degrees of freedom and the probability level.
3. The number of participants, the degrees of freedom, X2, and the probability level
4. The X2 and the probability level.
Answer: The Pearson's X2, degrees of freedom and the probability level.
20 :: If the assumption mentioned in question 10 is not met for a 2x2 chi square test, you should proceed to conducting _________?
1. A Pearson's correlation coefficient
2. A 2x2 test of independence
3. One variable chi square test (goodness of fit)
4. A Fisher's Exact Probability Test
Answer: a Fisher's Exact Probability Test
2. A 2x2 test of independence
3. One variable chi square test (goodness of fit)
4. A Fisher's Exact Probability Test
Answer: a Fisher's Exact Probability Test
21 :: A fundamental assumption of chi square tests is that no more than ____ % of cells can have an expected frequency of less than?
1. 25; 5
2. 75; 95
3. 4; 1
4. 5; n
Answer: 25; 5
2. 75; 95
3. 4; 1
4. 5; n
Answer: 25; 5
22 :: Which of the below statements is false of chi square testing?
1. Chi square tests can be used to check how well a model fits the data
2. Chi square can be applied to continuous variables; it just means that a larger contingency table is needed.
3. Chi square is used in research to measure the association between two categorical variables.
4. None of these statements are false, it is a trick question.
Answer: Chi square can be applied to continuous variables; it just means that a larger contingency table is needed.
2. Chi square can be applied to continuous variables; it just means that a larger contingency table is needed.
3. Chi square is used in research to measure the association between two categorical variables.
4. None of these statements are false, it is a trick question.
Answer: Chi square can be applied to continuous variables; it just means that a larger contingency table is needed.
23 :: Although in one variable chi square testing each participant cannot be in more than one group, in a 2x2 chi square test, this rule does not apply?
* True
* False
Answer: FALSE
* False
Answer: FALSE
24 :: Examine the output on p. 268. How would these results be reported?
1. The chi square value of 10.490 (DF=317) achieved an associated p value of <.001. There was a significant difference between the expected and the observed frequencies. We can conclude that there is a greater prevalence of right handedness in women with IBS.
2. The chi square value of 317 (DF=1) achieved an associated p value of <.001. There was a significant difference between the expected and the observed frequencies. We can conclude that there is a greater prevalence of left handedness in women with IBS.
3. The chi square value of 10.490 (DF=1) achieved an associated p value of .001. There was no significant difference between the expected and the observed frequencies. We can conclude that being left or right handed is unrelated to IBS in women.
4. The chi square value of 10.490 (DF=1) achieved an associated p value of <.001. There was a significant difference between the expected and the observed frequencies. We can conclude that there is a greater prevalence of left handedness in women with IBS.
Answer: The chi square value of 10.490 (DF=1) achieved an associated p value of <.001. There was a significant difference between the expected and the observed frequencies. We can conclude that there is a greater prevalence of left handedness in women with IBS.
2. The chi square value of 317 (DF=1) achieved an associated p value of <.001. There was a significant difference between the expected and the observed frequencies. We can conclude that there is a greater prevalence of left handedness in women with IBS.
3. The chi square value of 10.490 (DF=1) achieved an associated p value of .001. There was no significant difference between the expected and the observed frequencies. We can conclude that being left or right handed is unrelated to IBS in women.
4. The chi square value of 10.490 (DF=1) achieved an associated p value of <.001. There was a significant difference between the expected and the observed frequencies. We can conclude that there is a greater prevalence of left handedness in women with IBS.
Answer: The chi square value of 10.490 (DF=1) achieved an associated p value of <.001. There was a significant difference between the expected and the observed frequencies. We can conclude that there is a greater prevalence of left handedness in women with IBS.
25 :: You are conducting a one variable chi square test to test the hypothesis that there are equal numbers of vegetarians, meat eaters, and vegans eating at the student union. The categories are vegetarian, meat eaters, and vegans. Having conducted a survey, you found 85 individuals were vegetarian, 122 ate meat, and 32 followed a vegan diet. What would the expected frequencies be in each cell?
1. 239
2. 85, 122, and 32.
3. 79.67
4. There is insufficient information provided to calculate the expected frequencies.
Answer: 79.67
2. 85, 122, and 32.
3. 79.67
4. There is insufficient information provided to calculate the expected frequencies.
Answer: 79.67
