What is a Chi Square Test?
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How to do a Chi-square test for independence in SPSS? 1. Using Chi-Square Statistic in Research This easy tutorial will show you how to run the Chi-Square test in SPSS, and how to interpret the result. The chi-square test of independence uses to investigate the relationship between two categorical variables that have two or more categories. In addition, the test compares the proportions, that is, the frequency of cases observed in each of the categories with values that would be expected to have if there is no relationship between the variables. First of all, the chi-square test based on a contingency table that shows the intersection of each category of one variable with each group of the other variable.
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The ppt cover General Introduction to the topic,
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The Chi-Square (ฯ2) test is a nonparametric test used to test hypotheses about frequency distributions across categories. It can test for independence and compare variances. The key steps are: 1) Calculate expected frequencies under the null hypothesis, 2) Calculate the differences between observed and expected frequencies, 3) Sum the squared differences divided by the expected values to get the chi-square statistic, 4) Compare this statistic to critical values from the chi-square distribution to determine significance. An example tests if vaccination prevents smallpox attacks, finding the results do not support independence between vaccination status and attacks, suggesting vaccination is effective.
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The document discusses how to use a chi-squared (x2) test to examine differences between observed and expected frequencies of categorical data. It provides guidelines for when a chi-squared test is appropriate, how to perform the calculation, and how to interpret the results. A case study example is presented of a student analyzing questionnaire responses about the 2012 Olympics using a chi-squared test to determine if response frequencies differed significantly between demographic groups.
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Formulas considering costs and mark up will help you to do the Problem set assignment:
1. Markup for setting prices (Mark up $ = SP-CP); MARK UP % = (MU $/SP) X100)
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SP = Cost/(1- Markup %)
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= $25.00
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Analysis of Variance is a method of testing the equality of three or more population
means by analyzing sample variance.
One-Way ANOVA
The one-way ANOVA is used to compare three or more population means when there is
one factor of interest.
Requirements
๏ถ The populations have distributions that are approximately normal.
๏ถ The populations have the same variance.
๏ถ The samples are simple random samples of quantitative data.
๏ถ The samples are independent of each other.
๏ถ The different samples are from populations that are categorized in only one.
way
One-Way ANOVA is a hypothesis test. There are seven steps for a hypothesis test.
Example
A professor at a local University believes there is a relationship between head size and
the major of the students in her biostatistics classes. She takes a random sample of 20
students from each of three classes and records their major and head circumference.
The data are shown in the following table.
Step 1: State the null hypothesis.
Mean 1 equals mean 2 equals mean 3 equals mean 4.
Step 2: State the Alternative hypothesis.
At least one mean is different.
Step 3: State the Level of Significance.
The level of significance is 0.05.
Step 4: State the test statistic.
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variance within samples
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the numerator and one for the denominator.
The calculations for the test statistic are complicated, so a software program is
generally used for the calculations. We will be using Microsoft Excel for this example.
Step 5: Calculate
The calculations are done in Microsoft Excel using the data analysis toolpak. Enter the
data into the spread sheet as shown here. Click on data and the data analysis tookpak
button is on the right.
When you click on the button a dialogue box appears.
Choose ANOVA One Factor. Then another dialogue box appears.
Input range is where the data is in the table. Be sure to put a check in the box for labels
in.
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Tests of statistical significance : chi square and spss
This document discusses tests of statistical significance, specifically chi square tests. It defines key concepts like p-values, type I and type II errors, and null and alternative hypotheses. It explains how to calculate chi square manually and in SPSS. Chi square can be used to test for differences in proportions between groups and associations between variables. The document provides examples of chi square tests and interpreting their results.
