Hypothesis testng

1. T-test: Difference in means to test the statistical significance in the difference in

means ex. income by gender, the num. of years at work by gender

2. T-test: Difference in proportionsto test the statistical significance in the difference in proportions ex. the proportion employed in government jobs by gender

3. Contingency Table/Chi-Square Analysis to test whether all categories contain the same

proportion of values or not by comparing expected and actual values.

ex. the proportion employed in government jobs by gender

Hypothesis Testing

1. A Research Question2. The Null Hypothesis

usually assumes NO difference 2 tailed-test

3. Select Cases4. T-test or Contingency/Chi-Square Analysis5. Interpret Test Results

t-score, significance level, confidence interval,

likelihood ratio (for Chi-Square Analysis)

6. “Reject” or “Not reject” the null hypothesis

Hypothesis Testing Procedure

• Research Question: Are there differences in income between male and female graduates and if so, what factors might explain this difference?

1. Is there a difference in average income between male and female graduates?

2. Is there a significant difference in average length of time on the job, between male and female graduates?

3. Is there a difference in the proportion employed in government jobs between males and females?

Hypothesis Testing

Research Question: Is there a difference in average income

between male and female graduates?

H0: There is NO difference in average income between male and female graduates

Note: Limit the data to full-time employees or self- employed with income more than $20,000

and less than $400,000.

1. T-test: Difference in Means

Step 1: Data/Select Cases

• Select Data/Select Cases

Data/Select Cases

• In a Select Cases dialogue box, you specify logical expressions to select cases.– Select the “If condition is

satisfied” option

– Click on the If… button

Data/Select Cases

Specifying fullself and income range

Type logical expression: fullself = 1 & income > 20000 & income < 400000

to limit cases to alumni who work full-time or are self-employed and make more than $20,000 and less than $400,000.

Data/Select Cases

Data/Select Cases

Step 2: Independent T-Test

Analyze/Compare Means/Independent-Samples T-Test


income

gender(? ?)


Group 1: 1 for FemaleGroup 2: 2 for male

Note: The grouping variable can only have two categories.


gender(1 2)

T-test: Results

Using the Unequal Variance model, we REJECT H0 and conclude that there is a significant difference in average income between male and female graduates.

Group Statistics

128 79868.22 35165.875 3108.254

137 98606.49 47980.995 4099.293

GenderFemale

Male

IncomeN Mean Std. Deviation

Std. ErrorMean

Independent Samples Test

10.443 .001 -3.605 263 .000 -18738.270 5197.537 -28972.4 -8504.190

-3.642 249.145 .000 -18738.270 5144.458 -28870.4 -8606.100

Equal variancesassumed

Equal variancesnot assumed

IncomeF Sig.

Levene's Test forEquality of Variances

t df Sig. (2-tailed)Mean

DifferenceStd. ErrorDifference Lower Upper

95% ConfidenceInterval of the

Difference

t-test for Equality of Means

>-1.96 < 0.05 -18,738 Doesn’t include 0

Possible explanation for the difference in income:

Male income is higher because men have been on the job longer than women.

Research Question:

Is there a difference in average length of time on the job (YEARS) between male and female graduates?

H0: There is NO difference in length of time on the job between male and female graduates

1-2. T-test: Difference in Means




Years at Current Position [years]

gender(1 2)

T-test: Results

Using the Unequal Variance model, we REJECT H0 and conclude that there is a significant difference in average length of time on the job between male and female graduates.

Group Statistics

128 4.15 4.315 .381

137 5.90 5.764 .492

GenderFemale

Male

Years at Current PositionN Mean Std. Deviation

Std. ErrorMean


13.386 .000 -2.786 263 .006 -1.752 .629 -2.991 -.514

-2.813 251.276 .005 -1.752 .623 -2.979 -.525



Years at Current PositionF Sig.





Difference


Does not include 0

>-1.96 < 0.05 -1.752

Possible explanation for the difference in income: Male income is higher because more females work for government than males.

Research Question:

Is there a difference in the proportion employed in government jobs between male and female graduates?

