2 Tests of Significance Throughout this assignment you will review mock studies You will needs to follow the directions outlined in the section using SPSS and decide whether there is significance between the variables You will need to list the five steps of hypothesis testing (as covered in the lesson for Week 6) to see how every question should be formatted You will complete all of the problems Be sure to cut and past the appropriate test result boxes from SPSS under each problem and explain what you will do with your research hypotheses All calculations should be coming from your SPSS You will need to submit the SPSS output file to get credit for this assignment This file will save as a spv file and will need to be in a single file In other words, you are not allowed to submit more than one output file for this assignment The five steps of hypothesis testing when using SPSS are as follows 1 State your research hypothesis (H1) and null hypothesis (H0) 2 Identify your significance level ( 05 or 01) 3 Conduct your analysis using SPSS 4 Look for the valid score for comparison This score is usually under 'Sig 2 tail' or 'Sig 2' We will call this p 5 Compare the two and apply the following rule a If p is or significance level, than you reject the null Be sure to explain to the reader what this means in regards to your study (Ex will you recommend counseling services ) Be sure that your answers are clearly distinguishable Perhaps you bold your font or use a different color This assignment is due no later than Sunday of Week 6 by 11 55 pm ET Save the file in the following format your last name SOCI332 A2 The file must be a word file t Tests t Test for a Single Sample (20 points) Open SPSS Enter the number of activities of daily living performed by the depressed clients studied in 1 in the Data View window In the Variable View window, change the variable name to ADL and set the decimals to zero Click Analyze Compare Means One Sample T test the arrow to move ADL to the Variable(s) window Enter the population mean (17) in the Test Value box Click OK 1 Researches are interested in whether depressed people undergoing group therapy will perform a different number of activities of daily living after group therapy The researchers have randomly selected 12 depressed clients to undergo a 6 week group therapy program Use the five steps of hypothesis testing to determine whether the average number of activities of daily living (shown below) obtained after therapy is significantly different from a mean number of activities of 17 that is typical for depressed people (Clearly indicate each step) Test the difference at the 05 level of significance and at the 01 level (in SPSS this means you change the confidence level from 95 to 99 ) As part of Step 5, indicate whether the behavioral scientists should recommend group therapy for all depressed people based on evaluation of the null hypothesis at both levels of significance CLIENT A B C D E F G H I J K L AFTER THERAPY 18 14 11 25 24 17 14 10 23 11 22 19 t Test for Dependent Means (20 points) Open SPSS Enter the number of activities of daily living performed by the depressed clients studied in Problem 2 in the Data View window Be sure to enter the before therapy scores in the first column and the after therapy scores in the second column In the Variable View window, change the variable name for the first variable to ADLPRE and the variable name for the second variable to ADLPOST Set the decimals for both variables to zero Click Analyze Compare Means Paired Samples T Test the arrow to move ADLPRE to the Paired Variable(s) window ADLPOST and then click the arrow to move the variable to the Paired Variable(s) window Click OK 2 Researchers are interested in whether depressed people undergoing group therapy will perform a different number of activities of daily living before and after group therapy The researchers have randomly selected 8 depressed clients in a 6 week group therapy program Use the five steps of hypothesis testing to determine whether the observed differences in numbers of activities of daily living (shown below) obtained before and after therapy are statistically significant at the 05 level of (Clearly indicate each step) As part of Step 5, indicate whether the researchers should recommend group therapy for all depressed people based on evaluation of the null hypothesis at the 05 level of significance and calculate the measure of association CLIENT A B C D E F G H BEFORE THERAPY 11 7 10 13 9 8 13 12 AFTER THERAPY 17 12 12 21 16 17 17 8 The t Test for Independent Samples (20 points) Once you have entered the data, click on Analyze, then on Compare Means, and then click on Independent Samples T Test A dialog box will appear, with your variables (student, condition, score) on the left Your options are (a) move one or more variables into the Test Variable(s) box to select your dependent variables(s) and (b) move one of your variables into the Grouping Variable box to select the independent variables (or identify the groups to be compared) Make the dependent variable by moving it to the Test Variable(s) box Then make your independent variable by moving it to the Grouping Variable box Now, the Define Groups button is functioning, click on Define Groups and another dialog box appears Here you must specify the two values of the condition variable that represent the two groups you are comparing Click in the box next to Group 1 and type the number 1, then click in the box next to Group 2 and type the number 2 Now you can click Continue to return to the IndependentSamples T Test dialog box, and click on OK to run the analysis 3 Six months after an industrial accident, a researcher has been asked to compare the job satisfaction of employees who participated in counseling sessions with the satisfaction of employees who chose not to participate The scores on a job satisfaction inventory for both groups are listed in the table below Use the five steps of hypothesis testing to determine whether the job satisfaction scores of the group that participated in counseling are statistically higher than the scores of employees who did not participate in counseling at the 01 level of significance As part of Step 5, indicate whether the researcher should recommend counseling as a method to improve job satisfaction following industrial accidents based on evaluation of the null hypothesis and calculate the measure of association NOTE Do not forget to give a numeric value for those who participated in counseling (e g 0) and those who did not participate in counseling (e g 1) PARTICIPATED IN COUNSELING 36 39 41 36 37 35 37 39 42 DID NOT PARTICIPATE IN COUNSELING 38 36 36 32 30 39 41 35 33 ANOVA (20 points) Open SPSS Analyze the data for 1 Remember that SPSS assumes that all the scores in a row are from the same participant In this study, there are 15 participants divided into three groups of five Therefore, each of the 15 participants will be described by two variables, type of therapy and the number of activities of daily living performed If 1 represents the group receiving individual therapy for 1 hour every 2 weeks, 2 represents the group receiving 1 hour of individual therapy each week, and 3 indicates the group receiving 2 hours of individual therapy each week, the first participant will be described by entering 1 in the top cell of the first column in the Data View window and 16 in the top cell of the second column to indicate that the participant underwent 1 hour of therapy every 2 weeks and performed 16 activities of daily living The second participant will be described by 1 and 15 Chi Square Chi Square Test for Goodness of Fit Open SPSS Remember that SPSS assumes that all the scores in a row are from the same participant In the study presented in 1, there are 20 students, some of whom have been suspended for misbehavior The primary conflictresolution style used by each student is also entered Ignore the first variable in this analysis When you have entered the data for all 20 students, move to the Variable View window and change the first variable name to SUSPEND and the second to STYLE Set the number of decimals for both variables to zero Click Analyze Non Parametric Tests Chi Square Click the variable STYLE and then the arrow next to the box labeled Test Variable List to indicate that the chi square for goodness of fit should be conducted on the conflict resolution style variable Note that All categories equal is the default selection in the Expected Values box, which means that SPSS will conduct the goodness of fit test using equal expected frequencies for each of the four styles, in other words, SPSS will assume that the proportions of students each style are equal Click OK Chi Square Test for Independence Open SPSS For 2, you need to add the variable SUSPEND to the analysis Remember that in this problem, we are interested in whether there was an association between conflict resolution style and having been suspended from school for misbehavior Since the analysis will involve two nominal variables, the appropriate test is a chi square test for independence Click Analyze Descriptive Statistics Crosstabs Since SUSPEND is already selected, click the arrow next to the box labeled Rows Click the variable STYLE and click the arrow next to the box labeled Columns Click Statistics and click the box labeled Chi Square Click Continue Click Cells and click the box labeled Expected Click Continue Click OK 1 The following table includes resolution used by 20 students the primary method of conflict Method N of Students Aggressive 8 Manipulative 2 Passive 2 Assertive 8 a Following the five steps of hypothesis testing, conduct the appropriate chi square test to determine whether the observed frequencies are significantly different from the frequencies expected by change at the 05 level of significance Clearly identify each of the five steps The null and alternative hypotheses for this test are, H 0 The observed frequencies are identical with the frequencies expected by change H A The observed frequencies are significantly different from the frequencies expected by change Levelof significance, 0 05 The outputs of Chi