The chi โ square test
The chi-square test is used to determine if there is a relationship between two categorical variables in two or more independent groups. It can be used when data is arranged in a contingency table with observed and expected frequencies. A sample problem demonstrates how to calculate chi-square by finding the difference between observed and expected counts, squaring these differences, dividing by the expected counts, and summing across all cells. Degrees of freedom and critical values from tables determine whether to reject or fail to reject the null hypothesis of independence. Larger tables can be partitioned into subtables to identify where differences lie. Guidelines are provided for when chi-square or Fisher's exact test should be used based on sample size and expected cell counts.
Chi squared test
This document provides an overview of the chi-square test, including how to calculate it and interpret the results. It gives an example of using a chi-square test to analyze data from an observational study of 100 people to see if littering behavior differs by gender. A 2x2 contingency table is constructed, expected values are calculated under the null hypothesis of no association, and the chi-square statistic is found to be less than the critical value, so the null hypothesis is not rejected - the data do not provide evidence that gender is associated with littering behavior.
Chi square
This document provides an overview of chi square tests as a type of non-parametric statistic. It explains that chi square tests are used to analyze nominal or categorical data by comparing observed frequencies to expected frequencies if the null hypothesis is true. An example is provided to demonstrate how to calculate chi square values from a contingency table and determine significance based on degrees of freedom and p-values. The document also outlines other types of non-parametric tests that can be used depending on whether the data is nominal, ordinal, independent, or matched samples.
Chisquare Test
The chi-square test is used to determine if an observed frequency distribution differs from an expected theoretical distribution. It can test for independence and goodness of fit. Karl Pearson introduced the chi-square test to compare observed and expected frequencies across categories. The test calculates a chi-square statistic and compares it to a critical value to determine if the null hypothesis that the distributions are the same can be rejected. Examples demonstrated how to calculate expected frequencies, the chi-square statistic, degrees of freedom, and compare to critical values to test independence between variables and goodness of fit to theoretical distributions.
Chi square test
This document discusses the chi-square test, a statistical hypothesis test where the test statistic follows a chi-squared distribution if the null hypothesis is true. It was developed by Karl Pearson in 1900. The chi-square test can be used to examine differences between observed and expected results in a random pattern. It is applicable to testing independence between attributes and goodness of fit to theoretical distributions. Common mistakes in applying the chi-square test include small theoretical frequencies, neglecting non-occurrences, and incorrect determination of degrees of freedom. The chi-square test is used to test significance of sample variance, independence in contingency tables, comparisons of frequency distributions, and goodness of fit.
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The document discusses the chi-square test, which is used to determine if there is a statistically significant relationship between two categorical variables. It originated from the sum of squares of independent normal variables. The chi-square test can be used for goodness of fit tests to assess differences between observed and expected values. It has a characteristic distribution represented by the symbol ฯ2. The test assesses how likely observed categorical data distributions are by chance alone.
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Chi square
The ppt cover General Introduction to the topic,
Description of CHI-SQUARE TEST, Contingency table, Degree of Freedom, Determination of Chi โ square test, Assumption for validity of chi - square test, Characteristics , Applications, Limitations
The median test
I. The median test is used to determine if two independent groups have been drawn from populations with the same median. It requires at least ordinal scale data.
II. The combined median of both groups is calculated. Scores from each group are then split based on whether they are above or below the combined median. These frequencies are entered into a 2x2 contingency table.
III. The median test statistic (chi-square) is calculated and compared to a critical value based on the significance level and degrees of freedom to determine whether to reject or fail to reject the null hypothesis that the two groups have the same median.
Chi sqyre test
The document discusses the chi-square test, a non-parametric statistical test used to compare observed data with expected frequencies. It can be used to test independence and association between two categorical variables. The chi-square test involves calculating a quantity called chi-square from a contingency table of observed and expected frequencies and comparing it to critical values from a chi-square distribution table. If the calculated value exceeds the critical value, the null hypothesis of no relationship can be rejected.
