Introduction to Quantitative Techniques in Business

QUANTITATIVE TECHNIQUES IN BUSINESS QTB

INFERENTIAL STATISTICS


INFERENTIAL STATISTICS1. Lesson Objectives

After studying this session you would be able to

1. Understand and infer results from data in order to answer the associational and differential research questions using different parametric and non parametric tests.

2. Understand, implement and interpret the chi-square, phi and cramer’s V

3. understand, implement and interpret the correlation statistics

4. understand, implement and interpret the regression statistics

5. understand, implement and interpret the T-test statistics

Lesson Outline

1. Non parametric test.

1. Chi square /Fisher exact

2. Phi and cramer’s v

3. Kendall tau-b

4. Eta

2. Parametric test

1. Correlation

1. Pearson correlation

2. Spearman correlation

2. Regression

1. Simple regression

2. Multiple regression

3. T-Test

1. One-sample T-test

2. Independent sample T-test

3. Paired sample T-test


INFERENTIAL STATISTICS

Inferential statistics are used to make inferences (conclusions) about a population from a sample based on the statistical relationships or differences between two or more variables using statistical tests with the assumption that sampling is random in order to generalize or make predictions about the future.

Why we use inferential Statistics:-Inferential statistics are used

1. To test some hypothesis either to check relationship between variables (two/more) or to compare two groups to measure the differences among them.

2. To generalize the results about a population from a sample

3. To make predictions about the future.

4. To make conclusions

You don't need to understand the underlying calculus, but you do need to know which inferential statistic is appropriate to use and how to interpret it.

Some basic concepts about inferential statistics

1. Statistical significance (The p value)

Statistical significance test is the test of a null hypothesis Ho which is a hypothesis that we attempt to reject or nullify. i.e.

Ho =There is no relationship /Difference between variable 1 and variable 2

When we apply any inferential statistic, it gives us significance value (called p value). If the p value is less than 5% then the test result is said to be significant at the 5% level. The term significant means that the test signifies or points to the conclusion that there is evidence against the truth of the null hypothesis. The comparison of p with 5% is a standard method often used by researchers, but it is better to report and interpret the actual values of p.

Interpretation

If the p value is greater than 0.05 than it means that Ho is accepted and H1 is rejected. It relates that there is no relationship/difference between the variables/groups.


If the p value is less than or equal to 0.05 than it means that Ho is rejected and H1 is accepted. It relates that there is relationship/difference between the variables/groups. A higher p value means that the relationship is lesser significant and a smaller p value means that the relationship is highly significant.

2. Confidence Interval

Confidence interval is a range of values constructed for a variable of interest so that this range has a specified probability of including the true value of the variable. The specified probability is called the confidence level, and the end points of the confidence interval are called the confidence limits’.

It is one of the alternatives to null hypothesis significance testing (NHST). These intervals provide more information then NHST and may provide more practical information. For example, suppose one knew that an increase in reading scores of five points, obtained on a particular instrument, would lead to a functional increase in reading performance. Two different methods of instruction were compared. The result showed that students who used this new method scored significantly higher statistically than those who used the other method. According to NHST, we would reject the null hypothesis of no difference between methods and conclude that the new method is better. If we apply confidence intervals to this same study, we can determine an interval that contains the population mean difference 95% of the time. If the lower bound of that interval is greater than five points, we can conclude that using this method of instruction would lead to a practical or functional increase in reading levels. If, however, the confidence interval ranged from say 1 to 11, the result would be statistically significant, but the mean difference in the population could be as little as 1 point, or as big as 11 points. Given these results, we could not be confident that there would be a practical increase in reading using the new method.

3. The effect size (weak, moderate or strong)

Effect size is the strength of the relationship between the independent variable and the dependent variable, and/or the magnitude of the difference between levels of the independent variable with respect to the dependent variable.

A statistically significant outcome does not give information about the strength or size of the outcome. Therefore, it is important to know, the size of the effect. Statisticians have proposed many effect size measures that fall mainly into two types of families, the r family and the d family.

Interpreting Effect Sizes

Effect sizes always have an absolute value between -1.0 and +1.0. According to Cohen (1988) we can interpret the effect size (r/d) as follows


0 No effect No relationship

>0 – 0.33 Small effect Weak relationship

>0.33 – 0.70 Medium/typical effect Moderate relationship

>0.70 – <1 Large effect Strong relationship

1 Maximum effect Perfect relationship

Steps in interpreting inferential statistics

1. Relate why a test is applied

2. Discuss for which variable the test is applied

3. Elaborate whether the null hypothesis is rejected or accepted w.r.t. p value

As discussed above if the significance (p) value is less than 0.05 then HO is rejected and H1 is accepted, conversely if the significance value is greater than 0.05 then HO is accepted and H1 is rejected

4. State what is the direction of the effect

5. For associational research question indicate whether the association or relationship is positive or negative

6. For differential research question state which group performed better?

7. Conclude the results

Types of tests used in Inferential Statistics

Inferential statistics include a wide variety of tests to infer the results. This variety of tests can be classified in two broader categories that are

1. Non parametric tests

2. Parametric tests

Following is the detailed discussion related to both types of tests.

