Two-Samples T-Test about U

Case 1: Independent Samples & Equal Variances
Case 2: Independent Samples & Unequal Variances
Case 3: Dependent Samples: Matched/Paired Observations 

Case 1: Assumptions, Problem Description & Data, and                Discussion of the Results.
I. Underlying Assumptions
1. The samples (n₁ and n₂) from two normal populations are independent
2. One or both sample sizes are less than 30
3. The appropriate sampling distribution of the test statistic is the t distribution
4. The unknown variances of the two populations are equal

II. Problem Description and Data
Women who are union members earn $2.50 per hour more than women who are not union members. (The Wall Street Journal, July 26, 1994). Suppose independent samples of 15 unionized women and 20 nonunionized women in manufacturing have been selected and the following hourly wage rates are found (Anderson, et al., 1998, p. 394).

Union Workers (n₁= 15):

Nonunion Workers (n₂= 20):

Question: Does there appear to be any difference in the mean wage rate between these                           groups?

Data Entry
Note that HOURLY WAGES of female workers in manufacturing is the variable of interest (X); it is a continuous variable. To enter the values, double-click on var in column one; this action opens the Define Variable window. Type wages in the Variable Name box.  Open the Type window and set Decimal Places to two ( i.e.; accept the default value of 2). Then open the Labels window and type Hourly Wages of Females in the Variable Label box. Click on Continue option and then Okay to return to the data entry screen. Next, define a nominal variable member and label it Union Membership. In the Value box type yes and in the Value Label box type union. Then click on Add; notice that yes = "union" appears meaning that the variable "member" will take on the value union corresponding to the wages of unionized female workers. Similarly, declare another value no = "nonunion" to identify the wages of nonunionized female workers.  The datasheet should basically look like this.

Notes:
(1) The assumption of equal population variance Ó² (= Ó_u² = Ó_non²) means that a pooled sample variance S_p² must be used to compute the standard error of the sample mean difference S^*.
(2) Unlike the one sample t-test, there is no option for setting the test value before executing the procedure using the Command Sequence stated earlier. SPSS/win automatically tests the null that there is no difference in the average hourly wages of the two groups of female workers in manufacturing against the alternative that they differ significantly.
(3) Because the data are quantitative, the variable Type is automatically set to Numeric.
(4) In the lab, select FILE/PRINT or the Printer Icon to send your output to the local printer.

III. Discussion of the Results and Testing Procedure
A. The Outputs/Results 
                                                                T-Test

The first output table contains summary statistics for the two groups. The second table contains all the statistics that are needed to perform the test. Notice that SPSS/win actually performs the test under two alternative assumptions about the population variance -- equal and unequal variance. Because equal variance is assumed in this example we will use the results in that row for the ensuing discussion. 

B. Testing Procedure
I will summarize the testing procedure under Hypotheses, Decision Rules, and Conclusions, with the understanding that the six logical steps presented earlier in the one sample t-test are carried out mentally in the process. The goal here is to show that there is no difference between the mean wage of union workers µ_u and of nonunion workers µ_non who are women -- where the subscript 'u' denotes Union and 'non' denotes Nonunion.

Hypotheses
Ho: µ_u = µ_non or Ho: µ_u - µ_non = 0 (i.e., there is no significant difference in the average hourly                                                                   wage rates of female workers in the two groups)
H_a: µ_u is not equal µ_non or Ho: µ_u - µ_non is not equal to zero (i.e., there is a significant                                                                   difference in their average hourly wage rates)
Note that this is a t-test because the distribution of hourly wages in the populations of unionized and nonunionized women is assumed to be normal and also both n₁ and n₂ are small (i.e., < 30). Thus, the test statistic X_bar_u - X_bar_non can only be transformed into t-score using the equation t_ov = [(X_bar_u - X_bar_non) - (µ_non - µ_non)]/S^* where S^*denotes the standard error of (X_bar_u - X_bar_non).   Because n₁ = 15, n₂ = 20 it follows that v = n₁ + n₂ - 2 = 33, and given alpha = .05, the critical t-value is t_cv= t_.025,33 = t_.025,30= ±2.042 (conservative value by using v = 30 instead of a value larger than 33 -- which is not in the table). 