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Estimating the difference in a quantitative/ continuous parameter between more than 2 independent groups - ANOVA TEST
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Chi squared test
The chi-square test is used to compare observed data to expected data. It determines if differences between the observed and expected numbers are due to chance or something more significant. The chi-square test has several key steps: stating the null and alternative hypotheses, choosing a significance level, finding the critical value, calculating the test statistic by summing the squared differences between observed and expected values divided by the expected value, and making a conclusion by comparing the test statistic to the critical value. The chi-square test has assumptions of adequate sample sizes and independence of data. It is useful for testing goodness of fit, independence of attributes, and homogeneity.
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This document discusses Chi Square and related procedures for analyzing categorical data. It explains that Chi Square can be used for goodness of fit tests to check if a sample follows a particular distribution, and for tests of association to check if two categorical variables are related. It provides examples of how to conduct and interpret Chi Square goodness of fit and association tests using SPSS. Other related procedures discussed include Fisher's Exact Test for small sample sizes and McNamer's Test for analyzing changes in paired categorical data.
Chi square test final presentation
The Chi-Square (ฯ2) test is a nonparametric test used to test hypotheses about frequency distributions across categories. It can test for independence and compare variances. The key steps are: 1) Calculate expected frequencies under the null hypothesis, 2) Calculate the differences between observed and expected frequencies, 3) Sum the squared differences divided by the expected values to get the chi-square statistic, 4) Compare this statistic to critical values from the chi-square distribution to determine significance. An example tests if vaccination prevents smallpox attacks, finding the results do not support independence between vaccination status and attacks, suggesting vaccination is effective.
Research Methodology anova
ANOVA (analysis of variance) is a statistical technique used to compare differences between group means. It involves calculating the F ratio, which is the ratio of variance between groups to variance within groups. If the calculated F value is greater than the critical F value from statistical tables, then the difference between group means is considered statistically significant. The document provides steps for conducting a one-way ANOVA, including calculating sums of squares, mean squares, and the F ratio to determine if differences between three varieties of wheat are statistically significant based on per acre production data.
Chi Square Worked Example
The document discusses how to use a chi-squared (x2) test to examine differences between observed and expected frequencies of categorical data. It provides guidelines for when a chi-squared test is appropriate, how to perform the calculation, and how to interpret the results. A case study example is presented of a student analyzing questionnaire responses about the 2012 Olympics using a chi-squared test to determine if response frequencies differed significantly between demographic groups.
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Chapter 18 โ Pricing Setting in the Business WorldThere are few .docx
Chapter 18 โ Pricing Setting in the Business World
There are few Methods for setting pricing โ costs methods vs demand methods
Formulas considering costs and mark up will help you to do the Problem set assignment:
1. Markup for setting prices (Mark up $ = SP-CP); MARK UP % = (MU $/SP) X100)
Formula for setting price with the markup method
SP = Cost/(1- Markup %)
Example - retailer buys A hat for $15 and wants a 40% markup, his selling price would beโฆ.
SP = 15/(1-.40)=.60
= $25.00
2. Understand Role of different costs โ fixed, variable, total costs and average costs
3. What is the breakeven point? Formula for calculating the Break Even point.
BE = Total Fixed Cost/Fixed cost contribution
Fixed Cost Contribution=Price โ variable cost
4. Average Cost = when there are many flavors/types of the same product, producer determines average cost and then add the mark up to set a common selling price.
.ANOVA
Analysis of Variance is a method of testing the equality of three or more population
means by analyzing sample variance.
One-Way ANOVA
The one-way ANOVA is used to compare three or more population means when there is
one factor of interest.
Requirements
๏ถ The populations have distributions that are approximately normal.
๏ถ The populations have the same variance.
๏ถ The samples are simple random samples of quantitative data.
๏ถ The samples are independent of each other.
๏ถ The different samples are from populations that are categorized in only one.
way
One-Way ANOVA is a hypothesis test. There are seven steps for a hypothesis test.