1. Non parametric test

Non parametric tests are the statistical tests that are used


1. When the level of measurement is nominal or ordinal. E.g. chi-square test or Kendall’s tau-b.

2. When assumptions about normal distribution in the population is not met e.g. spearman correlation

Non parametric tests involve

1. Chi-Square test

2. Kend

3. all’s tau-b

4. Eta

5. Spearman correlation (will be discussed in correlation section)

Let’s see these tests in detail.

Chi-Squared Test

Chi-Squared test is the most commonly used non-parametric test to check the association between two nominal variables in order to accept or reject the null hypothesis.

Hypothesis for Chi-Square Test

Ho = there is no association between gender and geometry in h.s.

H1 = There is association between gender and geometry in h.s.

It is used to check

1. The association between two nominal variables

2. Compare two or more groups if they are categorical in nature

Assumptions and Conditions for the Chi-Squared test

1. The data of the variables must be independent. Each subject is assessed only once.

2. Both the variables are nominal.

3. All the expected counts are greater than 1 for chi-square.

4. At least 80% of the expected frequencies should be greater than or equal to 5.

Checking the assumptions for the Chi-Squared test


The assumptions for Chi-squared test are checked through cross tabulation of the categorical variables. It can be drawn by

1. Click the analyze menu

2. Select the descriptive statistics option

3. Select crosstabs option in the sub menu

4. Put geometry in h.s. in rows section and gender in columns section

5. Check chi-squared, phi and Cramer’s v from statistics tab

6. Check observed, expected and total from cells tab

7. Click continue then ok to get the following crosstabs in output window

geometry in h.s. * gender Crosstabulation

gender

Totalmale female

geometry in h.s. not taken Count 10 29 39

Expected Count 17.7 21.3 39.0

% of Total 13.3% 38.7% 52.0%

Taken Count 24 12 36


% of Total 32.0% 16.0% 48.0%

Total Count 34 41 75


% of Total 45.3% 54.7% 100.0%

1. Check if all the values of expected counts are greater than one (excluding total column and the total row)

2. Check if the 80% values of expected counts are greater than 5. You can calculate the percentage using following formula

Number of cells with expected counts greater than 5 × 100Total number of cells


3. If the assumptions are fulfilled then use significance value of Pearson chi-square as highlighted below

4. If the assumptions for chi-square are not fulfilled then select the significance value of Fisher’s exact test

5. To check the strength of the relationship (effect size) use the value of Phi for 2x2 crosstab and value of Cramer’s V for 3x3 crosstab. Remember that both Phi and Cramer’s v have similar values for 2x3 and 3x2 crosstabs


Interpretation:

To check the association between gender and geometry in h.s. chi-square test is conducted. The case processing summary table indicates that there is no participant with missing value. The

Case Processing Summary

Cases

Valid Missing Total

N Percent N Percent N Percent

geometry in h.s. * gender 75 100.0% 0 .0% 75 100.0%

Chi-Square Tests

Value df

Asymp. Sig. (2-

sided)

Exact Sig. (2-

sided)

Exact Sig. (1-

sided)

Pearson Chi-Square 12.714a 1 .000

Continuity Correctionb 11.112 1 .001

Likelihood Ratio 13.086 1 .000

Fisher's Exact Test .000 .000

Linear-by-Linear Association 12.544 1 .000

N of Valid Casesb 75

a. 0 cells (.0%) have expected count less than 5. The minimum expected count is 16.32.

b. Computed only for a 2x2 table

Symmetric Measures

Value Approx. Sig.

Nominal by Nominal Phi -.412 .000

Cramer's V .412 .000

N of Valid Cases 75


assumptions are checked through crosstabs. The Crosstabulation table includes the Counts and Expected Counts, and their relative percentages within gender. The result shows that there are 24 males who had taken geometry which is 71% of total 34 male students. On the other hand, 12 of 41 females took geometry; that is only 29% of the females. It looks like a higher percentage of males took geometry than female students. The Ch-Square Test table tell us whether we can be confident that this apparent difference is not due to chance.

Note: it is noted very carefully that, we use the Pearson Chi-Square or (for small samples) the Fisher’s exact test to interpret the results of the test.Note, in the Cross Tabulation table, that the Expected Count of the number of male students who didn’t take geometry is 17.7 and the observed or actual Count is 10. Thus, there are 7.7 fewer males who didn’t take geometry than would be expected by chance, given the Totals shown in the Table. There are also the same discrepancies between observed and expected counts in the other three cells of the table. A question answered by the chi-square test is whether these discrepancies between observed and expected counts are bigger than one might expect by chance.

The Chi-Square Tests table is used to determine if there is a statistically significant relationship between two dichotomous or nominal variables. It tells you whether the relationship is statistically significant but does not indicate the strength of the relationship, like phi or a correlation does. In output, we use the Pearson Chi-Square or (for small samples) the Fisher’s exact test to interpret the results of the test. They are statistically significant (p < .001), which indicates that we can be quite certain that males and females are different on whether they take geometry.