squaretest for goodness of fit using SPSS are shown below, STYLE Observed N Expected N Residual Aggressive 8 5 0 3 0 Manipulative 2 5 0 3 0 Passive 2 5 0 3 0 Assertive 8 5 0 3 0 Total 20 Test Statistics STYLE Chi Square 7 200a df Asymp Sig 3 066 a 0 cells ( 0 ) have expected frequencies less than 5 The minimum expected cell frequency is 5 0 Test Statistic, 2 7 200 Degrees of freedom, 3 p value for a two tailed test is 0 066 Since p value is greater than 0 05, we fail to reject our null hypothesis at 5 level Therefore, at 5 level data doesn't provide sufficient evidence to support the claim that the observed frequencies are significantly different from the frequencies expected by change b Explain your response to some who has never had a course in statistics We found out that the data of 20 students who suspended from school due to various reasons of misbehavior To check whether all the four reasons are equally likely to happen or not, conducted a Chi square test for goodness of fit that is appropriate here The test results provide the conclusions that all the four reasons are equally likely to happen 2 Next, researchers categorized the students based on the primary method of conflict resolution used and whether the student had been suspended from school for misbehavior These data are presented below Suspende d Yes No Total Aggressiv e 7 1 8 Method Manipulativ Passive e 1 1 1 1 2 2 Assertive Total 1 7 8 10 10 20 a Following the five steps of hypothesis testing, conduct the appropriate chi square test to determine whether the observed frequencies are significantly different from the frequencies expected by change at the 05 level of significance Clearly identify each of the five steps The null and alternative hypotheses for this test are, H 0 The observed frequencies are identical with the frequencies expected by change H A The observed frequencies are significantly different from the frequencies expected by change Levelof significance, 0 05 The outputs of Chi square test for independence of attributesusing SPSS are shown below, SUSPEND STYLE Crosstabulation STYLE Aggressive SUSPEND Yes Count Expected Count No Count Expected Count Total Count Expected Count Manipulative Passive Assertive 7 1 1 1 10 4 0 1 0 1 0 4 0 10 0 1 1 1 7 10 4 0 1 0 1 0 4 0 10 0 8 2 2 8 20 8 0 2 0 2 0 8 0 20 0 Chi Square Tests Asymp Sig (2Value df sided) Pearson Chi Square 9 000a 3 029 Likelihood Ratio 10 124 3 018 Linear by Linear 8 319 1 004 Association N of Valid Cases 20 a 8 cells (100 0 ) have expected count less than 5 The minimum expected count is 1 00 Test Statistic, 2 9 000 Total Degrees of freedom, 3 p value for a two tailed test is 0 029 Since p value is less than 0 05, we have to reject our null hypothesis at 5 level Therefore, at 5 level data provides sufficient evidence to support the claim that the observed frequencies are significantly different from the frequencies expected by change b Calculate the effect size Effect size can be measured through either Phi or Cramer's V The output of Phi or Cramer's Vusing SPSS are shown below, Symmetric Measures Value Nominal by Nominal N of Valid Cases 0 671 Approx Sig Phi 671 029 Cramer's V 671 029 20 Cramer's V 0 671 The effect size is 0 671 and it is significantly high It indicates we have strong evidence from data to conclude that the observed frequencies are significantly different from the frequencies expected by change c Explain your response to someone who has never had a course in statistics We have studied the data of 20 students who are classified by two attributes the type of misbehavior committed and whether suspended from school or not To check whether the two attributes under study are independent or not, conducted a Chi square test for independence of attributes that is appropriate here The test results provide the conclusions that the two attributes under study are not independent Suspension from School Yes No Total Conflict Resolution Assertive Other 1 6 7 Total 9 4 13 10 10 20 A Following the five steps of hypothesis testing, conduct the appropriate chi square test to determine whether the observed frequencies are significantly different from the frequencies expected by change at the 05 level of significance Clearly identify each of the five steps The null and alternative hypotheses for this test are, H 0 The observed frequencies are identical with the frequencies expected by change H A The observed frequencies are significantly different from the frequencies expected by change Levelof significance, 0 05 The outputs of Chi square test for independence of attributesusing SPSS are shown below, SUSPENSION CONFLICT Crosstabulation CONFLICT Assertive SUSPENSION Yes Count Expected Count No Count Expected Count Total Count Expected Count Other Total 1 9 10 3 5 6 5 10 0 6 4 10 3 5 6 5 10 0 7 13 20 7 0 13 0 20 0 Chi Square Tests Value df Asymp Sig (2 Exact Sig (2 Exact Sig (1 sided) sided) sided) 5 495a 1 019 Continuity Correctionb 3 516 1 061 Likelihood Ratio 5 936 1 015 Pearson Chi Square Fisher's Exact Test 057 Linear by Linear 5 220 1 029 022 Association N of Valid Cases 20 a 2 cells (50 0 ) have expected count less than 5 The minimum expected count is 3 50 b Computed only for a 2x2 table Test Statistic, 2 5 495 Degrees of freedom, 1 p value for a two tailed test is 0 019 Since p value is lessthan 0 05, we have to reject our null hypothesis at 5 level Therefore, at 5 level data provides sufficient evidence to support the claim that the observed frequencies are significantly different from the frequencies expected by change a Calculate the effect size Effect size can be measured through either Phi or Cramer's V The output of Phi or Cramer's Vusing SPSS are shown below, Symmetric Measures Value Nominal by Nominal Phi Cramer's V N of Valid Cases 0 524 Approx Sig 524 019 524 019 20 Cramer's V 0 524 The effect size is 0 524 and it is significantly high It indicates we have strong evidence from data to conclude that the observed frequencies are significantly different from the frequencies expected by change b Explain your results We have studied the data of 20 students who are classified by two attributes the type of misbehavior committed and whether suspended from school or not To check whether the two attributes under study are independent or not, conducted a Chi square test for independence of attributes that is appropriate here The test results provide the conclusions that the two attributes under study are not independent The deadline for this assignment is 11 59 PM EST on Sunday of Week 8 SPSS ASSIGNMENT 5 Single Sample Dependent Samples t Tests SPSS instructions (For more details, check the links provided under Course Materials in the Course Overview Folder (under Lessons) t Test for a Single Sample Open SPSS Enter the number of activities of daily living performed by the depressed clients studied in 1 in the Data View window In the Variable View window, change the variable name to ADL and set the decimals to zero Click Analyze Compare Means One Sample T test the arrow to move ADL to the Variable(s) window Enter the population mean (14) in the Test Value box Click OK t Test for Dependent Means Open SPSS Enter the number of activities of daily living performed by the depressed clients studied in Problem 2 in the Data View window Be sure to enter the before therapy scores in the first column and the after therapy scores in the second column In the Variable View window, change the variable name for the first variable to ADLPRE and the variable name for the second variable to ADLPOST Set the decimals for both variables to zero Click Analyze Compare Means Paired Samples T Test the arrow to move ADLPRE to the Paired Variable(s) window ADLPOST and then click the arrow to move the variable to the Paired Variable(s) window Click OK Review the five steps of hypothesis testing and complete the following problems Be sure to cut and past the appropriate result boxes from SPSS under each problem 1 Researchers are interested in whether depressed people undergoing group therapy will perform a different number of activities of daily living after group therapy The researchers have randomly selected 12 depressed clients to undergo a 6 week group therapy program Use the five steps of hypothesis testing to determine whether the average number of activities of daily living (shown below) obtained after therapy is significantly different from a mean number of activities of 14 that is typical for depressed people (Clearly indicate each step) Test the difference at the 05 level of significance and, for practice, at the 01 level (in SPSS this means you change the confidence level from 95 to 99 ) In Step 2, show all calculations As part of Step 5, indicate whether the behavioral scientists should recommend group therapy for all depressed people based on evaluation of the null hypothesis at both levels of significance and calculate the effect size CLIENT A B C D E F G H I J K L AFTER THERAPY 17 15 12 21 16 18 17 14 13 15 12 19 Step 1 Restate the question as a research hypothesis and a null hypothesis about the populations Population 1 Depressed people who undergo therapy Population2 Depressed people in general (those that do no undergo therapy) Suppose denotes the mean number of activities of daily living obtained after therapy for depressed people (Ho 14) (Ha 14) The null and alternative hypotheses are N The populations of depressed people, who go through group therapy, perform the same number of activities of daily living, than the population of depressed people who do not get the therapy A The populations of depressed people, who go through group therapy, perform a greater number of activities of daily living, than the population of depressed people who do not get the therapy Step 2 Determine the characteristics of the comparison distribution For this example population mean 14 (number of activities typical for a depressed person) 12,12,13,14,15,15,16,17,17,18,19,21 N 12 Mean N 12 12 13 14 15 15 16 17 17 18 19 21 189 189 12 15 75 Mean 15 75 Have to determine the variance of the population of individuals Use formula for estimated population variance S (X M) df (X M) (17 15 75) (15 15 75) (12 15 75) (21 15 