Example
A professor at a local University believes there is a relationship between head size and
the major of the students in her biostatistics classes. She takes a random sample of 20
students from each of three classes and records their major and head circumference.
The data are shown in the following table.
Step 1: State the null hypothesis.
Mean 1 equals mean 2 equals mean 3 equals mean 4.
Step 2: State the Alternative hypothesis.
At least one mean is different.
Step 3: State the Level of Significance.
The level of significance is 0.05.
Step 4: State the test statistic.
variance between samples
variance within samples
F ๏ฝ
The test statistic follows the F distribution which has two degrees of freedom, one for
the numerator and one for the denominator.
The calculations for the test statistic are complicated, so a software program is
generally used for the calculations. We will be using Microsoft Excel for this example.
Step 5: Calculate
The calculations are done in Microsoft Excel using the data analysis toolpak. Enter the
data into the spread sheet as shown here. Click on data and the data analysis tookpak
button is on the right.
When you click on the button a dialogue box appears.
Choose ANOVA One Factor. Then another dialogue box appears.
Input range is where the data is in the table. Be sure to put a check in the box for labels
in.
test 3 AAMultiple ChoiceIdentify the choice that best comple.docx
test 3 AA
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____ 1. For an ANOVA comparing three treatment conditions, what is stated by the alternative hypothesis (H1)?
a.
There are no differences between any of the population means.
b.
At least one of the three population means is different from another mean.
c.
All three of the population means are different from each other.
d.
None of the other choices is correct.
____ 2. On average, what value is expected for the F-ratio if the null hypothesis is true?
a.
0
c.
k โ 1
b.
1.00
d.
N โ k
____ 3. A research study comparing three treatments with n = 5 in each treatment produces T1 = 5, T2ย =ย 10, T3 = 15, with SS1 = 6, SS2 = 9, SS3 = 9, and SX2 = 94. For this study, what is SSbetween?
a.
10
c.
34
b.
24
d.
68
____ 4. An analysis of variance produces SSbetween = 30, SSwithin = 60, and an F-ratio with df = 2, 15. For this analysis, what is the F-ratio?
a.
30/60 = 0.50
c.
15/4 = 3.75
b.
60/30 = 2.00
d.
4/15 = 0.27
____ 5. If an analysis of variance is used for the following data, what would be the effect of changing the value of M1 to 20?
Sample Data
M1 = 15 M2 = 25
SS1 = 90 SS2 = 70
a.
Increase SSbetween and increase the size of the F-ratio
b.
Increase SSbetween and decrease the size of the F-ratio
c.
Decrease SSbetween and increase the size of the F-ratio
d.
Decrease SSbetween and decrease the size of the F-ratio
____ 6. In general, what factors are most likely to reject the null hypothesis for an ANOVA?
a.
Small mean differences and small variances
b.
Small mean differences and large variances
c.
Large mean differences and small variances
d.
Large mean differences and large variances
____ 7. A two-factor study with two levels of factor A and three levels of factor B uses a separate group of n = 5 participants in each treatment condition. How many participants are needed for the entire study?
a.
5
c.
25
b.
10
d.
30
____ 8. The following data represent the means for each treatment condition in a two-factor experiment. Note that one mean is not given. What value for the missing mean would result in no AยดB interaction?
B1 B2
A1
20
30
A2
10
?
a.
10
c.
30
b.
20
d.
40
____ 9. If a two-factor analysis of variance produces a statistically significant interaction, what can you conclude about the main effects?
a.
Either the main effect for factor A or the main effect for factor B is also significant
b.
Both the man effect for factor A and the main effect for factor B are significant
c.
Neither the main effect for factor A nor the main effect for factor B is significant
d.
The significance of the main effects is not related to the significance of the interaction
____ 10. What is indicated by a positive value for a correlation?
a.
Increases in X tend to be accompanied by increases in Y
b.
Increases in X tend to be accompanied by decreases in Y
c.
A much ...