Phi is -.412, and like the chi-square, it is statistically significant. Phi is also a measure of effect size for an associational statistic and, in this case, effect size is moderate according to Cohen (1988)

KENDALL’S TAU-B

If the variables are ordered (i.e. ordinal), you have several other choices. We will use Kendall’s tau-b in this problem.

Example: What is the relationship or association between father’s education and mother’s education?

1. Analyze Descriptive Statistics Crosstabs.2. Click on Reset to clear the previous entries.3. Put mother’s education revised in the Rows box and father’s education revised in the

columns box.


4. Click on Cells and ask that the Observed and Expected cell counts and Total percentages be printed in the table. Click on Continue and then Statistics.

5. Request the following Statistics: Kendall’s tau-b coefficient under ordinal, and Phi and Cramer’s V under nominal (for comparison purposes). Do not check Chi-Square.

6. Click on Continue 7. Click on OK.


Cases

Valid Missing Total


mother education revised * father education revised

73 97.3% 2 2.7% 75 100.0%

mother education revised * father education revised Crosstabulation

father education revised

Total1 2 3

mother education revised

1 Count 43 8 2 53

Expected Count 35.6 13.1 4.4 53.0

% of Total 58.9% 11.0% 2.7% 72.6%

2 Count 6 10 2 18

Expected Count 12.1 4.4 1.5 18.0

% of Total 8.2% 13.7% 2.7% 24.7%


3 Count 0 0 2 2

Expected Count 1.3 .5 .2 2.0

% of Total .0% .0% 2.7% 2.7%

Total Count 49 18 6 73

Expected Count 49.0 18.0 6.0 73.0

% of Total 67.1% 24.7% 8.2% 100.0%

Symmetric Measures

ValueAsymp.

Std. Errora

Approx. Tb

Approx. Sig.

Ordinal by Ordinal Kendall's tau-b .494 .108 3.846 .000

N of Valid Cases 73

a. Not assuming the null hypothesis.

b. Using the asymptotic standard error assuming the null hypothesis.Interpretation:

To investigate the relationship between father’s education and mother’s education, Kendall’s tau-b was used. The analysis indicated a significant positive association between father’s education and mother’s education, tau =.494, p<.001. This means that more highly educated fathers were married to more highly educated mothers and less educated fathers were married to less educated mothers. This tau is considered to be a large effect size (Cohen, 1988).ETA

If one variable is nominal and the other is scale then ETA is the appropriate test used to check the relationship between the two variables. Eta is calculated for both variables. First you should decide the dependent variable and consider the Eta value of that variable.Example: What is the association between gender and number of math courses taken? How strong is it?8. Analyze Descriptive Statistics Crosstabs.

9. Click on Reset to clear the previous entries.

10. Put math courses taken in the Rows box and gender in the columns box.


11. Click the Statistics and select Eta.

12. Click Continue

13. Click OK to get following results


Cases

Valid Missing Total


math courses taken * gender 75 100.0% 0 .0% 75 100.0%

math courses taken * gender Crosstabulation

Gender

TotalMale female

math courses taken

0 Count 4 12 16


1 Count 3 13 16


2 Count 9 6 15


3 Count 6 2 8


4 Count 7 5 12


5 Count 5 3 8

Expected Count 3.6 4.4 8.0Total Count 34 41 75




Cases

Valid Missing Total


Directional Measures

Value

Nominal by Interval

Eta math courses taken Dependent

.328

gender Dependent .419

InterpretationEta was used to investigate the strength of the association between gender and number of

math courses taken (eta=.33). This is a weak to medium effect size (Cohen, 1988). Males were

more likely to take several or all the math courses than females.


Class Activity Session 6

Please show all work and explain your answers.

1. Popcorn sales in movie theaters break down as 40% plain popcorn and 60% buttered popcorn. While 65% of the plain popcorn is purchased by adults, 80% of the buttered popcorn is purchased by children. If a child purchases popcorn, what is the probability that it is buttered popcorn?

(guidelines: develop a joint-probability table. Note that the problem is asking that you compute a conditional probability)

2. A process follows the binomial distribution with n = 7 and p = .4. Find

a. P(x = 3)

b. P(x > 5)

c. P(x 2)

1. Scores on an endurance test for cardiac patients are normally distributed with mean = 200 and standard deviation = 30.

a. What is the probability a patient will score above 206?

b. What percentage of patients score below 155?

c. What score does a patient at the 25th percentile receive?


2. A calculus instructor uses computer aided instruction and allows students to take the midterm exam as many times as needed until a passing grade is obtained. Following is a record of the number of students in a class of 20 who took the test each number of times.

Students Number of tests

7 1

6 2

4 3

2 4

1 5

a. use the relative frequency approach to construct a probability distribution

b. show that it satisfies the required condition for being a probability distribution.

c. Find the expected value of the number of tests taken.