H0: There is NO difference in the proportion employed in government jobs between male and female graduates

2. T-test: Difference in Proportions

• Create a new variable GOV that – has the value 1 if the EMPLOYER (1-6) indicates the

alumnus works for a government organization.

– has the value 0 if the EMPLOYER is not 1-6.

1. Use Transform/Compute to convert the EMPLOYER variable into a new categorical variable GOV.

2. Use Transform/Recode/Into Different Variables to create a new categorical variable GOV.

Step 1: Create a new variable (GOV)

OUTPUT:Analyze/Descriptive Statistics/Frequencies

Employer

8 2.9 2.9 2.9

12 4.3 4.3 7.2

13 4.7 4.7 11.9

45 16.2 16.2 28.2

17 6.1 6.1 34.3

3 1.1 1.1 35.4

5 1.8 1.8 37.2

5 1.8 1.8 39.0

20 7.2 7.2 46.2

11 4.0 4.0 50.2

47 16.9 17.0 67.1

51 18.3 18.4 85.6

4 1.4 1.4 87.0

25 9.0 9.0 96.0

11 4.0 4.0 100.0

277 99.6 100.0

1 .4

278 100.0

Gov: Federal

Gov: State

Gov: County

Gov: City

Gov: Special Agency

Gov: Non U.S.

Private: Single Person

Private: 2-4 Persons



Private: >= 50 Persons

Non-Profit (U.S.)

International Org.

Educational Inst.

Other

Total

Valid

SystemMissing

Total

Frequency Percent Valid PercentCumulative

Percent

7-11 Private

Missing Values

1-6 Government

Transform/Recode/Into Different Variables


Select the income variable, type “GOV”, click the “Change” button, click the “Old and New Values” button…






Save the data file!!

• Analyze/Descriptive Statistics/Frequencies

Step 2: Create a frequency table for GOV

Thirty five percent of the graduates employed full time or self-employed and making more than $20,000 and less than $400,000 work in government jobs.

Government Job

179 64.4 64.6 64.6

98 35.3 35.4 100.0

277 99.6 100.0

1 .4

278 100.0

No

Yes

Total

Valid

SystemMissing

Total


Percent




gov

gender(1 2)

T-test: Results

Using the Unequal Variance model, we CANNOT REJECT H0 and cannot conclude that there is a significant difference between male and female graduates with respect to the proportion working in the government sector.

Group Statistics

127 .3543 .48020 .04261

137 .3650 .48319 .04128

GenderFemale

Male

Government JobN Mean Std. Deviation

Std. ErrorMean


.129 .720 -.179 262 .858 -.01063 .05934 -.12748 .10622

-.179 260.726 .858 -.01063 .05933 -.12746 .10619



Government JobF Sig.





Difference


<-1.96 > 0.05 Includes 0

3. Contingency Table/Chi-Square Analysis

The same question can be analyzed by a contingency table with GOV and GENDER and testing using the Chi-Square statistic.

H0: There is NO relationship between employment sector and gender.

Analyze/Descriptive statistics/Crosstabs


Counts : ObservedPercentages : Row

Column

Select “gov” for “Row” & “Gender” for “column.”

Contingency tableAnalyze/Descriptive statistics/Crosstabs

Government Job * Gender Crosstabulation

82 87 169

48.5% 51.5% 100.0%

64.6% 63.5% 64.0%

45 50 95

47.4% 52.6% 100.0%

35.4% 36.5% 36.0%

127 137 264

48.1% 51.9% 100.0%

100.0% 100.0% 100.0%

Count

% within Government Job

% within Gender

Count


% within Gender

Count


% within Gender

No

Yes

GovernmentJob

Total

Female Male

Gender

Total


Chi-Square value = 0.032 < 3.84 (1.962 = Cutoff value at 95% confidence level at 1 df).We CANNOT REJECT the null hypothesis and cannot concludethere is a statistically significant relationship between gender and whether or not a person works for the government.

> 0.05

Chi-Square Tests

.032b 1 .857

.003 1 .959

.032 1 .857

.898 .480

264

Pearson Chi-Square

Continuity Correctiona

Likelihood Ratio

Fisher's Exact Test

N of Valid Cases

Value dfAsymp. Sig.