75) (16 15 75) (18 15 75) (17 15 75) (14 15 75) (13 15 75) (15 15 75) (12 15 75) (19 15 75) (1 25) ( 0 75) ( 3 75) (5 25) (0 25) (2 25) (1 25) ( 1 75) ( 2 75) (0 75) ( 3 75) (3 25) (1 5625) (0 56) (14 0625) (27 5625) (0 0625) (5 0625) (1 5625) (3 06) (7 56) (0 56) (14 0625) (10 5625) 86 24 df N 1 12 1 11 df 11 S (X M) df S 86 24 11 7 84 Estimated population variance is 7 84 (standard deviation 7 84 2 80) Variance of the distribution of means SM S N S 7 84 N 12 S M 7 84 12 S M 0 6533 Now we can calculate standard deviation of the distribution of means using formula SM S M SM 0 93167 0 808 The shape is a t distribution with 11 degrees of freedom Step3 Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected Here the level of significance is assumed to be 0 05 and 0 01 The null hypothesis should be rejected at cutoff score 2 201 or 2 201 at 5 and 3 106 or 3 106 at 1 Step 4 Determination your sample's score on the comparison distribution Used formula The test statistic is t (M PopM) SM (15 75 14) 0 808 2 165 Step 5 Decide whether to reject the null hypothesis For the example, the cutoff t score is 2 201 and the actual t score was 2 165, since the t value (2 165) is less than the level of significance 05, we fail to reject the null hypothesis Ho So we don't have sufficient evidence to conclude that the average number of activities of daily living obtained after therapy is significantly different from a mean number of activities of 14 that is typical for depressed people Since the null hypothesis is not rejected at 05 level of significance, the null hypothesis is also not rejected at the 01 level of significance The effect size is Effect size (Population 1M Population 2M) Population SD (15 75 14) 2 80 0 625 a medium effect SPSS Output for 1 One Sample T Test at 95 One Sample Statistics ADL N 12 Mean 15 75 Std Error Std Deviation Mean 2 800 808 One Sample Test Test Value 14 ADL t 2 165 df 11 Mean Sig (2 tailed) Difference 053 1 750 95 Confidence Interval of the Difference Lower Upper 03 3 53 SPSS Output for the One Sample T Test at 99 One Sample Statistics ADL N 12 Mean 15 75 Std Error Std Deviation Mean 2 800 808 One Sample Test Test Value 14 ADL 2 t 2 165 df 11 Mean Sig (2 tailed) Difference 053 1 750 99 Confidence Interval of the Difference Lower Upper 76 4 26 Researchers are interested in whether depressed people undergoing group therapy will perform a different number of activities of daily living before and after group therapy The researchers have randomly selected 8 depressed clients in a 6 week group therapy program Use the five steps of hypothesis testing to determine whether the observed differences in numbers of activities of daily living (shown below) obtained before and after therapy are statistically significant at the 05 level of significance and, for practice, at the 01 level (Clearly indicate each step) In Step 2, show all calculations As part of Step 5, indicate whether the researchers should recommend group therapy for all depressed people based on evaluation of the null hypothesis at both levels of significance and calculate the effect size CLIENT A B C D E F G H BEFORE THERAPY 12 7 10 13 9 8 14 11 AFTER THERAPY 17 15 12 21 16 18 17 8 Step 1 Restate the question as a research hypothesis and a null hypothesis about the populations Population 1 Depressed people who receive group therapy Population2 Depressed people whose daily activity changes from before and after group therapy Suppose D denote the mean observed differences in numbers of activities of daily living obtained before and after therapy Ho D 0 (The mean difference is zero) Ha D 0 (There is a mean difference) The null and alternative hypotheses are N Depressed people who receive group therapy, do not change their daily activities from before or after therapy A Population 1's mean difference score is different from Population 2's mean difference score of zero Step 2 Determine the characteristics of the comparison distribution BEFORE THERAPY AFTER THERAPY 12 7 10 13 9 8 14 11 17 15 12 21 16 18 17 8 Difference (AfterBefore) 5 8 2 8 7 10 3 3 N 8 Mean N Pre 12,7,10,13,9,8,14,11 84 Post 17,15,12,21,16,18,17,8 124 Diff 5, 8, 2, 8, 7, 10, 3,3 40 Pre Mean 84 8 10 5 Mean 10 5 Post Mean 124 8 15 5 Mean 15 5 Diff Mean 40 8 5 Mean 5 Assume a mean of the distribution of the means of difference scores of 0 Have to determine the variance of the population of difference scores using formula for estimated population variance S (X M) df PRE (X M) (12 10 5) (7 10 5) (10 10 5) (13 10 5) (9 10 5) (8 10 5) (14 10 5) (11 10 5) (1 5) ( 3 5) ( 3 75) ( 0 5) (2 5) ( 1 5) ( 2 5) (3 5) ( 0 5) (2 25) (12 25) (0 25) (6 25) (2 25) (6 25) (12 25) (0 25) 42 POST (X M) (17 15 5) (15 15 5) (12 15 5) (21 15 5) (16 15 5) (18 15 5) (1715 5) (8 15 5) (1 5) ( 0 5) ( 3 5) (5 5) (0 5) (2 5) (1 5) ( 7 5) (2 25) (0 25) (12 25) (30 25) (0 25) (6 25) (2 25) (56 25) 110 Diff (X M) ( 5 5) ( 8 5) ( 2 5) ( 8 5) ( 7 5) ( 10 5) ( 3 5) (3 5) (0) ( 3) (3) ( 3) ( 2) ( 5) (2) (8) (0) (9) (9) (9) (4) (25) (4) (64) 124 df N 1 8 1 7 df 7 S (X M) df PRE S 42 7 6 Estimated PRE population variance is 6 (standard deviation 6 2 449) POST S 110 7 15 714 Estimated POST population variance is 15 714 (standard deviation 15 714 3 964) Diff S 124 7 17 714 Estimated Diff population variance is 17 714 (standard deviation 17 714 4 209) Variance of the distribution of means SM S N PRE S 6 N 8 S M 6 8 S M 0 75 POST S 15 714 N 8 S M 15 714 8 S M 1 9675 Diff S 17 714 N 8 S M 17 714 8 S M 2 21425 Now we can calculate standard deviation of the distribution of means using formula SM S M PRE SM 0 75 0 866 POST SM 1 9675 1 402 Diff SM 2 21425 1 488 The shape is a t distribution with 7 degrees of freedom Step3 Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected Here the level of significance is assumed to be 0 05 and 0 01 The null hypothesis should be rejected at cutoff score 2 365 or 2 365 at 5 and 3 500 or 3 500 at 1 Step 4 Determination your sample's score on the comparison distribution Used formula The test statistic is t (M PopM) SM ( 5 0) 1 488 3 360 Step 5 Decide whether to reject the null hypothesis For the example, the cutoff t score is 2 365 and the actual t score was 3 360, since the t value ( 3 360) is less than the level of significance 05, reject the null hypothesis Ho Therefore the observed differences in numbers of activities of daily living obtained before and after therapy are statistically significant at the 05 level of significance The t score is greater than 01, we fail to reject the null hypothesis Ho if the level of significance is 01 Therefore, the observed differences in numbers of activities of daily living obtained before and after therapy are not statistically significant at the 01 level of significance The effect size is Effect size r2 t2 (t2 df) ( 3 360)2 ( 3 360)2 7 (11 289 (11 289) (7)) 11 289 18 289 0 61725 0 7856 a medium effect The SPSS output for 2 Paired Samples T Test at 95 Paired Samples Statistics Mean N Std Deviation Std Error Mean ADLPRE 10 50 8 2 449 866 ADLPOST 15 50 8 3 964 1 402 Pair 1 Paired Samples Correlations N Pair 1 ADLPRE ADLPOST Correlation 8 Sig 206 625 Paired Samples Test Paired Differences Mean Std Std 95 Confidence Interval Deviation Error of the Difference Mean Pair ADLPRE 1 ADLPOST 5 000 t 4 209 1 488 Lower 8 519 df Sig (2tailed) Upper 1 481 3 360 7 012 The SPSS output of the Paired Samples T Test at 99 Paired Samples Statistics Pair 1 ADLPRE ADLPOST Pair 1 Mean 10 50 15 50 N 8 8 Std Deviation 2 449 3 964 Paired Samples Correlations N Correlation ADLPRE ADLPOST 8 206 Std Error Mean 866 1 402 Sig 625 Paired Samples Test Paired Differences 99 Confidence Interval of the Difference Mean Std Deviation Std Error Mean Lower Upper t df Pair 1 ADLPRE ADLPOST 5 000 4 209 1 488 10 207 207 3 360 7 The deadline for this assignment is 11 59 PM EST on Sunday of Week 5 Sig (2tailed) 012 SPSS ASSIGNMENT 5 Single Sample Dependent Samples t Tests SPSS instructions (For more details, check the links provided under Course Materials in the Course Overview Folder (under Lessons) t Test for a Single Sample Open SPSS Enter the number of activities of daily living performed by the depressed clients studied in 1 in the Data View window In the Variable View window, change the variable name to ADL and set the decimals to zero Click Analyze Compare Means One Sample T test the arrow to move ADL to the Variable(s) window Enter the population mean (14) in the Test Value box Click OK t Test for Dependent Means Open SPSS Enter the number of activities of daily living performed by the depressed clients studied in Problem 2 in the Data View window Be sure to enter the before therapy scores in the first column and the after therapy scores in the second column In the Variable View window, change the variable name for the first variable to ADLPRE and the variable name for the second variable to ADLPOST Set the decimals for both variables to zero Click Analyze Compare Means Paired Samples T Test the arrow to move ADLPRE to the Paired Variable(s) window ADLPOST and then click the arrow to move the variable to the Paired Variable(s) window Click OK Review the five steps of hypothesis testing and complete the following problems Be sure to cut and past the appropriate result boxes from SPSS under each problem 1 Researchers are interested in whether depressed people undergoing group therapy will perform a different number of activities of daily living after group therapy The researchers have randomly selected 12 depressed clients to undergo a 6 week group therapy program Use the five steps of hypothesis testing to determine whether the average number of activities of daily living (shown below) obtained after therapy is significantly different from a mean number of activities of 14 that is typical for depressed people (Clearly indicate each step) Test the difference at the 05 level of significance and, for practice, at the 01 level (in SPSS this means you change the confidence level from 95 to 99 ) In Step 2, show all calculations As part of Step 5, indicate whether the behavioral scientists should recommend group therapy for all depressed people based on evaluation of the null hypothesis at both levels of