5. For the payoff table below, the decision maker will use P(s1) = .15, P(s2) = .5, and P(s3) = .35.

s1 s2 s3

d1 -5000 1000 10,000

d2 -15,000 -2000 40,000

a. What alternative would be chosen according to expected value?


b. For a lottery having a payoff of 40,000 with probability p and -15,000 with probability (1-p), the decision maker expressed the following indifference probabilities.

Payoff Probability

10,000 .85

1000 .60

-2000 .53

-5000 .50

Let U(40,000) = 10 and U(-15,000) = 0 and find the utility value for each payoff.

c. What alternative would be chosen according to expected utility?


CORRELATION & REGRESSION

Inferential Statistics


Correlation

Correlation is a statistical process that determines the mutual (reciprocal) relationship between two (or more) variables which are thought to be mutually related in a way that systematic changes in the value of one variable are accompanied by systematic changes in the other and vice versa.

It is used to determine

1. The existence of mutual relationship that is defined by the significance (p) value.

2. The direction of relationship that is defined by the sign (+,-) of the test value

3. The strength of relationship that is defined by the test value

Correlation Coefficient (r)

The correlation coefficient measures the strength of linear relationship between two or more numerical variables. The value of correlation coefficient can vary from -1.0 (a perfect negative correlation or association) through 0.0 (no correlation) to +1.0 (a perfect positive correlation). Note that +1 and -1 are equally high or strong, but they lead to different interpretations. A high positive correlation between anxiety and grades would mean that students with higher anxiety tended to have higher grades, those with lower anxiety had lower grades, and those in between had grades that were neither especially high nor especially low. A high negative correlation would mean that students with high anxiety tended to have low grades; also high grades would be associated with low anxiety. With a zero correlation there are no consistent associations. A student with high anxiety might have low, high or medium grades.

There are two types of correlation

1. Pearson Correlation

2. Spearman Correlation

1. Pearson Correlation

The Pearson Correlation is used when you have two variables that are normal/scale An assumption of the Pearson correlation is that the variables are related in a linear (straight line) way so we will examine the scatter plots to see if that assumption is reasonable. Second, the Pearson Correlation, and the Spearman correlation will be computed. and the Spearman is used when one or both is ordinal.


1. Assumptions and conditions for Pearson

1. The two variables have a linear relationship.

2. Scores on one variable are normally distributed for each value of the other variable and vice versa.

3. Outliers (i.e. extreme scores) can have a big effect on the correlation.

1. Checking the assumptions for Pearson Correlation

The assumptions for correlation test are checked through normal curve (normality assumption) and the scatter plot (linearity assumption)

Normality assumption

1. Click on the analyze menu

2. Select the descriptive statistics option

3. Select frequency option in the sub menu

4. Put math achievement and Satmath in variables box

5. Check skewness in statistics tab and histogram in charts tab

6. Click continue and then ok

7. You will get skewness values showing that the variables are

approximately normally distributed further check the

normality of data through normal curve in histograms using

chart editor


Statistics

math achievement test scholastic aptitude test - math

N Valid 75 75

Missing 0 0

Skewness .044 .128

Std. Error of Skewness .277 .277

Linearity assumption

8. Click on the graph menu

9. Select legacy dialogue, interactive and then scatter plot

10. Put math achievement in y-axis and satmath in x-axis

11. Click ok to get scatter plot in output window

12. Double click on the scatter plot to get into chart editor

13. Click on the “add fit line at total” button in tool bar to get linear line and R square linear = 0.62 close window

14. Repeat the previous step for quadratic line and get R square = 0.621

15. click apply and close the window

16. Calculate the difference between the two R square (0.621 – 0.62 = 0.001)

17. If the difference is less than 0.05 (the p value) then the relation is linear (0.001>0.05) hence apply Pearson correlation

How to apply Pearson Correlation

1. Select analyze then correlate and then bivariate

2. Put math achievement and Satmath in variable box


3. Ensure that Pearson, two tailed, and flag relationships are

checked

4. Click ok to get follow results in output window

Correlations

math

achievement test

scholastic

aptitude test -

math

math achievement test Pearson Correlation 1 .788**

Sig. (2-tailed) .000

N 75 75

scholastic aptitude test –

math

Pearson Correlation .788** 1

Sig. (2-tailed) .000

N 75 75

**. Correlation is significant at the 0.01 level (2-tailed).

Interpretation To investigate if there was a statistically significant association between Scholastic aptitude test and math achievement, a correlation was computed. Both the variables were approximately normal there is linear relationship between them hence fulfilling the assumptions for Pearson's correlation. Thus, the Pearson’s r is calculated, r= 0.79, p < .001 relating that there is highly significant relationship between the variables. The positive sign of the Pearson's test value shows that there is positive relationship, which means that students who have high scores in math achievement test do have high scores in scholastic aptitude test and vice versa. Using Cohen’s (1988) guidelines’ the effect size is large relating that there is strong relationship between math achievement and scholastic aptitude test.