Decision Rules: Reject Ho if |t_ov | > | t_cv |; Retain Ho if  |t_ov | < | t_cv |

From the "Group Statistics" table the pooled sample variance can be derived as
S_p² = [(n₁ -1) S_u² + (n₂-1)S_non²] /(n₁ + n₂ + -2) = [14(2.24)² + 19(1.99)²]/(15 + 20 - 2) = 4.41. Although SPSS/win does not report this result, it computes and uses it internally to determine the the value of S^* = [S_p²(1/n₁ + 1/n₂)]^½.  From the "Independent Samples test" table (along the Equal Variances Assumed row) we can obtain the following summary statistics: X_bar_u - X_bar_non = 2.1767 is the sample mean difference, its standard error is S^* = .7169, and t_ov = [(X_bar_u - X_bar_non) - 0]/S^* = 3. 036.

Conclusions: Since t_ov = 3.036 vis-a-vis t_cv = 2.042 is in the rejection region (right tail), the sample data do not provide sufficient evidence to support H_o. Thus, we must reject H_o in favor of H_a. Hence, union membership does make a significant difference in the average hourly wages of women working in manufacturing. On average, unionized women make $2.18 more per hour than nonunionized women because the observed mean wage of  X_bar_u = $17.54 per hour for unionized women vis-a-vis the sample mean wage of X_bar_non =  $15.36 for the nonunionized women is not a chance outcome; the difference is indeed real. 

Extension: A One-Tailed Test: Suppose the goal of the test was to validate the claim that average hourly wages of unionized women in manufacturing are generally higher than those of their nonunionized counterparts.  In this case, both the H_o and  H_a would be stated as follows:
H_o: µ_u is less than or equal µ_non or H_o: µ_u - µ_non is at the most zero                        
H_a: µ_u > µ_non or H_a: µ_u - µ_non > 0
Because this is a one-tailed test (to the right of the sampling and/or 't' distribution), alpha = .05 must not be divided by 2 when determining the value of the t_cv; indeed t_cv = t_{.05, 30} = 1.697. However, the value of t_ov = 3.036 is still valid for the test.  Since t_ov = 3.036 vis-a-vis t_cv = 1.697 is in the rejection region (right tail), the sample data do not provide sufficient evidence to support H_o.  Hence, we will have to conclude that unionized women working in manufacturing make more in hourly wages, on average, than their nonunuionized counterparts.

Case 2: Assumptions, Problem Description & Data, and                Discussion of the Results.
I. Underlying Assumptions
1. The samples (n₁ and n₂) from two normal populations are independent
2. One or both sample sizes are less than 30
3. The appropriate sampling distribution of the test statistic is the t distribution
4. The unknown variances of the two populations are not equal

II. Problem Description and Data
Starting annual salary for individuals entering the public accounting and financial planning professions were presented in Fortune, June 26, 1995.  The starting salaries for a sample of 12 public accountants and 14 financial planners are below. Data are in thousands of dollars (ASW, 1998, p. 403).

Public Accountant (n₁ = 12)

Financial Planner (n₂ = 14)

Question:                        
Using alpha = .05, test for any difference between the population mean starting annual salaries for the two professions. What is your conclusion?

Data Entry
Note that the starting annual SALARY for persons entering public accounting and financial planning professions is the variable of interest (X); it is a continuous variable. To enter the values, double-click on var in column one; this action opens the Define Variable window. Type salary in the Variable Name box.  Open the Type window and set Decimal Places to one ( since there is only one trailing decimal point for all the data). Then open the Labels window and type Starting Salary of Public Accountants & Financial Planners in the Variable Label box. Click on Continue option and then Okay to return to the data entry screen. Next, define a nominal variable person and label it Profession of an Individual. In the Value box type pa and in the Value Label box type public accountant. Then click on Add; notice that pa = "public accountant" appears meaning that the variable "person" will take on the value pa corresponding to the salary of those individuals in public accounting profession. Similarly, declare another value fp = "financial planner" to identify the salary of those individuals in financial planning profession. The datasheet should basically look like this.

Notes:
(1) The assumption of unequal population variance (i.e., Ó_pa² is not equal to Ó_fp²) means that both sample variances -- S_pa²  and S_fp² -- must be used to compute standard error of the sample mean difference S^*.
(2) Unlike the one sample t-test, there is no option for setting the test value before executing the procedure using the Command Sequence stated earlier. SPSS/win automatically tests the null that there is no difference in the average salary of public accountants and financial planners against the alternative of a significant difference.
(3) Because the data are quantitative, the variable Type is automatically set to Numeric.
(4) In the lab, select FILE/PRINT or the Printer Icon to send your output to the local printer.


III. Discussion of the Results and Testing Procedure
A. The Outputs/Results

T-Test

The first output table contains summary statistics for the two groups. The second table contains all the statistics that are needed to perform the test. As in the Case 1 above, SPSS/win actually performs the test under two alternative assumptions about the population variance -- equal and unequal variance.  Because unequal variance is assumed in this example we will use the results in that row for the ensuing discussion. 