significance and calculate the effect size CLIENT A B C D E F G H I J K L AFTER THERAPY 17 15 12 21 16 18 17 14 13 15 12 19 Step 1 Restate the question as a research hypothesis and a null hypothesis about the populations Population 1 Depressed people who undergo therapy Population2 Depressed people in general (those that do no undergo therapy) Suppose denotes the mean number of activities of daily living obtained after therapy for depressed people (Ho 14) (Ha 14) The null and alternative hypotheses are N The populations of depressed people, who go through group therapy, perform the same number of activities of daily living, than the population of depressed people who do not get the therapy A The populations of depressed people, who go through group therapy, perform a greater number of activities of daily living, than the population of depressed people who do not get the therapy Step 2 Determine the characteristics of the comparison distribution For this example population mean 14 (number of activities typical for a depressed person) 12,12,13,14,15,15,16,17,17,18,19,21 N 12 Mean N 12 12 13 14 15 15 16 17 17 18 19 21 189 189 12 15 75 Mean 15 75 Have to determine the variance of the population of individuals Use formula for estimated population variance S (X M) df (X M) (17 15 75) (15 15 75) (12 15 75) (21 15 75) (16 15 75) (18 15 75) (17 15 75) (14 15 75) (13 15 75) (15 15 75) (12 15 75) (19 15 75) (1 25) ( 0 75) ( 3 75) (5 25) (0 25) (2 25) (1 25) ( 1 75) ( 2 75) (0 75) ( 3 75) (3 25) (1 5625) (0 56) (14 0625) (27 5625) (0 0625) (5 0625) (1 5625) (3 06) (7 56) (0 56) (14 0625) (10 5625) 86 24 df N 1 12 1 11 df 11 S (X M) df S 86 24 11 7 84 Estimated population variance is 7 84 (standard deviation 7 84 2 80) Variance of the distribution of means SM S N S 7 84 N 12 S M 7 84 12 S M 0 6533 Now we can calculate standard deviation of the distribution of means using formula SM S M SM 0 93167 0 808 The shape is a t distribution with 11 degrees of freedom Step3 Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected Here the level of significance is assumed to be 0 05 and 0 01 The null hypothesis should be rejected at cutoff score 2 201 or 2 201 at 5 and 3 106 or 3 106 at 1 Step 4 Determination your sample's score on the comparison distribution Used formula The test statistic is t (M PopM) SM (15 75 14) 0 808 2 165 Step 5 Decide whether to reject the null hypothesis For the example, the cutoff t score is 2 201 and the actual t score was 2 165, since the t value (2 165) is less than the level of significance 05, we fail to reject the null hypothesis Ho So we don't have sufficient evidence to conclude that the average number of activities of daily living obtained after therapy is significantly different from a mean number of activities of 14 that is typical for depressed people Since the null hypothesis is not rejected at 05 level of significance, the null hypothesis is also not rejected at the 01 level of significance The effect size is Effect size (Population 1M Population 2M) Population SD (15 75 14) 2 80 0 625 a medium effect SPSS Output for 1 One Sample T Test at 95 One Sample Statistics ADL N 12 Mean 15 75 Std Error Std Deviation Mean 2 800 808 One Sample Test Test Value 14 ADL t 2 165 df 11 Mean Sig (2 tailed) Difference 053 1 750 95 Confidence Interval of the Difference Lower Upper 03 3 53 SPSS Output for the One Sample T Test at 99 One Sample Statistics ADL N 12 Mean 15 75 Std Error Std Deviation Mean 2 800 808 One Sample Test Test Value 14 ADL 2 t 2 165 df 11 Mean Sig (2 tailed) Difference 053 1 750 99 Confidence Interval of the Difference Lower Upper 76 4 26 Researchers are interested in whether depressed people undergoing group therapy will perform a different number of activities of daily living before and after group therapy The researchers have randomly selected 8 depressed clients in a 6 week group therapy program Use the five steps of hypothesis testing to determine whether the observed differences in numbers of activities of daily living (shown below) obtained before and after therapy are statistically significant at the 05 level of significance and, for practice, at the 01 level (Clearly indicate each step) In Step 2, show all calculations As part of Step 5, indicate whether the researchers should recommend group therapy for all depressed people based on evaluation of the null hypothesis at both levels of significance and calculate the effect size CLIENT A B C D E F G H BEFORE THERAPY 12 7 10 13 9 8 14 11 AFTER THERAPY 17 15 12 21 16 18 17 8 Step 1 Restate the question as a research hypothesis and a null hypothesis about the populations Population 1 Depressed people who receive group therapy Population2 Depressed people whose daily activity changes from before and after group therapy Suppose D denote the mean observed differences in numbers of activities of daily living obtained before and after therapy Ho D 0 (The mean difference is zero) Ha D 0 (There is a mean difference) The null and alternative hypotheses are N Depressed people who receive group therapy, do not change their daily activities from before or after therapy A Population 1's mean difference score is different from Population 2's mean difference score of zero Step 2 Determine the characteristics of the comparison distribution BEFORE THERAPY AFTER THERAPY 12 7 10 13 9 8 14 11 17 15 12 21 16 18 17 8 Difference (AfterBefore) 5 8 2 8 7 10 3 3 N 8 Mean N Pre 12,7,10,13,9,8,14,11 84 Post 17,15,12,21,16,18,17,8 124 Diff 5, 8, 2, 8, 7, 10, 3,3 40 Pre Mean 84 8 10 5 Mean 10 5 Post Mean 124 8 15 5 Mean 15 5 Diff Mean 40 8 5 Mean 5 Assume a mean of the distribution of the means of difference scores of 0 Have to determine the variance of the population of difference scores using formula for estimated population variance S (X M) df PRE (X M) (12 10 5) (7 10 5) (10 10 5) (13 10 5) (9 10 5) (8 10 5) (14 10 5) (11 10 5) (1 5) ( 3 5) ( 3 75) ( 0 5) (2 5) ( 1 5) ( 2 5) (3 5) ( 0 5) (2 25) (12 25) (0 25) (6 25) (2 25) (6 25) (12 25) (0 25) 42 POST (X M) (17 15 5) (15 15 5) (12 15 5) (21 15 5) (16 15 5) (18 15 5) (1715 5) (8 15 5) (1 5) ( 0 5) ( 3 5) (5 5) (0 5) (2 5) (1 5) ( 7 5) (2 25) (0 25) (12 25) (30 25) (0 25) (6 25) (2 25) (56 25) 110 Diff (X M) ( 5 5) ( 8 5) ( 2 5) ( 8 5) ( 7 5) ( 10 5) ( 3 5) (3 5) (0) ( 3) (3) ( 3) ( 2) ( 5) (2) (8) (0) (9) (9) (9) (4) (25) (4) (64) 124 df N 1 8 1 7 df 7 S (X M) df PRE S 42 7 6 Estimated PRE population variance is 6 (standard deviation 6 2 449) POST S 110 7 15 714 Estimated POST population variance is 15 714 (standard deviation 15 714 3 964) Diff S 124 7 17 714 Estimated Diff population variance is 17 714 (standard deviation 17 714 4 209) Variance of the distribution of means SM S N PRE S 6 N 8 S M 6 8 S M 0 75 POST S 15 714 N 8 S M 15 714 8 S M 1 9675 Diff S 17 714 N 8 S M 17 714 8 S M 2 21425 Now we can calculate standard deviation of the distribution of means using formula SM S M PRE SM 0 75 0 866 POST SM 1 9675 1 402 Diff SM 2 21425 1 488 The shape is a t distribution with 7 degrees of freedom Step3 Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected Here the level of significance is assumed to be 0 05 and 0 01 The null hypothesis should be rejected at cutoff score 2 365 or 2 365 at 5 and 3 500 or 3 500 at 1 Step 4 Determination your sample's score on the comparison distribution Used formula The test statistic is t (M PopM) SM ( 5 0) 1 488 3 360 Step 5 Decide whether to reject the null hypothesis For the example, the cutoff t score is 2 365 and the actual t score was 3 360, since the t value ( 3 360) is less than the level of significance 05, reject the null hypothesis Ho Therefore the observed differences in numbers of activities of daily living obtained before and after therapy are statistically significant at the 05 level of significance The t score is greater than 01, we fail to reject the null hypothesis Ho if the level of significance is 01 Therefore, the observed differences in numbers of activities of daily living obtained before and after therapy are not statistically significant at the 01 level of significance The effect size is Effect size r2 t2 (t2 df) ( 3 360)2 ( 3 360)2 7 (11 289 (11 289) (7)) 11 289 18 289 0 61725 0 7856 a medium effect The SPSS output for 2 Paired Samples T Test at 95 Paired Samples Statistics Mean N Std Deviation Std Error Mean ADLPRE 10 50 8 2 449 866 ADLPOST 15 50 8 3 964 1 402 Pair 1 Paired Samples Correlations N Pair 1 ADLPRE ADLPOST Correlation 8 Sig 206 625 Paired Samples Test Paired Differences Mean Std Std 95 Confidence Interval Deviation Error of the Difference Mean Pair ADLPRE 1 ADLPOST 5 000 t 4 209 1 488 Lower 8 519 df Sig (2tailed) Upper 1 481 3 360 7 012 The SPSS output of the Paired Samples T Test at 99 Paired Samples Statistics Pair 1 ADLPRE ADLPOST Pair 1 Mean 10 50 15 50 N 8 8 Std Deviation 2 449 3 964 Paired Samples Correlations N Correlation ADLPRE ADLPOST 8 206 Std Error Mean 866 1 402 Sig 625 Paired Samples Test Paired Differences 99 Confidence Interval of the Difference Mean Std Deviation Std Error Mean Lower Upper t df Pair 1 ADLPRE ADLPOST 5 000 4 209 1 488 10 207 207 3 360 7 The deadline for this assignment is 11 59 PM EST on Sunday of Week 5 Sig (2tailed) 012

Question