Spearman Correlation:


If the assumptions for Pearson correlation are not fulfilled then consider the Spearman correlation with the assumption that the Relationship between two variables is monotonically non-linear Example: what is the association between mother’s education and math achievement?1. Analyze Correlate Bivariate.2. Move math achievement and mother’s education to the variables box3. Next ensure that the spearman and Pearson boxes are checked.4. Make sure that the two-tailed (under test of significance), flag significant correlations

and two-tailed are checked5. Now click on options and check means and standard deviations and click on exclude

cases list wise.6. Click on continue and click on Ok

Correlationsa

mother's education

math achieveme

nt test

Spearman's rho

mother's education

Correlation Coefficient

1.000 3.15**

Sig. (2-tailed) . .006

math achievement test

Correlation Coefficient

.315** 1.000

Sig. (2-tailed) .006 .

Interpretation

To investigate if there was a statistically significant association between mother’s education and math achievement, a correlation was computed. Mother’s education was skewed (skewness=1.13), which violated the assumption of normality. Thus, the spearman rho statistic was calculated, r, (73) = .32, p = .006. The direction of the correlation was positive, which means that students who have highly educated mothers tend to have higher math achievement test scores and vice versa. Using Cohen’s (1988) guidelines’ the effect size is medium for studies in his area. The r2 indicates that approximately 10% of the variance in math achievement test score can be predicted from mother’s education.

REGRESSION ANALYSIS


Regression analysis is used to measure the relationship between two or more variables. One variable is called dependent (response, or outcome) variable and the other is called Independent (explanatory or predictor) variables.

It is used to check that due to one unit change in the independent variable(s) how much change occurs in dependent variable.

Regression Equation

It is the equation representing the relation between selected values of one variable (x:the independent variable) and observed values of the other (y: the dependent variable); it permits the prediction of the most probable values of y. The standard form of this equation for two variables and for more than two variables respectively is as follows

Y = a + bx Y = a + bx1 + cx2 + dx3 + ex4

Y = dependent variable

a = Constant

b, c, d, e, = slope coefficients

x1, x2, x3, x4 = Independent variables

Types of Regression

There are two types of regression analysis that are

7. Simple Regression

8. Multiple regression

9. Simple Regression

Simple regression is used to check the contribution of independent variable(s) in the dependent

variable if the independent variable is one.


10. Assumptions and conditions of simple regression

1. Dependent variable should be scale 2. The relationship of variables should be linear 3. Data should be independent

Example: Can we predict math achievement from grades in high school

Commands

1. Analyze Regression Linear2. Highlight math achievement. Click the arrow to move it into the dependent box3. Highlight grades in high school and click on the arrow to move it into the independent

(s) box. 4. Click on Ok

Variables Entered/Removedb

ModelVariables Entered

Variables Removed Method

1 grades in h.s.a . Enter

a. All requested variables entered.

Model Summary

Model R R SquareAdjusted R

SquareStd. Error of the Estimate

1 .504a .254 .244 5.80018

a. Predictors: (Constant), grades in h.s.

ANOVAb

ModelSum of Squares Df Mean Square F Sig.

1 Regression 836.606 1 836.606 24.868 .000a

Residual 2455.875 73 33.642

Total 3292.481 74


Variables Entered/Removedb

ModelVariables Entered

Variables Removed Method

1 grades in h.s.a . Enter

a. Predictors: (Constant), grades in h.s.

b. Dependent Variable: math achievement test

Coefficientsa

Model

Unstandardized Coefficients

Standardized Coefficients

t Sig.B Std. Error Beta

1 (Constant) .397 2.530 .157 .876

grades in h.s. 2.142 .430 .504 4.987 .000

a. Dependent Variable: math achievement testRegression equation is Y = 0.40 + 2.14X

InterpretationSimple regression was conducted to investigate how well grades in high school predict math achievement scores. The results were statistically significant F (1, 73) = 24.87, p<.001. The indentified equation to understand this relationship was math achievement = .40 + 2.14* (grades in high school). The adjusted R2 value was .244. This indicates that 24% of the variance in math achievement was explained by the grades in high school. According to Cohen (1988), this is a large effect.

Multiple RegressionMultiple regressions is used to check the contribution of independent variable(s) in the

dependent variable if the independent variables are more than one.

5. Assumptions and conditions of Multiple regression

1. Dependent variables should be scale.

Example: How well can you predict math achievement from a combination of four variables:

grades in high school, father’s education, mother education and gender


Commands

6. Analyze Regression Linear7. Highlight math achievement. Click the arrow to move it into the dependent box8. Highlight grades in high school, father’s education, mother education and gender and click

on the arrow to move them into the independent (s) box. 9. Under method, be sure that enter is selected.10. Click on continue and then ok to get the following results in output window

Descriptive Statistics

MeanStd.

Deviation N


12.6621 6.49659 73

grades in h.s. 5.70 1.552 73

father's education 4.73 2.830 73

mother's education 4.14 2.263 73

Gender .55 .501 73

Model Summary

Model R R SquareAdjusted R

Square

Std. Error of the

Estimate

1 .616a .379 .343 5.26585

ANOVAb

ModelSum of Squares df

Mean Square F Sig.