B. Testing Procedure
Again, I will summarize the testing procedure under Hypotheses, Decision Rules, and Conclusions, with the understanding that the six logical steps presented earlier in the one sample t-test are carried out mentally in the process. The goal here is to determine whether the average annual starting salary in the public accounting profession µ_pa differs significantly from the average annual starting salary in the financial planning profession µ_fp -- where the subscript 'pa' denotes Public Accounting and 'fp' denotes Financial Planning professions, respectively.

Hypotheses
Ho: µ_pa = µ_fp or Ho: µ_pa - µ_fp = 0 (i.e., there is no significant difference in the average annual                                                                   starting salary in the two professions)
H_a: µ_pa is not equal µ_fp or Ho: µ_pa - µ_fp is not equal to zero (i.e., a significant difference does                                                                    exist)
Note that this is a t-test because the distribution of salaries in the populations of public accounting and financial planning professions is assumed to be normal and also both n₁ and n₂ are small (i.e., < 30).   Thus, the test statistic X_bar_pa - X_bar_fp can only be transformed into t-score using the equation t_ov = [(X_bar_pa - X_bar_fp) - (µ_pa - µ_fp)]/S^* where S^*denotes the standard error of (X_bar_u - X_bar_non).   Because Ó_pa² is not equal to Ó_fp², it follows that v = n₁ + n₂ - 2 = 12 + 14 - 2 = 24 is not applicable for determining the critical t-value t_cv.   Indeed, the rule for determining the value of v yields a value that is smaller than 24.  This rule, which can be found in almost any introductory statistics text, is built into SPSS/win and yields the value v = 20.848 or 21. Thus, given alpha = .05, the t_cv= t_.025,21 = ±2.080. 

Decision Rules: Reject Ho if |t_ov | > | t_cv |; Retain Ho if  |t_ov | < | t_cv |

From the "Group Statistics" table the sample variance of the public accountants' salaries is given as S_pa² = 3.35, and that of the financial planners' salaries is S_fp² = 2.64. These values are used to determine the value of S^* = [S_pa²/n₁ + S_fp²/n₂)]^½.  From the "Independent Samples test" table (along the Equal Variances Not Assumed row) we can obtain the following summary statistics: X_bar_pa - X_bar_fp = 3.517 or  $3517 is the sample mean difference, its standard error is S^* =1.197, and t_ov = [(X_bar_pa - X_bar_fp) - 0]/S^* = 2.939.

Conclusions: Since t_ov = 2.939 vis-a-vis t_cv = 2.080 is in the rejection region (right tail), the sample data do not provide sufficient evidence to support H_o.  Instead, evidence appears to support the existence of a significant difference in the average salary of the two groups of professionals. Thus, it can be concluded that the average annual staring salary in public accounting profession is generally higher by $3517 (point estimate of the difference between µ_pa and µ_fp) than the average annual starting salary in financial planning profession.

Extension:Interval Estimation: The value $3517 is a point estimate of the difference between µ_pa and µ_fp. From the output, we can also obtain a 95% confidence interval estimate of the difference between the two population means.  Note that the rule for doing so is given as (X_bar_pa - X_bar_fp) ± t_cv . S^*.  The lower limit value is reported as 1.027  and the upper limit value is reported as 6.006.  Thus, we can conclude with a 95% confidence that the true mean difference in the annual starting salary in the two professions is between or $1027 and $6006.

Case 3: Assumptions, Problem Description & Data, and                Discussion of the Results.
I. Underlying Assumptions
This case considers a research situation in which two samples are not independent. This situation occurs when each individual observation (i) within a sample is related (matched or paired) to an individual observation in the second sample. The relatedness may be the result of the individual observations in the two samples
1. representing before and after results (which is presented in this example),
2. having matching characteristics,
3. being matched by location, or
4. being matched by time.

If there are definite reasons for pairing (or matching) the individual observations in the two samples, the two samples are dependent rather than being independent. Generally, the precision from an analysis of dependent samples is greater than that from the analysis of independent samples. Thus, if paired analysis is appropriate, it is the preferred approach. 

II. Problem Description and Data
Figure Perfect, Inc., is a women's figure salon that specializes in weight reduction programs. Weights for a sample of clients before and after a six-week introductory program are shown below (ASW, p. 419).

Question: Using alpha = .05, test to determine whether the introductory program provides a 
                         statistically significant weight loss. What is your conclusion? 