2  Tests of Significance Throughout this assignment you will review mock studies  You will needs to follow the directions outlined in the section using SPSS and decide whether there is significance between the variables  You will need to list the five steps of hypothesis testing (as covered in the lesson for Week 6) to see how every question should be formatted  You will complete all of the problems  Be sure to cut and past the appropriate test result boxes from SPSS under each problem and explain what you will do with your research hypotheses  All calculations should be coming from your SPSS  You will need to submit the SPSS output file to get credit for this assignment  This file will save as a  spv file and will need to be in a single file  In other words, you are not allowed to submit more than one output file for this assignment  The five steps of hypothesis testing when using SPSS are as follows  1  State your research hypothesis (H1) and null hypothesis (H0)  2  Identify your significance level ( 05 or  01) 3  Conduct your analysis using SPSS  4  Look for the valid score for comparison  This score is usually under 'Sig 2 tail' or 'Sig  2'  We will call this   p    5  Compare the two and apply the following rule  a  If   p   is   or   significance level, than you reject the null  Be sure to explain to the reader what this means in regards to your study  (Ex  will you recommend counseling services )   Be sure that your answers are clearly distinguishable  Perhaps you bold your font or use a different color  This assignment is due no later than Sunday of Week 6 by 11 55 pm ET  Save the file in the following format   your last name SOCI332 A2   The file must be a word file  t Tests t Test for a Single Sample (20 points) Open SPSS Enter the number of activities of daily living performed by the depressed clients studied in  1 in the Data View window  In the Variable View window, change the variable name to   ADL   and set the decimals to zero  Click Analyze Compare Means One Sample T test the arrow to move   ADL   to the Variable(s) window  Enter the population mean (17) in the   Test Value   box  Click OK  1  Researches are interested in whether depressed people undergoing group therapy will perform a different number of activities of daily living after group therapy  The researchers have randomly selected 12 depressed clients to undergo a 6 week group therapy program  Use the five steps of hypothesis testing to determine whether the average number of activities of daily living (shown below) obtained after therapy is significantly different from a mean number of activities of 17 that is typical for depressed people  (Clearly indicate each step)  Test the difference at the  05 level of significance and at the  01 level (in SPSS this means you change the   confidence level   from 95  to 99 )  As part of Step 5, indicate whether the behavioral scientists should recommend group therapy for all depressed people based on evaluation of the null hypothesis at both levels of significance  CLIENT A B C D E F G H I J K L AFTER THERAPY 18 14 11 25 24 17 14 10 23 11 22 19 t Test for Dependent Means (20 points) Open SPSS Enter the number of activities of daily living performed by the depressed clients studied in Problem 2 in the Data View window  Be sure to enter the   before therapy   scores in the first column and the   after therapy   scores in the second column  In the Variable View window, change the variable name for the first variable to   ADLPRE   and the variable name for the second variable to   ADLPOST    Set the decimals for both variables to zero  Click Analyze Compare Means Paired Samples T Test the arrow to move   ADLPRE   to the Paired Variable(s) window   ADLPOST   and then click the arrow to move the variable to the Paired Variable(s) window  Click OK  2  Researchers are interested in whether depressed people undergoing group therapy will perform a different number of activities of daily living before and after group therapy  The researchers have randomly selected 8 depressed clients in a 6 week group therapy program  Use the five steps of hypothesis testing to determine whether the observed differences in numbers of activities of daily living (shown below) obtained before and after therapy are statistically significant at the  05 level of  (Clearly indicate each step)  As part of Step 5, indicate whether the researchers should recommend group therapy for all depressed people based on evaluation of the null hypothesis at the  05 level of significance and calculate the measure of association  CLIENT A B C D E F G H BEFORE THERAPY 11 7 10 13 9 8 13 12 AFTER THERAPY 17 12 12 21 16 17 17 8 The t Test for Independent Samples (20 points) Once you have entered the data, click on Analyze, then on Compare Means, and then click on Independent Samples T Test    A dialog box will appear, with your variables (student, condition, score) on the left  Your options are (a) move one or more variables into the   Test Variable(s)   box to select your dependent variables(s) and (b) move one of your variables into the   Grouping Variable   box to select the independent variables (or identify the groups to be compared)  Make       the dependent variable by moving it to the   Test Variable(s)   box  Then make       your independent variable by moving it to the   Grouping Variable   box  Now, the   Define Groups   button is functioning, click on Define Groups and another dialog box appears  Here you must specify the two values of the condition variable that represent the two groups you are comparing  Click in the box next to Group 1 and type the number 1, then click in the box next to Group 2 and type the number 2  Now you can click Continue to return to the   IndependentSamples T Test   dialog box, and click on OK to run the analysis  3  Six months after an industrial accident, a researcher has been asked to compare the job satisfaction of employees who participated in counseling sessions with the satisfaction of employees who chose not to participate  The scores on a job satisfaction inventory for both groups are listed in the table below  Use the five steps of hypothesis testing to determine whether the job satisfaction scores of the group that participated in counseling are statistically higher than the scores of employees who did not participate in counseling at the  01 level of significance  As part of Step 5, indicate whether the researcher should recommend counseling as a method to improve job satisfaction following industrial accidents based on evaluation of the null hypothesis and calculate the measure of association  NOTE  Do not forget to give a numeric value for those who participated in counseling (e g  0) and those who did not participate in counseling (e g  1)  PARTICIPATED IN COUNSELING 36 39 41 36 37 35 37 39 42 DID NOT PARTICIPATE IN COUNSELING 38 36 36 32 30 39 41 35 33 ANOVA (20 points) Open SPSS Analyze the data for  1  Remember that SPSS assumes that all the scores in a row are from the same participant  In this study, there are 15 participants divided into three groups of five  Therefore, each of the 15 participants will be described by two variables, type of therapy and the number of activities of daily living performed  If   1   represents the group receiving individual therapy for 1 hour every 2 weeks,   2   represents the group receiving 1 hour of individual therapy each week, and   3   indicates the group receiving 2 hours of individual therapy each week, the first participant will be described by entering   1   in the top cell of the first column in the Data View window and   16   in the top cell of the second column to indicate that the participant underwent 1 hour of therapy every 2 weeks and performed 16 activities of daily living  The second participant will be described by   1   and   15 Chi Square Chi Square Test for Goodness of Fit  Open SPSS Remember that SPSS assumes that all the scores in a row are from the same participant  In the study presented in  1, there are 20 students, some of whom have been suspended for misbehavior  The primary conflictresolution style used by each student is also entered   Ignore the first variable in this analysis   When you have entered the data for all 20 students, move to the Variable View window and change the first variable name to   SUSPEND   and the second to   STYLE    Set the number of decimals for both variables to zero  Click Analyze Non Parametric Tests Chi Square Click the variable   STYLE   and then the arrow next to the box labeled   Test Variable List   to indicate that the chi square for goodness of fit should be conducted on the conflict resolution style variable  Note that   All categories equal   is the default selection in the   Expected Values   box, which means that SPSS will conduct the goodness of fit test using equal expected frequencies for each of the four styles, in other words, SPSS will assume that the proportions of students each style are equal  Click OK  Chi Square Test for Independence  Open SPSS For  2, you need to add the variable   SUSPEND   to the analysis  Remember that in this problem, we are interested in whether there was an association between conflict resolution style and having been suspended from school for misbehavior  Since the analysis will involve two nominal variables, the appropriate test is a chi square test for independence  Click Analyze Descriptive Statistics Crosstabs Since   SUSPEND   is already selected, click the arrow next to the box labeled   Rows    Click the variable   STYLE   and click the arrow next to the box labeled   Columns    Click   Statistics   and click the box labeled   Chi Square    Click Continue  Click   Cells   and click the box labeled   Expected    Click Continue  Click OK  1  The following table includes resolution used by 20 students  the primary method of conflict Method N of Students Aggressive 8 Manipulative 2 Passive 2 Assertive 8 a  Following the five steps of hypothesis testing, conduct the appropriate chi square test to determine whether the observed frequencies are significantly different from the frequencies expected by change at the  05 level of significance  Clearly identify each of the five steps  The null and alternative hypotheses for this test are, H 0   The observed frequencies are identical with the frequencies expected by change  H A   The observed frequencies are significantly different from the frequencies expected by change  Levelof significance,   0 05  The outputs of Chi squaretest for goodness of fit using SPSS are shown below, STYLE Observed N Expected N Residual Aggressive 8 5 0 3 0 Manipulative 2 5 0  3 0 Passive 2 5 0  3 0 Assertive 8 5 0 3 0 Total 20 Test Statistics STYLE Chi Square 7 200a df Asymp  Sig  3  066 a  0 cells ( 0 ) have expected frequencies less than 5  The minimum expected cell frequency is 5 0  Test Statistic, 2   7 200 Degrees of freedom,    3   p value for a two tailed test is 0 066  Since p value is greater than 0 05, we fail to reject our null hypothesis at 5  level  Therefore, at 5  level data doesn't provide sufficient evidence to support the claim that the observed frequencies are significantly different from the frequencies