1 Regression 1153.222 4 288.305 10.397 .000a


Descriptive Statistics

MeanStd.

Deviation N


12.6621 6.49659 73

grades in h.s. 5.70 1.552 73

father's education 4.73 2.830 73

mother's education 4.14 2.263 73

Residual 1885.583 68 27.729

Total 3038.804 72

Coefficients

Model

Unstandardized Coefficients

Standardized Coefficients

T Sig.B Std. Error Beta

1 (Constant) 1.047 2.526 .415 .680

grades in h.s. 1.946 .427 .465 4.560 .000

father's education .191 .313 .083 .610 .544

mother's education .406 .375 .141 1.084 .282

Gender -3.759 1.321 -.290 -2.846 .006

a. Dependent Variable: math achievement test

Regression Equation:

Y = 1.047 + 1.95X1 + 0.19X2 + 0.41X3 – 3.76 X4

Interpretation

Simultaneously multiple regression was conducted to investigate the best predictors of math achievement test scores. The means, standard deviation, and inter correlations can be found in table. The combination of variables to predict math achievement from grades in high school,


father’s education, mother’s education and gender was statistically significant, F = 10.40, p <0.05. The beta coefficients are presented in last table. Note that high grades and male gender significantly predict math achievement when all four variables are included. The adjusted R2

value was 0.343. This indicates that 34 % of the variance in math achievement was explained by the model according to Cohen (1988), this is a large effect.



Correlation

Some studies are interested in whether two variables are related to each other.

Is there a relationship between birth order and IQ scores?

Is there a relationship between socioeconomic status (SES) and health?

The CORRELATION COEFFICIENT is a statistic that shows the strength of the relationship between the two variables. The correlation coefficient falls between -1.00 and +1.00. The statistic shows both the STRENGTH of the relationship between the variables, and the DIRECTION of the relationship. The numerical value indicates the strength of the relationship. The sign in front of the numerical value indicates the direction of the relationship. Let us consider each of these in more detail.

THE NUMBERICAL VALUE:

Correlation coefficient values that are close to zero (e.g., -.13, +.08) suggest that there is no relationship between the two variables. The closer the correlation is to one (e.g., -.97, +.83) the stronger the relationship between the two variables. Thus, we might expect that there would be no relationship between the height of college students and their SAT scores, and we would be correct. The correlation coefficient is very close to zero. However, we might expect a correlation between adult height and weight to be stronger, and again we would be correct.

THE SIGN:

The sign of the correlation coefficient tells us whether these two variable are directly related or inversely related.

Do the two variables increase and decrease in the same direction? The more time a student spends studying the better their grade, the less time spent studying the lower the grade. Notice how both study time and grade vary in the same direction. As studying increases grades increase, and when studying decreases grades decline. Grade and study time would be POSITIVELY correlated. The term POSITIVE does not necessarily mean its a good thing (when is getting a poor grade a "good" thing!). It simply means that there is a direct relationship, the variables are varying (changing) in the same direction.


Do the two variables vary in opposing directions? As the number of children in a family increase the lower the IQ scores of the children. Thus, family size and children's IQ scores vary in the opposite direction. As family size increases the IQ scores decline, as the family size decreases IQ scores increase. IQ and family size are NEGATIVELY correlated (inversely related).

Try the following exercise to see if you understand the concept of correlation. It is best if you have read both this section and the research method section on correlational studies before completing the exercise. EXERCISE


Inferential Statistics allow researchers to draw conclusions (inferences) from the data. There are several types of inferential statistics. The choice of statistic depends on the nature of the study. Covering the different procedures used is beyond the scope of this course. However, understanding why they are used is important.

A researcher asks two groups of children to complete a personality test. The researcher then wants to know whether the males scored differently than the females on certain measures of personality. We will create a fictitious personality trait "Z." Here are the scores for the girls and the boys:

Girls Boys

23 37 The mean score for the "Z" trait in boys was higher than the mean score for "Z" in the girls. But notice how within the two groups there was considerable fluctuation. By "chance" alone we might have obtained these different values. Thus, in order to conclude that "Z" shows a gender difference, we need to rule out that these differences were just

40 5637 1841 4141 42 33 3828 5025 22


a fluke. This is where inferential statistics come in to play.

24 33

13 47

28 25

44 46

Mean=31.42 Mean=37.92

SD=9.03 SD=11.14

An important concept in inferential statistics is STATISTICAL SIGNIFICANCE. When an inferential statistic reveals a statistically significant result the differences between the groups were unlikely due to chance. Thus, we can rule out chance with a certain degree of confidence. When the results of the inferential statistic are not statistically significant, chance could still be a reason why we obtained the observations that we did.


