Statistics II · Unit 3 · 8 hrs
Non parametric test
Exam-focused notes for Non parametric test (Statistics II, STA215): what the TU syllabus asks and how it has actually been tested, with 15 solved past questions from this unit.
What this unit covers
- Parametric vs. non-parametric test
- Needs of applying non-parametric tests
- One-sample test: Run test, Binomial test, Kolmogorov–Smirnov test
- Two independent sample test: Median test, Kolmogorov-Smirnov test, Wilcoxon Mann Whitney test, Chi-square test
- Paired-sample test: Wilcoxon signed rank test
- Cochran's Q test
- Friedman two way analysis of variance test
- Kruskal Wallis test
- Problems and illustrative examples related to computer Science and IT
Two independent sample test
Chi-Square Test of Independence
Based on interviews of couples seeking divorce, a social worker compiles the following data related to the period of acquaintanceship before marriage and the duration of marriage. Perform a test to determine if the data substantiate an association between the duration of a marriage and the acquaintanceship prior to marriage. Use 5% level of significance.
| Acquaintanceship before marriage | Less than or equal to 5 years | More than 5 years | Total |
|---|---|---|---|
| Below 0.5 year | 15 | 7 | 22 |
| 0.5–1.5 years | 26 | 22 | 48 |
| Over 1.5 years | 19 | 11 | 30 |
| Total | 60 | 40 | 100 |
[5]
Observed contingency table: Acquaintanceship ≤ 5 years 5 years Total ------------ Below 0.5 yr 15 7 22 0.5-1.5 yr 26 22 48 Over 1.5 yr 19 11 30 Total 60 40 100 Significance level $\alpha = 0.05$. - $H0$: No association between duration of marriage and acqua...
Full solved answer →Test of Independence: Opinion on Core Curriculum Change vs. Class Standing
A random sample of students is asked their opinion on proposed core curriculum change. The results are as follows. Test the hypothesis that opinion on the change is independent of class standing. Use 0.01 significance level.
$$\begin{array}{|c|c|c|} \hline \text{Class} & \text{Favoring} & \text{Opposing} \ \hline \text{Freshman} & 125 & 80 \ \text{Sophomore} & 60 & 140 \ \text{Junior} & 50 & 60 \ \text{Senior} & 40 & 55 \ \hline \end{array}$$
[5]
Observed frequencies: Class Favoring Opposing --------------------------- Freshman 125 80 Sophomore 60 140 Junior 50 60 Senior 40 55 Significance level: $\alpha = 0.01$ - $H0$: Opinion on the change is independent of class standing. - $H1$: Opinion is not i...
Full solved answer →A study showed the following results for the different age groups. At the 0.05 level of significance, is there evidence of a difference among the age groups with respect to use of mobile phones for accessing social networking?
$$\begin{array}{|c|c|c|c|} \hline \text{Use mobile phones to access social networking?} & \text{18-34} & \text{35-64} & \text{65+} \ \hline \text{Yes} & 60 & 37 & 14 \ \text{No} & 40 & 63 & 86 \ \hline \end{array}$$
[5]
18-34 35-64 65+ Row Total --------------- Yes 60 37 14 111 No 40 63 86 189 Col Total 100 100 100 300 Level of significance: $\alpha = 0.05$ - $H0$: There is no difference among age groups regarding use of mobile phones for social networking (independent). -...
Full solved answer →Apply the Mann-Whitney U test for examining the following knowledge score on IT among two groups of IT workers at a 5% level of significance.
| Group A: | 5 | 8 | 2 | 7 | 6 |
|---|---|---|---|---|---|
| Group B: | 9 | 12 | 4 | 6 |
[5]
Group A (n₁ = 5) 5 8 2 7 6 ------------------ Group B (n₂ = 4) 9 12 4 6 - H₀: No significant difference in knowledge scores between the two groups. - H₁: There is a significant difference between the two groups. - Significance level: α = 0.05 (two-tailed). ...
Full solved answer →Chi-Square Test for Association Between Email Account Type and Hacking Status
A survey was conducted to see the association between hacking status of the email and the type of email account. The survey has reported the following cross tabulation. Do the information provide sufficient evidence to conclude that the type email account and the hacking status is associated? Use Chi-square test at 1% level of significance.
$$\begin{array}{|c|c|c|} \hline \text{Type of e-mail account} & \text{Hacking status Yes} & \text{Hacking status No} \ \hline \text{Yahoo} & 60 & 15 \ \text{Gmail} & 20 & 120 \ \hline \end{array}$$
[5]
Email Account Yes (Hacked) No (Not Hacked) Row Total ------------ Yahoo 60 15 75 Gmail 20 120 140 Column Total 80 135 215 Level of significance: $\alpha = 0.01$ - $H0$: Email account type and hacking status are independent (not associated) - $H1$: Email acc...
Full solved answer →Chi-Square Test for Association Between Email Account Type and Hacking Status
A survey was conducted to see the association between hacking status of the email and the type of email account. The survey has reported the following cross tabulation. Do the information provide sufficient evidence to conclude that the type email account and the hacking status is associated? Use Chi-square test at 1% level of significance.
$$\begin{array}{|c|c|c|} \hline \text{Type of e-mail account} & \text{Hacking status Yes} & \text{Hacking status No} \ \hline \text{Yahoo} & 60 & 15 \ \text{Gmail} & 20 & 120 \ \hline \end{array}$$
[5]
Account Yes (Hacked) No (Not Hacked) Row Total -------------------------------------------------- Yahoo 60 15 75 Gmail 20 120 140 Column Total 80 135 215 Level of significance: $\alpha = 0.01$ - $H0$: Type of email account and hacking status are independent...