expected by change  b  Explain your response to some who has never had a course in statistics  We found out that the data of 20 students who suspended from school due to various reasons of misbehavior  To check whether all the four reasons are equally likely to happen or not, conducted a Chi square test for goodness of fit that is appropriate here  The test results provide the conclusions that all the four reasons are equally likely to happen  2  Next, researchers categorized the students based on the primary method of conflict resolution used and whether the student had been suspended from school for misbehavior  These data are presented below  Suspende d Yes No Total Aggressiv e 7 1 8 Method Manipulativ Passive e 1 1 1 1 2 2 Assertive Total 1 7 8 10 10 20 a  Following the five steps of hypothesis testing, conduct the appropriate chi square test to determine whether the observed frequencies are significantly different from the frequencies expected by change at the  05 level of significance  Clearly identify each of the five steps  The null and alternative hypotheses for this test are, H 0   The observed frequencies are identical with the frequencies expected by change  H A   The observed frequencies are significantly different from the frequencies expected by change  Levelof significance,   0 05  The outputs of Chi square test for independence of attributesusing SPSS are shown below, SUSPEND   STYLE Crosstabulation STYLE Aggressive SUSPEND Yes Count Expected Count No Count Expected Count Total Count Expected Count Manipulative Passive Assertive 7 1 1 1 10 4 0 1 0 1 0 4 0 10 0 1 1 1 7 10 4 0 1 0 1 0 4 0 10 0 8 2 2 8 20 8 0 2 0 2 0 8 0 20 0 Chi Square Tests Asymp  Sig  (2Value df sided) Pearson Chi Square 9 000a 3  029 Likelihood Ratio 10 124 3  018 Linear by Linear 8 319 1  004 Association N of Valid Cases 20 a  8 cells (100 0 ) have expected count less than 5  The minimum expected count is 1 00  Test Statistic, 2   9 000   Total Degrees of freedom,  3   p value for a two tailed test is 0 029  Since p value is less than 0 05, we have to reject our null hypothesis at 5  level  Therefore, at 5  level data provides sufficient evidence to support the claim that the observed frequencies are significantly different from the frequencies expected by change  b  Calculate the effect size  Effect size can be measured through either Phi or Cramer's V  The output of Phi or Cramer's Vusing SPSS are shown below, Symmetric Measures Value Nominal by Nominal N of Valid Cases   0 671 Approx  Sig  Phi  671  029 Cramer's V  671  029 20   Cramer's V   0 671  The effect size is 0 671 and it is significantly high  It indicates we have strong evidence from data to conclude that the observed frequencies are significantly different from the frequencies expected by change  c  Explain your response to someone who has never had a course in statistics  We have studied the data of 20 students who are classified by two attributes the type of misbehavior committed and whether suspended from school or not  To check whether the two attributes under study are independent or not, conducted a Chi square test for independence of attributes that is appropriate here  The test results provide the conclusions that the two attributes under study are not independent  Suspension from School Yes No Total Conflict Resolution Assertive Other 1 6 7 Total 9 4 13 10 10 20 A Following the five steps of hypothesis testing, conduct the appropriate chi square test to determine whether the observed frequencies are significantly different from the frequencies expected by change at the  05 level of significance  Clearly identify each of the five steps  The null and alternative hypotheses for this test are, H 0   The observed frequencies are identical with the frequencies expected by change  H A   The observed frequencies are significantly different from the frequencies expected by change  Levelof significance,   0 05  The outputs of Chi square test for independence of attributesusing SPSS are shown below, SUSPENSION   CONFLICT Crosstabulation CONFLICT Assertive SUSPENSION Yes Count Expected Count No Count Expected Count Total Count Expected Count Other Total 1 9 10 3 5 6 5 10 0 6 4 10 3 5 6 5 10 0 7 13 20 7 0 13 0 20 0 Chi Square Tests Value df Asymp  Sig  (2  Exact Sig  (2  Exact Sig  (1  sided) sided) sided) 5 495a 1  019 Continuity Correctionb 3 516 1  061 Likelihood Ratio 5 936 1  015 Pearson Chi Square Fisher's Exact Test  057 Linear by Linear 5 220 1  029  022 Association N of Valid Cases 20 a  2 cells (50 0 ) have expected count less than 5  The minimum expected count is 3 50  b  Computed only for a 2x2 table Test Statistic, 2 5 495 Degrees of freedom,   1   p value for a two tailed test is 0 019  Since p value is lessthan 0 05, we have to reject our null hypothesis at 5  level  Therefore, at 5  level data provides sufficient evidence to support the claim that the observed frequencies are significantly different from the frequencies expected by change  a  Calculate the effect size  Effect size can be measured through either Phi or Cramer's V  The output of Phi or Cramer's Vusing SPSS are shown below, Symmetric Measures Value Nominal by Nominal Phi Cramer's V N of Valid Cases   0 524 Approx  Sig    524  019  524  019 20   Cramer's V   0 524  The effect size is 0 524 and it is significantly high  It indicates we have strong evidence from data to conclude that the observed frequencies are significantly different from the frequencies expected by change  b  Explain your results  We have studied the data of 20 students who are classified by two attributes the type of misbehavior committed and whether suspended from school or not  To check whether the two attributes under study are independent or not, conducted a Chi square test for independence of attributes that is appropriate here  The test results provide the conclusions that the two attributes under study are not independent  The deadline for this assignment is 11 59 PM EST on Sunday of Week 8 SPSS ASSIGNMENT  5 Single Sample   Dependent Samples t Tests SPSS instructions  (For more details, check the links provided under   Course Materials   in the Course Overview Folder (under Lessons)  t Test for a Single Sample  Open SPSS Enter the number of activities of daily living performed by the depressed clients studied in  1 in the Data View window  In the Variable View window, change the variable name to   ADL   and set the decimals to zero  Click Analyze Compare Means One Sample T test the arrow to move   ADL   to the Variable(s) window  Enter the population mean (14) in the   Test Value   box  Click OK  t Test for Dependent Means  Open SPSS Enter the number of activities of daily living performed by the depressed clients studied in Problem 2 in the Data View window  Be sure to enter the   before therapy   scores in the first column and the   after therapy   scores in the second column  In the Variable View window, change the variable name for the first variable to   ADLPRE   and the variable name for the second variable to   ADLPOST    Set the decimals for both variables to zero  Click Analyze Compare Means Paired Samples T Test the arrow to move   ADLPRE   to the Paired Variable(s) window   ADLPOST   and then click the arrow to move the variable to the Paired Variable(s) window  Click OK  Review the five steps of hypothesis testing and complete the following problems  Be sure to cut and past the appropriate result boxes from SPSS under each problem  1  Researchers are interested in whether depressed people undergoing group therapy will perform a different number of activities of daily living after group therapy  The researchers have randomly selected 12 depressed clients to undergo a 6 week group therapy program  Use the five steps of hypothesis testing to determine whether the average number of activities of daily living (shown below) obtained after therapy is significantly different from a mean number of activities of 14 that is typical for depressed people  (Clearly indicate each step)  Test the difference at the  05 level of significance and, for practice, at the  01 level (in SPSS this means you change the   confidence level   from 95  to 99 )  In Step 2, show all calculations  As part of Step 5, indicate whether the behavioral scientists should recommend group therapy for all depressed people based on evaluation of the null hypothesis at both levels of significance and calculate the effect size  CLIENT A B C D E F G H I J K L AFTER THERAPY 17 15 12 21 16 18 17 14 13 15 12 19 Step 1  Restate the question as a research hypothesis and a null hypothesis about the populations  Population 1  Depressed people who undergo therapy Population2  Depressed people in general (those that do no undergo therapy) Suppose denotes the mean number of activities of daily living obtained after therapy for depressed people  (Ho    14) (Ha  14) The null and alternative hypotheses are  N  The populations of depressed people, who go through group therapy, perform the same number of activities of daily living, than the population of depressed people who do not get the therapy  A  The populations of depressed people, who go through group therapy, perform a greater number of activities of daily living, than the population of depressed people who do not get the therapy  Step 2  Determine the characteristics of the comparison distribution  For this example population mean 14 (number of activities typical for a depressed person) 12,12,13,14,15,15,16,17,17,18,19,21 N  12 Mean    N   12 12 13 14 15 15 16 17 17 18 19 21  189 189 12 15 75 Mean   15 75 Have to determine the variance of the population of individuals Use formula for estimated population variance S   (X M)  df (X M)   (17 15 75)   (15 15 75)   (12 15 75)   (21 15 75)   (16 15 75)   (18 15 75)   (17 15 75)   (14 15 75)   (13 15 75)   (15 15 75)   (12 15 75)   (19 15 75)   (1 25)   ( 0 75)   ( 3 75)   (5 25)   (0 25)   (2 25)   (1 25)   ( 1 75)   ( 2 75)   (0 75)   ( 3 75)   (3 25)   (1 5625)   (0 56)   (14 0625)   (27 5625)   (0 0625)  (5 0625)   (1 5625)   (3 06)   (7 56)   (0 56)   (14 0625)   (10 5625)   86 24 df  N 1 12 1  11 df   11 S   (X M)  df S   86 24 11  7 84 Estimated population