Week 4Confidence Intervals and Chi Squareย (Chs 11 - 12).docx
Week 4
Confidence Intervals and Chi Squareย (Chs 11 - 12)
For questions 3 and 4 below, be sure to list the null and alternate hypothesis statements.ย Use .05 for your significance level in making your decisions.
For full credit, you need to also show the statistical outcomes - either the Excel test result or the calculations you performed.
1
Using our sample data, construct a 95% confidence interval for the population's mean salary for each gender.ย ย ย
Interpret the results.ย How do they compare with the findings in the week 2 one sample t-test outcomes (Question 1)?
Mean
St errorย
t value
Lowย
toย
High
Males
Females
Interpretation:
2
Using our sample data, construct a 95% confidence interval for the mean salary difference between the genders in the population.ย ย ย ย
ย How does this compare to the findings in week 2, question 2?
Difference
St Err.
T value
Lowย
toย
High
Yes/No
Can the means be equal?
Why?
How does this compare to the week 2, question 2 result (2 sampe t-test)?
a.
Why is using a two sample tool (t-test, confidence interval) a better choice than using 2 one-sample techniques when comparing two samples?
3
We found last week that the degreeย values within the population do not impact compa rates.ย ย
ย This does not mean that degrees are distributed evenly across the grades and genders.
Do males and females have athe same distribution of degrees by grade?
(Note: while technically the sample size might not be large enough to perform this test, ignore this limitation for this exercise.)
What are the hypothesis statements:
Ho:ย
Ha:
Note:ย You can either use the Excel Chi-related functions or do the calculations manually.
Data input tables - graduate degrees by gender and grade level
OBSERVED
Aย
B
C
D
E
F
Total
If desired, you can do manual calculations per cell here.
M Grad
Aย
B
C
D
E
Fem Grad
M Grad
Male Und
Fem Grad
Female Und
Male Und
Female Und
Sum =
EXPECTED
M Grad
For this exercise - ignore the requirement for a correction factor
Fem Grad
for cells with expected values less than 5.
Male Und
Female Und
Interpretation:
What is the value of the chi square statistic:ย
What is the p-value associated with this value:ย
Is the p-value <0.05?
Do you reject or not reject the null hypothesis:ย
If you rejected the null, what is the Cramer's V correlation:
What does this correlation mean?
What does this .
Ash bus 308 week 2 problem set new
This document provides instructions for a problem set analyzing employee salary data. It instructs the user to copy salary and employee data into their assignment file. It then provides guidance on using descriptive statistics and t-tests to analyze differences in compensation ratios between male and female employees. The goal is to determine if there is statistical evidence that male and female employees are paid equally for equal work.
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Spss cross tab n chi sq bivariate analysis
The document discusses crosstabulation, which shows the relationship between two or more categorical variables through frequency tables. It explains how to compute crosstabs in SPSS by selecting variables for the row and column and choosing relevant statistics like chi-square. The chi-square test determines if the frequencies differ from what is expected by chance alone. It assumes categories are independent and expected counts are at least 5 in each cell.
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Ash bus 308 week 2 problem set new
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T test
This document provides an overview of a presentation on statistical hypothesis testing using the t-test. It discusses what a t-test is, how to perform a t-test, and provides an example of a t-test comparing spelling test scores of two groups that received different teaching strategies. The document outlines the six steps for conducting statistical hypothesis testing using a t-test: 1) stating the hypotheses, 2) choosing the significance level, 3) determining the critical values, 4) calculating the test statistic, 5) comparing the test statistic to the critical values, and 6) writing a conclusion.
Session1b.ppt
This document provides an introduction to concepts in biostatistics and hypothesis testing. It outlines the objectives of learning about study design, types of data, hypothesis testing, p-values, and choosing appropriate statistical tests. It also describes office hours, topics beyond the scope of the class, and objectives for learning about stages of research studies, hypothesis tests, t-tests, and Wilcoxon tests.