Full solved answer →Use Mann-Whitney U test to assess whether the following satisfaction scores based on the performance of two different special types of gadgets at 5% level of significance.
| 1 | 2 | 3 | 4 | |
|---|---|---|---|---|
| Gadget A | 50 | 40 | 30 | 20 |
| Gadget B | 40 | 30 | 10 | 40 |
[5]
Gadget A (n₁ = 4): 50, 40, 30, 20 Gadget B (n₂ = 4): 40, 30, 10, 40 Level of significance: α = 0.05 - H₀: No significant difference in satisfaction scores between Gadget A and Gadget B. - H₁: There is a significant difference (two-tailed test). Combined sor...
Full solved answer →Chi-Square Test for Independence
The following data related to the number of children classified according to the type of feeding and nature of teeth. Do the information provide sufficient evidence to conclude that type of feeding and nature of teeth are dependent? Use chi square test at 5% level of significance.
$$\begin{array}{|c|c|c|} \hline \text{Type of feed} & \text{Normal} & \text{Defective} \ \hline \text{Breast} & 18 & 12 \ \text{Bottle} & 2 & 13 \ \hline \end{array}$$
[5]
Type of Feed Normal Defective Row Total ------------ Breast 18 12 30 Bottle 2 13 15 Column Total 20 25 N = 45 --- - $H0$: Type of feeding and nature of teeth are independent. - $H1$: Type of feeding and nature of teeth are dependent. - $\alpha = 0.05$ --- $...
Full solved answer →Kruskal Wallis test
Kruskal-Wallis Test for Propellant Burning Rates
In an experiment to determine which of three different missile systems is preferable, the propellant burning rate is measured. The data after coding are given in the table. Use Kruskal-Wallis test (significance level of 0.01) to test the hypothesis that the propellant burning rates are same for three missile systems.
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | |
|---|---|---|---|---|---|---|---|
| Missile system I | 22.3 | 16.7 | 22.7 | 19.3 | 18.5 | ||
| Missile system II | 23.4 | 19.5 | 17.5 | 20.8 | 16.0 | 19.9 | |
| Missile system III | 18.4 | 19.5 | 17.8 | 18.0 | 19.6 | 22.8 | 17.1 |
[10]
System I (n₁=5) System II (n₂=6) System III (n₃=7) ------------------------------------------------------ 22.3 23.4 18.4 16.7 19.5 19.5 22.7 17.5 17.8 19.3 20.8 18.0 18.5 16.0 19.6 19.9 22.8 17.1 Total $N = 5 + 6 + 7 = 18$ - $H0$: The propellant burning rat...
Full solved answer →Kruskal-Wallis H Test for Catalyst Comparison
A chemist uses three catalysts for distilling alcohol and the results were tabulated below. Are there any significant differences between catalysts? Test at 5% level of significance. Use Kruskal-Wallis H test.
| Catalyst | Alcohol (in cc) | ||||
|---|---|---|---|---|---|
| C1 | 380 | 430 | 410 | ||
| C2 | 290 | 350 | 270 | 250 | 270 |
| C3 | 400 | 380 | 450 |
[5]
Catalyst Alcohol yield (cc) Sample size ------------------------------------------- C1 380, 430, 410 $n1 = 3$ C2 290, 350, 270, 250, 270 $n2 = 5$ C3 400, 380, 450 $n3 = 3$ Total: $N = 3 + 5 + 3 = 11$, significance level $\alpha = 0.05$. - $H0$: There is no ...
Full solved answer →Parametric vs. non-parametric test
Write short notes on any two: a. Difference between parametric and non-parametric test. b. Required assumptions for linear regression model. c. Stochastic process. [5]
--- Basis Parametric Test Non-Parametric Test --------- Assumption Requires strict assumptions about population distribution (usually normality) Does not require assumptions about population distribution Data Type Works on interval or ratio scale data Works...
Full solved answer →What do you mean by non parametric test? Write down advantages of non parametric test over parametric tests? [5]
A non-parametric test (also called a distribution-free test) is a statistical test that does not require any assumption about the population distribution from which the sample is drawn. Unlike parametric tests, these tests do not involve estimation or testi...
Full solved answer →One-sample test
Bank of Nepal recorded the sex of first 30 customers who appeared last Monday with notation M M F M M F M F M F M F M F M F M F M F M F M F M F M F M F. At the 0.005 level of significance, test the randomness of this sequence. [5]
Sequence (30 customers): M M F M M F M F M F M F M F M F M F M F M F M F M F M F M F - Total observations $n = 30$ - Level of significance $\alpha = 0.005$ --- Let me count carefully. Positions 1-30: 1:M 2:M 3:F 4:M 5:M 6:F 7:M 8:F 9:M 10:F 11:M 12:F 13:M 1...
Full solved answer →Needs of applying non-parametric tests
Write short notes on the following: I. The rationale of using the non-parametric statistical test II. Estimation of minimum size for the given proportion [5]
--- Non-parametric statistical tests (also called distribution-free tests) are statistical methods that do not require assumptions about the underlying population distribution (e.g., normality). 1. No assumption about population distribution: When the popul...
Full solved answer →Write short notes of the following: a. Need of non parametric statistical methods. b. Efficiency of Randomized Block Design relative to Completely Randomized Design [5]
--- Non-parametric statistical methods (also called distribution-free methods) are statistical tests that do not require assumptions about the population distribution from which the sample is drawn. The need for non-parametric methods arises due to the foll...
Full solved answer →Make Unit 3 stick
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