variance is 7 84 (standard deviation  7 84   2 80) Variance of the distribution of means  SM  S N S  7 84 N 12 S M   7 84 12 S M   0 6533 Now we can calculate standard deviation of the distribution of means using formula  SM  S M SM  0 93167   0 808 The shape is a t distribution with 11 degrees of freedom Step3  Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected  Here the level of significance is assumed to be 0 05 and 0 01 The null hypothesis should be rejected at cutoff score  2 201 or  2 201 at 5  and  3 106 or  3 106 at 1  Step 4  Determination your sample's score on the comparison distribution  Used formula  The test statistic is t  (M PopM) SM   (15 75   14) 0 808   2 165 Step 5  Decide whether to reject the null hypothesis  For the example, the cutoff t score is 2 201 and the actual t score was 2 165, since the t value (2 165) is less than the level of significance  05, we fail to reject the null hypothesis Ho  So we don't have sufficient evidence to conclude that the average number of activities of daily living obtained after therapy is significantly different from a mean number of activities of 14 that is typical for depressed people  Since the null hypothesis is not rejected at  05 level of significance, the null hypothesis is also not rejected at the  01 level of significance  The effect size is Effect size  (Population 1M Population 2M) Population SD   (15 75   14) 2 80   0 625 a medium effect SPSS Output for  1  One Sample T Test at 95   One Sample Statistics ADL N 12 Mean 15 75 Std  Error Std  Deviation Mean 2 800  808 One Sample Test Test Value   14 ADL t 2 165 df 11 Mean Sig  (2 tailed) Difference  053 1 750 95  Confidence Interval of the Difference Lower Upper   03 3 53 SPSS Output for the One Sample T Test at 99   One Sample Statistics ADL N 12 Mean 15 75 Std  Error Std  Deviation Mean 2 800  808 One Sample Test Test Value   14 ADL 2  t 2 165 df 11 Mean Sig  (2 tailed) Difference  053 1 750 99  Confidence Interval of the Difference Lower Upper   76 4 26 Researchers are interested in whether depressed people undergoing group therapy will perform a different number of activities of daily living before and after group therapy  The researchers have randomly selected 8 depressed clients in a 6 week group therapy program  Use the five steps of hypothesis testing to determine whether the observed differences in numbers of activities of daily living (shown below) obtained before and after therapy are statistically significant at the  05 level of significance and, for practice, at the  01 level  (Clearly indicate each step)  In Step 2, show all calculations  As part of Step 5, indicate whether the researchers should recommend group therapy for all depressed people based on evaluation of the null hypothesis at both levels of significance and calculate the effect size  CLIENT A B C D E F G H BEFORE THERAPY 12 7 10 13 9 8 14 11 AFTER THERAPY 17 15 12 21 16 18 17 8 Step 1  Restate the question as a research hypothesis and a null hypothesis about the populations  Population 1  Depressed people who receive group therapy Population2  Depressed people whose daily activity changes from before and after group therapy Suppose D denote the mean observed differences in numbers of activities of daily living obtained before and after therapy  Ho  D   0 (The mean difference is zero) Ha  D 0 (There is a mean difference) The null and alternative hypotheses are  N  Depressed people who receive group therapy, do not change their daily activities from before or after therapy  A  Population 1's mean difference score is different from Population 2's mean difference score of zero  Step 2  Determine the characteristics of the comparison distribution  BEFORE THERAPY AFTER THERAPY 12 7 10 13 9 8 14 11 17 15 12 21 16 18 17 8 Difference (AfterBefore)  5  8  2  8  7  10  3 3 N  8 Mean    N   Pre  12,7,10,13,9,8,14,11   84   Post  17,15,12,21,16,18,17,8   124   Diff   5, 8, 2, 8, 7, 10, 3,3   40 Pre Mean  84 8  10 5 Mean  10 5 Post Mean  124 8 15 5 Mean   15 5 Diff Mean   40 8  5 Mean    5 Assume a mean of the distribution of the means of difference scores of 0  Have to determine the variance of the population of difference scores using formula for estimated population variance S   (X M)  df PRE  (X M)   (12 10 5)   (7 10 5)   (10 10 5)   (13 10 5)   (9 10 5)   (8 10 5)   (14 10 5)   (11 10 5)   (1 5)   ( 3 5)   ( 3 75)   ( 0 5)   (2 5)   ( 1 5)   ( 2 5)   (3 5)   ( 0 5)   (2 25)   (12 25)   (0 25)   (6 25)   (2 25)  (6 25)   (12 25)   (0 25)   42 POST  (X M)   (17 15 5)   (15 15 5)   (12 15 5)   (21 15 5)   (16 15 5)   (18 15 5)   (1715 5)   (8 15 5)   (1 5)   ( 0 5)   ( 3 5)   (5 5)   (0 5)   (2 5)   (1 5)   ( 7 5)   (2 25)   (0 25)   (12 25)   (30 25)   (0 25)  (6 25)   (2 25)   (56 25)   110 Diff  (X M)   ( 5  5)   ( 8  5)   ( 2  5)   ( 8  5)   ( 7  5)   ( 10  5)   ( 3  5)   (3  5)   (0)   ( 3)   (3)   ( 3)   ( 2)   ( 5)   (2)   (8)   (0)   (9)   (9)   (9)   (4)   (25)   (4)   (64)   124 df  N 1  8 1  7 df   7 S   (X M)  df PRE  S   42 7  6 Estimated PRE population variance is 6 (standard deviation  6  2 449) POST  S   110 7   15 714 Estimated POST population variance is 15 714 (standard deviation  15 714  3 964) Diff  S   124 7   17 714 Estimated Diff population variance is 17 714 (standard deviation  17 714   4 209) Variance of the distribution of means  SM  S N PRE  S  6 N 8 S M   6 8 S M   0 75 POST  S  15 714 N 8 S M   15 714 8 S M   1 9675 Diff  S   17 714 N 8 S M   17 714 8 S M   2 21425 Now we can calculate standard deviation of the distribution of means using formula  SM  S M PRE  SM  0 75   0 866 POST  SM  1 9675   1 402 Diff  SM  2 21425   1 488 The shape is a t distribution with 7 degrees of freedom Step3  Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected  Here the level of significance is assumed to be 0 05 and 0 01 The null hypothesis should be rejected at cutoff score  2 365 or  2 365 at 5  and  3 500 or  3 500 at 1  Step 4  Determination your sample's score on the comparison distribution  Used formula  The test statistic is t  (M PopM) SM   ( 5   0) 1 488    3 360 Step 5  Decide whether to reject the null hypothesis  For the example, the cutoff t score is 2 365 and the actual t score was  3 360, since the t value ( 3 360) is less than the level of significance  05, reject the null hypothesis Ho  Therefore the observed differences in numbers of activities of daily living obtained before and after therapy are statistically significant at the  05 level of significance  The t score is greater than  01, we fail to reject the null hypothesis Ho if the level of significance is  01  Therefore, the observed differences in numbers of activities of daily living obtained before and after therapy are not statistically significant at the  01 level of significance  The effect size is Effect size   r2   t2 (t2   df)   ( 3 360)2  ( 3 360)2   7    (11 289  (11 289)   (7))  11 289 18 289  0 61725   0 7856 a medium effect The SPSS output for  2 Paired Samples T Test at 95   Paired Samples Statistics Mean N Std  Deviation Std  Error Mean ADLPRE 10 50 8 2 449  866 ADLPOST 15 50 8 3 964 1 402 Pair 1 Paired Samples Correlations N Pair 1 ADLPRE   ADLPOST Correlation 8 Sig   206  625 Paired Samples Test Paired Differences Mean Std  Std  95  Confidence Interval Deviation Error of the Difference Mean Pair ADLPRE 1 ADLPOST    5 000 t 4 209 1 488 Lower  8 519 df Sig  (2tailed) Upper  1 481  3 360 7  012 The SPSS output of the Paired Samples T Test at 99   Paired Samples Statistics Pair 1 ADLPRE ADLPOST Pair 1 Mean 10 50 15 50 N 8 8 Std  Deviation 2 449 3 964 Paired Samples Correlations N Correlation ADLPRE   ADLPOST 8  206 Std  Error Mean  866 1 402 Sig   625 Paired Samples Test Paired Differences 99  Confidence Interval of the Difference Mean Std  Deviation Std  Error Mean Lower Upper t df Pair 1 ADLPRE   ADLPOST  5 000 4 209 1 488  10 207  207  3 360 7 The deadline for this assignment is 11 59 PM EST on Sunday of Week 5 Sig  (2tailed)  012 SPSS ASSIGNMENT  5 Single Sample   Dependent Samples t Tests SPSS instructions  (For more details, check the links provided under   Course Materials   in the Course Overview Folder (under Lessons)  t Test for a Single Sample  Open SPSS Enter the number of activities of daily living performed by the depressed clients studied in  1 in the Data View window  In the Variable View window, change the variable name to   ADL   and set the decimals to zero  Click Analyze Compare Means One Sample T test the arrow to move   ADL   to the Variable(s) window  Enter the population mean (14) in the   Test Value   box  Click OK  t Test for Dependent Means  Open SPSS Enter the number of activities of daily living performed by the depressed clients studied in Problem 2 in the Data View window  Be sure to enter the   before therapy   scores in the first column and the   after therapy   scores in the second column  In the Variable View window, change the variable name for the first variable to   ADLPRE   and the variable name for the second variable to   ADLPOST    Set the decimals for both variables to zero  Click Analyze Compare Means Paired Samples T Test the arrow to move   ADLPRE   to the Paired Variable(s) window   ADLPOST   and then click the arrow to move the variable to the Paired Variable(s) window  Click OK  Review the five steps of hypothesis testing and complete the following problems  Be sure to cut and past the appropriate result boxes from SPSS under each problem  1  Researchers are interested in whether depressed people undergoing group therapy will perform a different number of activities of daily living after group therapy  The researchers have randomly selected 12 depressed clients to undergo a 6 week group therapy program  Use the five steps of hypothesis testing to determine whether the average number of activities of daily living (shown below) obtained after therapy is significantly different from a mean number of activities of 14 that is typical for depressed people  (Clearly indicate each step)  Test the difference at the  05 level of significance and, for practice, at the  01 level (in SPSS this means you change the   confidence level   from 95  to 99 )  In Step 2, show all calculations  As part