T-TEST STATISTICInferential Statistics


T-TEST Statistics

The t test is used to compare to groups to answer the differential research questions. Its values determines the difference by comparing means

Hypothesis for T-test

HO: there is no Difference between variable 1 and variable 2

H1: There is difference between variable 1 and variable 2

Types of T-test

There are three types of T-test

1. One sample t-test

2. Independent sample t-test

3. Paired sample t-test

1. ONE SAMPLE T-TEST

One sample t-test is used to determine if there is difference between population mean (Test value) and the sample mean (X)

Assumptions and conditions of 1 sample t-test

1. The dependent variable should be normally distributed within the population

2. The data are independent.(scores of one participant are not depend on scores of the other :participant are independent of one another )

Example: is the mean SAT-Math score in the modified HSB data set significantly different

from the presumed population mean of 500?

Commands

1. Analyze Compare means One sample t-test

2. Move scholastic aptitude test-math to the test variables box.

3. Type 500 in the test value box


4. Click on Ok

One-Sample Statistics

N Mean Std. DeviationStd. Error

Mean

scholastic aptitude test – math

75 490.53 94.553 10.918

One-Sample Test

Test Value = 500

t DfSig. (2-tailed)

Mean Difference

95% Confidence Interval of the Difference

Lower Upper

scholastic aptitude test – math -.867 74 .389 -9.467 -31.22 12.29

Interpretation:

To investigate the difference between population and the sample, one-sample t-test is conducted. The One-Sample Statistics table provides basic descriptive statistics for the variable under consideration. The Mean AT-Math for the students in the sample will be compared to the hypothesize population mean, displayed as the Test Value in the One-Sample Test table. On the bottom line of this table are the t value, df, and the two-tailed sig. (p) value, which are circled. Note that p=.389 so we can say that the sample mean (490.53) is not significantly different from the population mean of 500. The table also provides the difference (-9.47) between the sample and population mean and the 95% Confidence Interval. The difference between the sample and the population mean is likely to be between +12.29 and -31.22 points. Notice that this range includes the value of zero, so it is possible that there is no difference. Thus, the difference is not statistically significant.

1. INDEPENDENT SAMPLE T-TEST

Independent sample T-test is used to compare two independent groups (Male and Female)with respect to there effect on same dependent variable.


Assumptions and conditions of Independent T-test

1. Variance of the dependent variable for two categories of the independent variable should be equal to each other

2. Dependent variable should be scale

3. Data on dependent variable should be independent.

Example: Do male and female students differ significantly in regard to their average math

achievement scores

Commands

1. Analyze Compare means independent sample t-test

2. Move math achievement scores to the test variables box.

3. Move gender to the grouping variable box

4. Click on define groups

5. Type 0 for males in the group 1 box and 1 for females in the group 2 box

6. Click on continue

7. Click on Ok

Interpretation


The first table, Group Statistics, shows descriptive statistics for the two groups (males and females) separately. Note that the means within each of the three pairs look somewhat different. This might be due to chance, so we will check the t test in the next table.

The second table, Independent Sample Test, provides two statistical tests. The left two columns of numbers are the Levene’s test for the assumption that the variances of the two groups are equal. This is not the t test; it only assesses an assumption! If this F test is not significant (as in the case of math achievement and grades in high school), the assumption is not violated, and one uses the Equal variances assumed line for the t test and related statistics. However, if Levene’s F is statistically significant (Sig. <.05), as is true for visualization, then variances are significantly different and the assumption of equal variances is violated. In that case, the Equal variances not assumed line used; and SSPS adjusts t, df, and Sig. The appropriate lines are circled.

Thus, for visualization, the appropriate t=2.39, degree of freedom (df) = 57.15, p=.020. This t is statistically significant so, based on examining the means, we can say that boys have higher visualization scores than girls. We used visualization to provide an example where the assumption of equal variances was violated (Levene’s test was significant). Note that for grades in high school, the t is not statistically significant (p=.369) so we conclude that there is no evidence of a systematic difference between boys and girls on grades. On the other hand, math achievement is statistically significant because p<.05; males have higher means.

The 95% Confidence Interval of the Difference is shown in the two right-hand column of the output. The confidence interval tells us if we repeated the study 100 times, 95 of the times the true (population) difference would fall within the confidence interval, which for math achievement is between 1.05 points and 6.97 points. Note that if the Upper and Lower bounds have the same sign (either + and + or – and -), we know that the difference is statistically significant because this means that the null finding of zero difference lies outside of the confident interval. On the other hand, if zero lies between the upper or lower limits, there could be no difference, as is the case of grades in h.s. The lower limit of the confidence interval on math achievement tells us that the difference between males and females could be as small as 1.05 points out 25, which are the maximum possible scores.

Effects size measures for t tests are not provided in the printout but can be estimated relatively easily. For math achievement, the difference between the means (4.01) would be divided by about 6.4, an estimate of the pooled (weighted average) standard deviation. Thus, d would be approximately .60, which is, according to Cohen (1988), a medium to large sized “effect.” Because you need means and standard deviations to compute the effect size, you should


include a table with means and standard deviations in your results section for a full interpretation of t tests.

8. PAIRED SAMPLE T-TEST

Paired sample T-test is used to compare two paired groups (e.g. Mothers and fathers) with respect to there effect on same dependent variable.