(2-sided)Exact Sig.(2-sided)

Exact Sig.(1-sided)

Computed only for a 2x2 tablea.

0 cells (.0%) have expected count less than 5. The minimum expected count is 45.70.

b.

< 3.84 > 0.05

OUTPUT:Analyze/Descriptive Statistics/Frequencies

Missing Values

Employer

8 2.9 2.9 2.9

12 4.3 4.3 7.2

13 4.7 4.7 11.9

45 16.2 16.2 28.2

17 6.1 6.1 34.3

3 1.1 1.1 35.4

5 1.8 1.8 37.2

5 1.8 1.8 39.0

20 7.2 7.2 46.2

11 4.0 4.0 50.2

47 16.9 17.0 67.1

51 18.3 18.4 85.6

4 1.4 1.4 87.0

25 9.0 9.0 96.0

11 4.0 4.0 100.0

277 99.6 100.0

1 .4

278 100.0

Gov: Federal

Gov: State

Gov: County

Gov: City

Gov: Special Agency

Gov: Non U.S.

Private: Single Person




Private: >= 50 Persons

Non-Profit (U.S.)

International Org.

Educational Inst.

Other

Total

Valid

SystemMissing

Total


Percent

7-11. Private

3-2. Contingency Table/Chi-Square Analysis

How about analyzing the difference in the proportion of males and females in the private sector by a contingency table with PRIVATE and GENDER.

H0: There is NO relationship between employment sector and gender.

• Create a new variable PRIVATE that – has the value 1 if the EMPLOYER (7-11) indicates the

alumnus works for a government organization.

– has the value 0 if the EMPLOYER is not 7-11 (else).

Method 2.

Use Transform/Recode/Into Different Variables to create a new categorical variable PRIVATE.

Step1: Create a new variable (PRIVATE)


Counts: ObservedPercentages: Row

Column

Select “private” for “Row” & “Gender” for “column.”


Private Sector Job * Gender Crosstabulation

92 87 179

51.4% 48.6% 100.0%

72.4% 63.5% 67.8%

35 50 85

41.2% 58.8% 100.0%

27.6% 36.5% 32.2%

127 137 264

48.1% 51.9% 100.0%

100.0% 100.0% 100.0%

Count

% within PrivateSector Job

% within Gender

Count


% within Gender

Count


% within Gender

.00

1.00

Private SectorJob

Total

Female Male

Gender

Total


Chi-Square value = 2.411 < 3.84 (1.962).We CANNOT REJECT the null hypothesis and cannot conclude that the difference in the proportion of males and females in the private sector is statistically significant.

Chi-Square Tests

2.411b 1 .120

2.019 1 .155

2.422 1 .120

.147 .077

264

Pearson Chi-Square

Continuity Correctiona

Likelihood Ratio

Fisher's Exact Test

N of Valid Cases

Value dfAsymp. Sig.

(2-sided)Exact Sig.(2-sided)

Exact Sig.(1-sided)

Computed only for a 2x2 tablea.

0 cells (.0%) have expected count less than 5. The minimum expected count is 40.89.

b.

< 3.84 > 0.05

The degrees of freedom in the chi-square test of a contingency table:

d.o.f = (r-1)*(c-1)

where

r & c are the number of rows and columns (or the number of categories of two variables) in a table.

The number of d.o.f is the number of comparisons between actual and expected frequencies minus the number of restrictions imposed on these frequencies.

Since the number of cells in a contingency tables is r*c, there are r*c actual frequencies to be compared with the corresponding expected frequencies. Because the sum (total) of the frequencies in each row and each column are given, there are r+c-1 restrictions.

Therefore, the number of d.o.f is: r*c - (r+c-1) = (r-1)*(c-1).

The degrees of freedom in the chi-square test

• What other factors may influence income?• Control for job sector (government, private, non-profit),

and examine a difference in average income between males and females within each sector.– Select cases: Data/Select Casesif STATUS =1 & INCOME >20000 & INCOME > 400000 & GOV = 1

if STATUS =1 & INCOME >20000 & INCOME > 400000 & PRIVATE = 1

– Compare means/Independent Sample T-test

• If we see differences within each sector, other factors besides job sector are influencing income.

Extensions to the Analysis