Lesson-7-Statistical-Tools-1_for-lecture.pdf

Lesson 7: STATISTICAL TOOLS  I recall that _______________________________________________ _______________________________________________ _______________________________________________ Done? Let’s go to today’s lesson! At the end of the lesson, you are expected to:...

Lesson 7: STATISTICAL TOOLS  I recall that _______________________________________________ _______________________________________________ _______________________________________________ Done? Let’s go to today’s lesson! At the end of the lesson, you are expected to: ▪ Define statistics; ▪ Determine the level of measurement of a given set of data; ▪ Compare and contrast the basic statistical tools; ▪ Solve a given problem by applying the statistical tools. ACTIVITY: Riddle Read the riddle and answer the question that follows. Write your answer on the space provided. You are a prisoner sentenced to death. The King offers you a chance to live by playing a simple game. He gives you 50 black marbles, 50 white marbles and 2 empty bowls. He then says, "Divide these 100 marbles into these 2 bowls. You can divide them any way you like as long as you use all the marbles. Then I will blindfold you and mix the bowls around. You then can choose one bowl and remove one marble. If the marble is white you will live, but if the marble is black... you will die.“ How would you divide the marbles up so that you have the greatest probability of choosing a white marble? Downie and Heath (1984) said that statistics may be considered as collections of data associated with human enterprises. It is a method that can be used to analyze data, that is to organize and make sense out of a large amount of material. It is used to show how an individual in the group stands in reference to others. It is used to determine relationship between two variables; one variable may be used to predict another. It is used to make inference about a population using samples. Statistics is basic to the intelligent reading. Levels or Scales of Measurement  Levels or Scales of Measurement RATIO Numerical INTERVAL ORDINAL Categorical NOMINAL 1. Nominal ▪Numbers are used as measures of identity. ▪Numbers may serve as labels to identify items or classes. ▪Numbers are classifications of individual into categories (e.g. jersey number). 2. Ordinal ▪Numbers reflect the rank order of the individuals or objects. ▪Numbers are arranged from the highest to the lowest or vice versa (e.g. top 10 of the class). 3. Interval ▪Numbers reflect differences among items. ▪Numbers show that a person or item is so many units larger or smaller, heavier or lighter, brighter or duller, etc. (e.g. temperature) 4. Ratio ▪Numbers that have absolute zero (e.g. test scores). ▪ It has the same properties as the interval level. The order and difference can be described, but it has a true zero and the ratio between two points have meanings. SUMMARY Measurement Named Order Difference True Zero Level 1.Nominal ̸ X X X 2. Ordinal ̸ ̸ X X 3. Interval ̸ ̸ ̸ X 4. Ratio ̸ ̸ ̸ ̸ Basic Statistical Tools (Dacanay of PNU) 1. Pearson r Pearson r (r) is used to determine correlation or (significant) relationship between two variables of interval/ratio type. Step 1 – Group the given data: N= number of respondents Σx= summation of values of x Σy= summation of values of y Σxy= summation of the product of x and y Σx2= summation of x2 Σy2= summation of y2  Step 1- Group the given data: Cases (N) Creativity Test Motivation Test x2 y2 xy Steps to follow: (x) (y) 1. To get the value of 1 9 9 81 81 81 x2, multiply the 2 8 7 64 49 56 value of x by itself. 2. To get the value of 3 7 9 49 81 63 y2, multiply the 4 4 6 16 36 24 value of y by itself. 5 3 6 9 36 18 3. To get the value of xy, multiply the 6 4 5 16 25 20 value of x and y. 7 5 3 25 9 15 4. To get all the value 8 9 6 81 36 54 of Σx, Σy, Σx2, Σy2 and Σxy, add all the 9 8 5 64 25 40 values in each 10 5 9 25 81 45 column N= 10 Σx= 62 Σy= 65 Σx2= 430 Σy2= 459 Σxy= 416 Step 2 – State the problem/research question: Is there a relationship between creativity test and motivation test scores in the performance of students? N= 10 Σx= 62 Σy= 65 Σx2= 430 Σy2= 459 Σxy= 416 Steps to follow: 1. Multiply 2. Subtract 3. Multiply 4. Get the square root 5. Last, Divide r= 0.319 Step 4 – State the interpretation of the result: The result is 0.319. Thus, there is low correlation between creativity test and motivation test in the performance of students. 2. Spearman Rank-Order Spearman Rank-Order (ρ) is used to determine correlation or (significant) relationship between two variables of the ordinal type.  