Assumptions and conditions of Paired sample T-test

9. The independent variable is dichotomous and its levels (or groups) are paired, or matched, in some way (husband-wife, pre-post etc)

10. The dependent variable is normally distributed in the two conditions

Example: Do students’ fathers or mothers have more education?

Commands

11. Analyze Compare means paired sample t-test12. Click on both of the variables, fathers education and mothers education, and move

them simultaneously to the paired variables box13. Click on Ok

Paired Samples Statistics

Mean NStd.

DeviationStd. Error

Mean

Pair 1 father's education 4.73 73 2.830 .331

mother's education

4.14 73 2.263 .265

Paired Samples Correlations

N Correlation Sig.

Pair 1 father's education & mother's education

73 .681 .000

Interpretation

The first table shows the descriptive statistics used to compare mother’s and father’s education levels. The second table Paired Samples Correlations, provides correlations between the two


paired scores. The correlation (r=.68) between mother’s and father’s education indicates that highly educate men tend to marry highly educated women and vice versa. It doesn’t tell you whether men or women have more education. That is what t in the third table tells you.

The last table shows the Paired Samples t Test. The Sig. for the comparison of the average education level of the students’ mothers and fathers was p=.019. Thus, the difference in educational level is statistically significant, and we can tell from the means in the first table that fathers have more education; however, the effect size is small (d=.28), which is computed by dividing the mean of the paired differences (.59) by the standard deviation (2.1) of the paired differences. Also, we can tell from the confidence interval that the difference in the means could be as small as .10 of a point or as large as 1.08 points on the 2 to 10 scale.




Inferential Statistics allow researchers to draw conclusions (inferences) from the data. There are several types of inferential statistics. The choice of statistic depends on the nature of the study. Covering the different procedures used is beyond the scope of this course. However, understanding why they are used is important.

A researcher asks two groups of children to complete a personality test. The researcher then wants to know whether the males scored differently than the females on certain measures of personality. We will create a fictitious personality trait "Z." Here are the scores for the girls and the boys:

Girls Boys

23 37 The mean score for the "Z" trait in boys was higher than the mean score for "Z" in the girls. But notice how within the two groups there was considerable fluctuation. By "chance" alone we might have obtained these different values. Thus, in order to conclude that "Z" shows a gender difference, we need to rule out that these differences were just a fluke. This is where inferential statistics come in to play.

40 56

37 18

41 41

41 42

33 38

28 50

25 22

24 33

13 47

28 25


44 46

Mean=31.42 Mean=37.92

SD=9.03 SD=11.14

An important concept in inferential statistics is STATISTICAL SIGNIFICANCE. When an inferential statistic reveals a statistically significant result the differences between the groups were unlikely due to chance. Thus, we can rule out chance with a certain degree of confidence. When the results of the inferential statistic are not statistically significant, chance could still be a reason why we obtained the observations that we did.

In the example above we would use an inferential statistic called a T-TEST. The t-test is used when we are comparing TWO groups. In this instance the t-test does not yield a statistically significant difference. In other words, the differences between the scores for the boys and the scores for the girls are not large enough for us to rule out chance as a possible explanation. We would have to conclude then that there is no gender difference for our hypothetical "Z" trait.

Inferential statistics do not tell you whether your study is accurate or whether your findings are important. Statistics cannot make up for an ill-conceived study or theory. They simply assess whether we can rule out the first "extraneous" variable of all research, CHANCE.


Final-Term Project Discussion

&

Lab Practice Session

Lab Practice Session


The students will be given two hours session in Lab revision of what they have learnt in Post-mid session.

1. The objectives of this session are to provide students an opportunity to

2. Revise the whole course that they have learnt throughout the post mid session

3. Have hands on practice on dealing with quantitative data using SPSS

4. Share their problems that they confront during revision and get the solution

5. Clarify if they have any ambiguity regarding understanding or application of any concept regarding QTB

Final Project Discussion

The students will be given one hour’s session to discuss about the final draft of their final projects.

The objectives of this session are to provide students an opportunity to

1. Share their problems that they confront during revision and get the solution

2. Clarify if they have any ambiguity regarding understanding or application of any concept regarding QTB

3. Get productive feedback on what they have done regarding their projects

The Drafts will be checked on the following criteriaThe drafts will be checked if the following components are covered

1. Whether the survey is appropriately designed to collect the primary data

2. Whether the following components are appropriately discussed in the report

1. An introduction explaining the background and objectives of your work.

2. The Justification of the topic selection

3. A description of the data – definitions of the variables, conclusions about data quality,

and so on.

4. A justification of the methods you have chosen to analyze the data.

5. Analysis and results descriptive as well as inferential with results

6. Conclusion – a discussion and interpretation of your results and a summary of what you

have achieved.

7. Length: 1500 to 2000 words


PRESENTATION

on

Final Term Project


Presentation

Students will be evaluated on the basis of following criteria.

1. Timing of presentation.2. Clarity of concepts.3. Structure of the presentation.4. Quality of overheads, handouts etc. 5. Application of theory to practice.6. Ability to answer questions effectively


Documents

Introduction to Quantitative Techniques in Business