Data Entry
Note that the Weight of the clients is actually the variable of interest (X); it is a continuous variable. However, to determine whether there is a significant reduction in the weight of the participants after joining the program we will have to declare two pseudo variables, viz., Xbefore, and Xafter for each client.  To enter the before weight values, double-click on var in column one; this action opens the Define Variable window. Type Xbefore in the Variable Name box.  Next, open the Type window and set Decimal Places to zero ( i.e.; type 0 to replace the default value of 2). Finally, open the Labels window and type Client's Weight Before in the Variable Label box. Click on Continue option and then Okay to return to the data entry screen. Repeat similar steps to define the Xafter weight values; use Client's Weight After as the variable label. The datasheet should basically look like this.

Notes:
(1) The key to the analysis of the matched/paired design sample design is to realize that we consider only the column of differences d_i where d_i = X_after,i - X_before,i for each client. If defined as d_i = X_before,i - X_after,i, the reported sample mean of the difference will not be preceded by a negative sign. 
(2) Unlike the one sample t-test, there is no option for setting the test value before executing the procedure using the Command Sequence stated earlier. SPSS/win automatically tests the null that the mean of the difference µ_d = µ_afer - µ_before in the average difference in the weight of the population of  program participants is zero, equal to or less than zero, or greater than or equal to zero (i.e., H_o: µ_d = 0, = or < 0, = or > 0) against H_a that it differs significantly from zero, significantly greater than 0, or significantly less than 0 (i.e., H_o: µ_d = 0, or > 0, or < 0).
(3) Because the data are quantitative, the variable Type is automatically set to Numeric.
(4) In the lab, select FILE/PRINT or the Printer Icon to send your output to the local printer.

III. Discussion of the Results and Testing Procedure
A. The Outputs/Results

T-Test

The first output table contains summary statistics for the paired samples. The second table reports the Pearson correlation coefficient for weight before and after the program.  The value of .979 suggests the existence of a strong positive association between the clients weight before and after the program; the p-value of .001 indicates that the observed association is statistically significant at 5% level. The third table contains all the statistics that are needed to perform the test.

B. Testing Procedure
Again, I will summarize the testing procedure under Hypotheses, Decision Rules, and Conclusions, with the understanding that the six logical steps presented earlier in the one sample t-test are carried out mentally in the process.

Hypotheses
Ho: µ_d = 0 or > 0 (i.e., the average weight after is the same as the average weight before, if anything,
                             the weight after is greater than the weight before)
H_a : µ_d < 0 (i.e., the average weight after is strictly less than the average weight before the introduction 
                            of the program)
Note that this is a t-test because the distribution of weight in the population of all possible clients is assumed to be normal; also the sample size 'n' is small (i.e., n < 30). Thus, the test statistic d-bar can only be transformed into t-score using the equation t_ov = (d-bar - µ_d)/S_d-bar where S_d-bar denotes the standard error of d-bar. Because n = 6 it follows that v = n -1 = 5, and given alpha = .05, the critical t-value is t_cv= t_.05,5 = ±2.015. 

Decision Rules: Reject Ho if |t_ov | > | t_cv |; Retain Ho if  |t_ov | < | t_cv |

From the "Paired Samples test" table we can obtain the following summary statistics: the mean weight loss after introduction of the program is d-bar = -6.17 (negative sign because the difference d_i is defined as d_i = X_after,i - X_before,i), the sample standard deviation of the difference d_i is S_d = 6.59; the standard error of d-bar is  S_d-bar = S_d/n^½ = 2.69, and t_ov = (d-bar - µ_d)/S_d-bar = -2.294.

Conclusions: Since, in absolute terms, t_ov = -2.294 is greater than t_cv= -2.015 (it is in the rejection region of the left tail), the sample of data does not provide sufficient evidence to retain H_o.  Hence, it can be concluded that the new program does provide weight loss.

Extension: Interval Estimation: From the output, we can also obtain a 95% confidence interval estimate of the difference between the two population (technically) means (µ_after - µ_before) where µ_after and µ_before are the average weight of all clients after and before joining the programs, respectively. Note that the rule for doing so is given as d-bar ± t_cv . S_d-bar. The lower limit is reported as -13.08 and the upper limit is reported as .74. Their interpretation is as follows.  Letting µ_after - µ_before = -13.08 implies that µ_after = µ_before - 13.08 which means that an individual can lose as much as 13.08Ibs,on average, after joining the program.  Similarly, µ_after - µ_before = .74 implies that µ_after =  µ_before + .74 which means that at the very worst an individual may not gain more than .74Ibs, on average, after joining the program.     


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Last revised: Sunday, August 05, 2001.