Step 1- Group the given data: N Reading Self-Concept RX Ry D D2 Steps to follow: Comprehension (x) (y) 1. Rank the sources. 1 125 129 1 2 -1 1 2. To get the value of D, subtract the 2 122 130 2 1 1 1 scores 3 118 114 3 7 -4 16 3. To get the value of 4 115 117 4 6 -2 4 D2, multiply the value of D by itself. 5 114 125 5 3 2 4 4. To get all the value 6 100 121 6 4 2 4 of ΣD2, add the 7 96 119 7 5 2 4 values of the column of D2. N= 7 ΣD2= 34 Step 2 – State the problem/research question: Is there a relationship between the ranking of reading comprehension scores and self-concept scores in English test? N= 7 ΣD2= 34 p= 1- ___6 (34)___ Steps to follow: 7 (72-1) 1. Multiply p= 1- ___204____ 2. Subtract 7 (49-1) 3. Multiply p= 1- ___204____ 7 (48) 4. Divide p= 1- __204____ 336 5. Last, Subtract p= 1- 0.607143 r= 0.393 Step 4 – State the interpretation of the result: The result is 0.393. There is a low correlation between the ranking of reading comprehension scores and self-concept scores. 3. Z-Test of Independent Proportions Z-Test of Independent Proportions is used to determine if there is significant difference between two independent or different groups in situations that call for two types (dichotomous) of nominal data/response (e.g. yes or no, agree or disagree, in favor or against, etc.). Sample Problem: Fifty (50) SPJ students and 50 SSC students are asked about the proposed implementation of the trimester scheme. Of the 50 SPJ students, 30 are in favor (IF) and 20 are against (A). Among the SSC students, 25 are IF and 25 are A.  SPJ SSC IF 30 25 A 20 25 TOTAL 50 50 Step 2 – State the problem/research question: Is there a significant difference between SPJ students and SSC students who are in favor of the implementation of the trimester scheme? Step 3 – State the null (Ho) hypothesis and the alternative (H 1) hypothesis: Ho (Null Hypothesis) – there is no significant difference between specified populations. H1 (Alternative Hypothesis)- is contrary to the null hypothesis because there is a significant difference between populations. Ho : P1 = P2 H1 : P1 ≠ P2 Step 4 – State the level of significance: Use.05 if you are likely to reject the Ho Use.01 if you are likely to accept the Ho Step 5 – State the decision rule: DR: if Z > 1.96 reject H0 if Z < 1.96 accept H0.05 1.96 Step 6 – Solve for Z: P1 = A/N1 (proportion 1) P2 = B/N2 (proportion 2) P (population proportion estimate) = (A+B)/N1+N2 Q=1–P  SPJ SSC IF 30 25 A 20 25 TOTAL 50 50 P1 = A/N1 (proportion 1) P2 = B/N2 (proportion 2) P (population proportion estimate) = (A+B)/N1+N2 Q= 1–P P1 = 30/ 50 = 0.6 Q = 1- 0.55 = 0.45 P2 = 25/ 50 = 0.5 N1 = 50 P = 30 + 25 = _55_ N2 = 50 50 + 50 100 P = 0.55 P1 = 0.6 P2 = 0.5 P = 0.55 Q = 0.45 N1 = 50 N2 = 50 Steps to follow: 1. Subtract the numerator 2. Add the fraction in the denominator 3. Multiply 4. Divide the fraction in the denominator 5. Multiply the denominator 6. Get the square root of the denominator 7. Divide the numerator and z= 1.01 denominator. Step 7 – State the decision based on the result/decision rule: Accept H0 Step 8 – State the interpretation of the result: There is no significant difference between SPJ students and SSC students who are in favor of the implementation of the trimester scheme. 4. Z-Test of Dependent Proportions Z-Test of Dependent Proportions is used to determine if there is a significant differences between pairs of nominal observations or responses (i.e., before and after) from a single group. Sample Problem: Fifty (50) SPJ students are asked if they are in favor (IF) or against (A) the implementation of online classes in the upcoming school year. Results showed that 30 SPJ students are IF and 20 are A. The same SPJ students are asked to attend a seminar on the said proposal. After the seminar, the same survey is administered among them. Of the 30 who are IF before, 22 remain IF and 8 go against. Of the 20 who are A before, 16 switch to IF and 4 remain A. BEFORE AFTER IF 30 A B A 20 16 22 C D 4 8 Step 2 – State the problem/research question: Is there a significant difference between SPJ students who are in favor of the implementation of online classes in the upcoming school year before and after the seminar? Step 3 – State the null (Ho) and the alternative (H1) hypothesis: Ho : P1 = P2 H1 : P1 ≠ P2 Step 4 – State the level of significance: α=.05 Use.05 if you are likely to reject the H o Use.01 if you are likely to accept the H o Step 5 – State the decision rule: DR: if Z > 1.96 reject H0 if