of Step 5, indicate whether the behavioral scientists should recommend group therapy for all depressed people based on evaluation of the null hypothesis at both levels of significance and calculate the effect size  CLIENT A B C D E F G H I J K L AFTER THERAPY 17 15 12 21 16 18 17 14 13 15 12 19 Step 1  Restate the question as a research hypothesis and a null hypothesis about the populations  Population 1  Depressed people who undergo therapy Population2  Depressed people in general (those that do no undergo therapy) Suppose denotes the mean number of activities of daily living obtained after therapy for depressed people  (Ho    14) (Ha  14) The null and alternative hypotheses are  N  The populations of depressed people, who go through group therapy, perform the same number of activities of daily living, than the population of depressed people who do not get the therapy  A  The populations of depressed people, who go through group therapy, perform a greater number of activities of daily living, than the population of depressed people who do not get the therapy  Step 2  Determine the characteristics of the comparison distribution  For this example population mean 14 (number of activities typical for a depressed person) 12,12,13,14,15,15,16,17,17,18,19,21 N  12 Mean    N   12 12 13 14 15 15 16 17 17 18 19 21  189 189 12 15 75 Mean   15 75 Have to determine the variance of the population of individuals Use formula for estimated population variance S   (X M)  df (X M)   (17 15 75)   (15 15 75)   (12 15 75)   (21 15 75)   (16 15 75)   (18 15 75)   (17 15 75)   (14 15 75)   (13 15 75)   (15 15 75)   (12 15 75)   (19 15 75)   (1 25)   ( 0 75)   ( 3 75)   (5 25)   (0 25)   (2 25)   (1 25)   ( 1 75)   ( 2 75)   (0 75)   ( 3 75)   (3 25)   (1 5625)   (0 56)   (14 0625)   (27 5625)   (0 0625)  (5 0625)   (1 5625)   (3 06)   (7 56)   (0 56)   (14 0625)   (10 5625)   86 24 df  N 1 12 1  11 df   11 S   (X M)  df S   86 24 11  7 84 Estimated population variance is 7 84 (standard deviation  7 84   2 80) Variance of the distribution of means  SM  S N S  7 84 N 12 S M   7 84 12 S M   0 6533 Now we can calculate standard deviation of the distribution of means using formula  SM  S M SM  0 93167   0 808 The shape is a t distribution with 11 degrees of freedom Step3  Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected  Here the level of significance is assumed to be 0 05 and 0 01 The null hypothesis should be rejected at cutoff score  2 201 or  2 201 at 5  and  3 106 or  3 106 at 1  Step 4  Determination your sample's score on the comparison distribution  Used formula  The test statistic is t  (M PopM) SM   (15 75   14) 0 808   2 165 Step 5  Decide whether to reject the null hypothesis  For the example, the cutoff t score is 2 201 and the actual t score was 2 165, since the t value (2 165) is less than the level of significance  05, we fail to reject the null hypothesis Ho  So we don't have sufficient evidence to conclude that the average number of activities of daily living obtained after therapy is significantly different from a mean number of activities of 14 that is typical for depressed people  Since the null hypothesis is not rejected at  05 level of significance, the null hypothesis is also not rejected at the  01 level of significance  The effect size is Effect size  (Population 1M Population 2M) Population SD   (15 75   14) 2 80   0 625 a medium effect SPSS Output for  1  One Sample T Test at 95   One Sample Statistics ADL N 12 Mean 15 75 Std  Error Std  Deviation Mean 2 800  808 One Sample Test Test Value   14 ADL t 2 165 df 11 Mean Sig  (2 tailed) Difference  053 1 750 95  Confidence Interval of the Difference Lower Upper   03 3 53 SPSS Output for the One Sample T Test at 99   One Sample Statistics ADL N 12 Mean 15 75 Std  Error Std  Deviation Mean 2 800  808 One Sample Test Test Value   14 ADL 2  t 2 165 df 11 Mean Sig  (2 tailed) Difference  053 1 750 99  Confidence Interval of the Difference Lower Upper   76 4 26 Researchers are interested in whether depressed people undergoing group therapy will perform a different number of activities of daily living before and after group therapy  The researchers have randomly selected 8 depressed clients in a 6 week group therapy program  Use the five steps of hypothesis testing to determine whether the observed differences in numbers of activities of daily living (shown below) obtained before and after therapy are statistically significant at the  05 level of significance and, for practice, at the  01 level  (Clearly indicate each step)  In Step 2, show all calculations  As part of Step 5, indicate whether the researchers should recommend group therapy for all depressed people based on evaluation of the null hypothesis at both levels of significance and calculate the effect size  CLIENT A B C D E F G H BEFORE THERAPY 12 7 10 13 9 8 14 11 AFTER THERAPY 17 15 12 21 16 18 17 8 Step 1  Restate the question as a research hypothesis and a null hypothesis about the populations  Population 1  Depressed people who receive group therapy Population2  Depressed people whose daily activity changes from before and after group therapy Suppose D denote the mean observed differences in numbers of activities of daily living obtained before and after therapy  Ho  D   0 (The mean difference is zero) Ha  D 0 (There is a mean difference) The null and alternative hypotheses are  N  Depressed people who receive group therapy, do not change their daily activities from before or after therapy  A  Population 1's mean difference score is different from Population 2's mean difference score of zero  Step 2  Determine the characteristics of the comparison distribution  BEFORE THERAPY AFTER THERAPY 12 7 10 13 9 8 14 11 17 15 12 21 16 18 17 8 Difference (AfterBefore)  5  8  2  8  7  10  3 3 N  8 Mean    N   Pre  12,7,10,13,9,8,14,11   84   Post  17,15,12,21,16,18,17,8   124   Diff   5, 8, 2, 8, 7, 10, 3,3   40 Pre Mean  84 8  10 5 Mean  10 5 Post Mean  124 8 15 5 Mean   15 5 Diff Mean   40 8  5 Mean    5 Assume a mean of the distribution of the means of difference scores of 0  Have to determine the variance of the population of difference scores using formula for estimated population variance S   (X M)  df PRE  (X M)   (12 10 5)   (7 10 5)   (10 10 5)   (13 10 5)   (9 10 5)   (8 10 5)   (14 10 5)   (11 10 5)   (1 5)   ( 3 5)   ( 3 75)   ( 0 5)   (2 5)   ( 1 5)   ( 2 5)   (3 5)   ( 0 5)   (2 25)   (12 25)   (0 25)   (6 25)   (2 25)  (6 25)   (12 25)   (0 25)   42 POST  (X M)   (17 15 5)   (15 15 5)   (12 15 5)   (21 15 5)   (16 15 5)   (18 15 5)   (1715 5)   (8 15 5)   (1 5)   ( 0 5)   ( 3 5)   (5 5)   (0 5)   (2 5)   (1 5)   ( 7 5)   (2 25)   (0 25)   (12 25)   (30 25)   (0 25)  (6 25)   (2 25)   (56 25)   110 Diff  (X M)   ( 5  5)   ( 8  5)   ( 2  5)   ( 8  5)   ( 7  5)   ( 10  5)   ( 3  5)   (3  5)   (0)   ( 3)   (3)   ( 3)   ( 2)   ( 5)   (2)   (8)   (0)   (9)   (9)   (9)   (4)   (25)   (4)   (64)   124 df  N 1  8 1  7 df   7 S   (X M)  df PRE  S   42 7  6 Estimated PRE population variance is 6 (standard deviation  6  2 449) POST  S   110 7   15 714 Estimated POST population variance is 15 714 (standard deviation  15 714  3 964) Diff  S   124 7   17 714 Estimated Diff population variance is 17 714 (standard deviation  17 714   4 209) Variance of the distribution of means  SM  S N PRE  S  6 N 8 S M   6 8 S M   0 75 POST  S  15 714 N 8 S M   15 714 8 S M   1 9675 Diff  S   17 714 N 8 S M   17 714 8 S M   2 21425 Now we can calculate standard deviation of the distribution of means using formula  SM  S M PRE  SM  0 75   0 866 POST  SM  1 9675   1 402 Diff  SM  2 21425   1 488 The shape is a t distribution with 7 degrees of freedom Step3  Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected  Here the level of significance is assumed to be 0 05 and 0 01 The null hypothesis should be rejected at cutoff score  2 365 or  2 365 at 5  and  3 500 or  3 500 at 1  Step 4  Determination your sample's score on the comparison distribution  Used formula  The test statistic is t  (M PopM) SM   ( 5   0) 1 488    3 360 Step 5  Decide whether to reject the null hypothesis  For the example, the cutoff t score is 2 365 and the actual t score was  3 360, since the t value ( 3 360) is less than the level of significance  05, reject the null hypothesis Ho  Therefore the observed differences in numbers of activities of daily living obtained before and after therapy are statistically significant at the  05 level of significance  The t score is greater than  01, we fail to reject the null hypothesis Ho if the level of significance is  01  Therefore, the observed differences in numbers of activities of daily living obtained before and after therapy are not statistically significant at the  01 level of significance  The effect size is Effect size   r2   t2 (t2   df)   ( 3 360)2  ( 3 360)2   7    (11 289  (11 289)   (7))  11 289 18 289  0 61725   0 7856 a medium effect The SPSS output for  2 Paired Samples T Test at 95   Paired Samples Statistics Mean N Std  Deviation Std  Error Mean ADLPRE 10 50 8 2 449  866 ADLPOST 15 50 8 3 964 1 402 Pair 1 Paired Samples Correlations N Pair 1 ADLPRE   ADLPOST Correlation 8 Sig   206  625 Paired Samples Test Paired Differences Mean Std  Std  95  Confidence Interval Deviation Error of the Difference Mean Pair ADLPRE 1 ADLPOST    5 000 t 4 209 1 488 Lower  8 519 df Sig  (2tailed) Upper  1 481  3 360 7  012 The SPSS output of the Paired Samples T Test at 99   Paired Samples Statistics Pair 1 ADLPRE ADLPOST Pair 1 Mean 10 50 15 50 N 8 8 Std  Deviation 2 449 3 964 Paired Samples Correlations N Correlation ADLPRE   ADLPOST 8  206 Std  Error Mean  866 1 402 Sig   625 Paired Samples Test Paired Differences 99  Confidence Interval of the Difference Mean Std  Deviation Std  Error Mean Lower Upper t df Pair 1 ADLPRE   ADLPOST  5 000 4 209 1 488  10 207  207  3 360 7 The deadline for this assignment is 11 59 PM EST on Sunday of Week 5 Sig  (2tailed)  012

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2: Tests of Significance Throughout this assignment you will review mock studies. You will needs to follow the directions outlined in the section using SPSS

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