Data Analysis for Graduate Studies Summary
This document provides guidance on analysing qualitative and quantitative data. For qualitative data, it discusses preparing the data, identifying concepts and themes, and ensuring quality analysis. Key strategies for qualitative analysis include open coding, classification, and conceptual frameworks. For quantitative data, the document outlines recording, describing, and managing the data using techniques such as frequency counts, cross-tabulation, t-tests, chi-squared tests, and measures of central tendency and correlation. Examples are provided for coding, entering, and presenting both types of data.
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Nonparametric tests assignment
This document provides information about non-parametric statistical tests. It discusses the Mann-Whitney U test, chi-square test, and how to perform chi-square tests in SPSS. Key points include:
- Non-parametric tests do not assume a specific data distribution and can be used for small sample sizes, ordinal data, or outliers. Examples include Mann-Whitney U, Kruskal-Wallis, and chi-square tests.
- Chi-square tests independence between two categorical variables. Assumptions include frequencies data and expected counts over 5 in 80% of cells.
- To perform a chi-square test in SPSS, select two categorical variables, choose crosstabs
Research Procedure
This part of the thesis describes the methodology section which provides details of the research activities, data collection strategies, and administration of questionnaires and interviews to achieve the study objectives and address the problem. It discusses preparing and testing questionnaires, identifying persons responsible for data collection, and approaches for administering questionnaires and conducting interviews.
Hyphotheses testing 6
The document discusses hypothesis testing, which is a statistical method used to determine if a claim about a population parameter is true. It involves setting a null and alternative hypothesis, choosing a test statistic, determining critical values, and making a conclusion about whether to reject or fail to reject the null hypothesis based on comparing the test statistic to critical values. Examples provided include testing claims about a motorcycle's mileage, a banana's weight, and the effectiveness of a medicine.
Week 5 Lecture 14 The Chi Square TestQuite often, patterns of .docx
Week 5 Lecture 14
The Chi Square Test
Quite often, patterns of responses or measures give us a lot of information. Patterns are generally the result of counting how many things fit into a particular category. Whenever we make a histogram, bar, or pie chart we are looking at the pattern of the data. Frequently, changes in these visual patterns will be our first clues that things have changed, and the first clue that we need to initiate a research study (Lind, Marchel, & Wathen, 2008).
One of the most useful test in examining patterns and relationships in data involving counts (how many fit into this category, how many into that, etc.) is the chi-square. It is extremely easy to calculate and has many more uses than we will cover. Examining patterns involves two uses of the Chi-square - the goodness of fit and the contingency table. Both of these uses have a common trait: they involve counts per group. In fact, the chi-square is the only statistic we will look at that we use when we have counts per multiple groups (Tanner & Youssef-Morgan, 2013). Chi Square Goodness of Fit Test
The goodness of fit test checks to see if the data distribution (counts per group) matches some pattern we are interested in. Example: Are the employees in our example company distributed equal across the grades? Or, a more reasonable expectation for a company might be are the employees distributed in a pyramid fashion โ most on the bottom and few at the top?
The Chi Square test compares the actual versus a proposed distribution of counts by generating a measure for each cell or count: (actual โ expected)2/actual. Summing these for all of the cells or groups provides us with the Chi Square Statistic. As with our other tests, we determine the p-value of getting a result as large or larger to determine if we reject or not reject our null hypothesis. An example will show the approach using Excel.
Regardless of the Chi Square test, the chi square related functions are found in the fx Statistics window rather than the Data Analysis where we found the t and ANOVA test functions. The most important for us are:
ยท CHISQ.TEST (actual range, expected range) โ returns the p-value for the test
ยท CHISQ.INV.RT(p-value, df) โ returns the actual Chi Square value for the p-value or probability value used.
ยท CHISQ.DIST.RT(X, df) โ returns the p-value for a given value.