Z < 1.96 accept H0.05 - 1.96 Step 6 – Solve for Z: P1= 16+22 = 0.76 50 P2 = 22+8 = 0.6 50 P1 = (A+B)/N P2 = (B+D)/N a = A/N a= 16 = 0.32 d = D/N 50 d= 8 = 0.16 50 P1 = 0.76 P2 = 0.6 a = 0.32 d = 0.16 N = 50 Z= 0.76 - 0.6 Steps to follow: √(0.32 + 0.16) 1. Subtract the numerator 50 2. Add the numerator Z= 0.16 3. Divide the denominator √ 0.48 50 4. Get the square root of the denominator Z= 0.16 √0.0096 5. Divide the numerator and the denominator Z= 0.16 0.098 Z= 1.633 Step 7 – State the decision based on the result/decision rule: Accept H0 Step 8 – State the interpretation of the result: The result is 1.633. Therefore, there is no significant difference between SPJ students who are in favor of the implementation of online classes in upcoming school year before and after the seminar. 5. Chi-Square Test of Goodness of Fit The Chi -Square Test of Goodness of Fit is used to determine if there is a significant difference between the observed and the expected nominal distribution/responses. Sample Problem: One hundred (100) high school students are asked about their opinions about Duterte’s impeachment. The result of the survey is as follows: Step 1 – Get the expected distribution:  Expected Distribution (E) Strongly Agree (SA) 20 Agree (A) 20 No opinion (NO) 20 Disagree (D) 20 Strongly Disagree (SD) 20 Step 2 – State the problem/research question: Is there a significant difference between the observed and expected values in terms of students' opinions about Duterte's impeachment? Step 3 – State the null (Ho) and the alternative (H1) hypothesis: Ho : P1 = P2= P3= P4= P5 H1 : P1 ≠ P2 ≠ P3 ≠ P4≠ P5 Step 4 – State the level of significance: α=.05 Use.05 if you are likely to reject the H o Use.01 if you are likely to accept the H o Step 5 – State the degree of freedom: df = number of category - 1 df= 5- 1 df= 4 Step 6 – State the decision rule: DR: if X² > 9.488 reject H0 if X² < 9.488 accept H0 Refer to the Chi-square Distribution Table Go to https://www.statology.org/wp- content/uploads/2018/09/chi_table_test.png for the complete list. Step 7 – Apply x2: X²= ∑(O - E)²/ E Strongly Agree (A) No opinion Disagree Strongly Agree (SA) (NO) (D) Disagree (SD) O 30 25 5 30 10 E 20 20 20 20 20 O-E 10 5 -15 10 -10 (O - E)² 100 25 225 100 100 (O - E)² 5 1.25 11.25 5 5 E X²= 27.5 Step 8 – State the decision based on the result/decision rule: Reject H0 Step 9 – State the interpretation of the result: The result is 27.5. Therefore, there is a significant difference between the observed and expected values in terms of students' opinion about Duterte's impeachment. 6. Chi-Square Test of Homogeneity The Chi -Square Test of Goodness of Fit is used to determine if there is a correlation or (significant) relationship between two variables of the nominal type. Sample Problem: Fifty (50) male (M) and 50 female (F) SHS students are asked about their drink preferences. Among the M, 20 prefer hard drinks (HD) and 30 prefer soft drinks (SD). Among the F, 25 prefer HD and 25 prefer SD.  Observed Distribution (O) Expected Distribution (E) HD SD Total HD SD Total M A B I M a b i 20 30 50 22.5 27.5 50 F C D H F c d h 25 25 50 22.5 27.5 50 Total E F G Total e f g 45 55 100 45 55 100 a = (E x I)/G a = (45 x 50)/ 100 = a= 22.5 b = (F x I) / G b = (55 x 50) / 100 = b= 27.5 c = (E x H) / G c= (45 x 50) / 100 = c= 22.5 d= (F x H) / G d= (55 x 50) / 100 = d= 27.5 Step 2 – State the problem/research question: Is there a significant correlation between male and female SHS students when it comes to their drink preferences? Step 3 – State the null (Ho) and the alternative (H1) hypothesis: Ho : P1 (Male-HD) = P2 (Female-HD) P1 (Male-SD) = P2 (Female-SD) H1 : P1 (Male-HD) ≠ P2 (Female-HD) P1 (Male-SD) ≠ P2 (Female-SD) Step 4 – State the level of significance: α=.05 Use.05 if you are likely to reject the H o Use.01 if you are likely to accept the H o Step 5 – State the degree of freedom: df = (row – 1)(column – 1) df = (2 - 1) (2 -1) = (1) (1) df = 1 Step 6 – State the decision rule: DR: if X² > 3.841 reject H0 if X² < 3.841 accept H0 Refer to the Chi-square Distribution Table Step 7 – Apply x2: X²= ∑(O - E)²/ E MALE FEMALE HD SD HD SD O 20 30 25 25 E 22.5 27.5 22.5 27.5 O-E -2.5 2.5 2.5 -2.5 (O - E) ² 6.25 6.25 6.25 6.25 (O - E) ² 0.28 + 0.23. + 0.28. + 0.23 E X² = 1.02 Step 8 – State the decision based on the result/decision rule: Accept H0 Step 9 – State the interpretation of the result: The result is 1.02 Therefore, there is no significant relationship between male and female SHS students when it comes to their drink preferences.