When we have a table of actual and expected results, using the =CHISQ.TEST(actual range, expected range) will provide us with the p-value of the calculated chi square value (but does not give us the actual calculated chi square value for the test). We can compare this value against our alpha criteria (generally 0.05) to make our decision about rejecting or not rejecting the null hypothesis.
If, after finding the p-value for our chi square test, we want to determine the calculated value of the chi square statistic, we can use the =CHISQ.INV.RT(probability, df) function, the value for probability is .
Thesigntest
The document discusses the sign test, a nonparametric hypothesis test that does not require assumptions about the population distribution. The sign test can be used to test claims involving matched pairs, nominal data with two categories, or the population median. The document provides guidelines for performing the sign test in each of these cases, including stating hypotheses, determining sample sizes and test statistics, and making conclusions. Examples are also given to illustrate the sign test for matched pairs, nominal data, and testing the population median.
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What is a Chi Square Test?
- 1. How to do a Chi-square test for independence in SPSS? What is a Chi-square test for independence? The chi-square test of independence is used to investigate the relationship between two categorical variables that have two or more categories. The chi-square test compares the proportions, that is, the frequency of cases observed in each of the categories with values that would be expected to have if there is no relationship between the variables. The chi-square test is based on a contingency table that shows the intersection of each category of one variable with each group of the other variable. We will test whether education (primary school, secondary school, BA, Master, PhD) is independent of gender (male, female). Null hypothesis: The education level is not independent of gender. Alternative hypothesis: The education level is independent of gender. How to do a Chi-square test for independence in SPSS? 1. From SPSS menu, choose Analyze โ Descriptive Statistics โ Crosstabs. 2. A new window will open. Transfer variables from the left box to the right boxes. One variable transfer into box Row(s) and other into box Column(s). Checkbox Display clustered bar charts.
- 2. 3. Click on Statistics tab. A new window will appear. Choose Chi-square and Phi and Cramer's V from Nominal box. Click Continue. 4. Click on Cellsโฆ tab and a new window will appear. Check Observed in the Counts box. Check Row, Column, Total in the Percentages box. Click Continue and then OK.
- 3. 5. The results of the Chi-square test for independence will appear in the output box.
- 4. How to report a Chi-square test for independence results? 1. The first table is Case processing Summary. It shows the number of valid, missing, and total cases. 2. The second table is Crosstabulation. It shows the number of cases in each group. For example, there is 11 male finished primary school, 11 male finished secondary school, 13 male finished BA, 8 male finished Master, and 7 male finished PhD. There is a female finished primary school, 15 females finished secondary school, 21 female finished BA, 10 female completed Master, and 10 female finished PhD. There are 50 males and 66 females in the sample. 21 respondents finished primary school, 26 finished secondary school, 34 finished BA, 18 completed Master, and 17 finished PhD. The total number of respondents is 116. 3. The third table shows the Chi-square test statistic and its significance. 4. If p-value (Asymp. Sig. (2-sided)) for Pearson Chi-square statistic is higher than .05 (p > .05), the education is not dependent of gender (we fail to reject the null hypothesis). If p-value (Asymp. Sig. (2-sided)) for Pearson Chi-square statistic is lower than .05 (p < .05), the education is dependent on gender (we reject the null hypothesis).
- 5. Since the results show that p = .892 > .05, we fail to reject the null hypothesis. 5. Following table shows the results of Cramer's V and Phi measures. If Cramer's V/Phi is 0.10, there is a small impact, 0.30 medium, and 0.50 high. 6. In the end, there are clustered bar chart.
- 6. We may report the results of the Chi-square test for independence in the following way: A chi-square test for independence was computed to determine whether education (primary school, secondary school, BA, Master, PhD) is independent of gender (male, female). The results are not significant, ฯ2(4) = 1.111, p = .892, Cramer's V/phi = .892. We fail to reject the null hypothesis that education is the same across gender (male, female) and conclude that education is not related to gender.