T Tests__SPSS
Assignment 4: t Tests
The one sample t Test for a mean
A market research company found that children in America between the ages of 5 and 12 years old watch on average 196 minutes of TV per day. A survey was conducted by randomly selecting 20 British children within this age group to see if British Children are statistically different in the amount of TV watched per night from their American counterparts.
Based on this example answer the following questions:
 Why is a one sample t Test the most appropriate technique to test the example?
 State (in words) the null and the alternate
 Use the data set provided and conduct a one sample t Test using SPSS.
Hint: Once you are in the One Sample t Test dialogue box in SPSS, your test value will be 196 (minutes).
 What is the Mean number of minutes watched per night?
 What is the t statistic?
 What is the p value or significance?
 Discuss the findings in regard to the null and alternate hypothesis using Morgan et al. (2002)
The two samples ttest for the equality of Means
A survey was conducted by randomly selecting 20 American children and 20 British children to determine if there is a statistical difference in amount of time they watch TV.Based on this example answer the following questions:
 Why is a two samples t test most appropriate technique for this research example?
 State (in words) the null and the alternate
 Use the data set provided and conduct a two samples ttest using SPSS. Use the SPSS command/actions listed below to conduct the analysis:
Hint: Once you are in the IndependentSamples t Test dialog box in SPSS, you will have to Define your groups.
 Click on Define Groups
 In the area next to Group 1: type 1, and in the next to Group 2: type 2
 Click on Continue
 Click on OK
 What is the mean number of minutes of TV watched by the British children?
 What is the mean number of minutes of TV watched by the American children?
 In the Equal Variances Assumed row, what is the t statistic?
 What is the p value or significance?
 Discuss the findings in regard to the null and alternate hypothesis using Morgan et al. (2002) pp. 1012.
 Why might these results be inaccurate due to the technique of Random Sampling (see Creswell, 2008)?
Calculate the Effect size (Cohen’s d) for the paired samples ttest output:
Use the following equation: d = M_{1} – M_{2}/σ_{D} (Mean of Posttotal Social Competency – Mean of Pretotal Social Competency)/Paired Differences Std. Deviation
Cohen (1988) defined effect sizes as “small, d = .2,” “medium, d = .5,” and “large, d = .8″, stating that “there is a certain risk in inherent in offering conventional operational definitions for those terms for use in power analysis in as diverse a field of inquiry as behavioral science”.
Discuss the meaning of the findings (12 sentences).
Solution
Assignment 4
1.
A one sample tTest is the most appropriate technique to test the research as they are studying one subject (American and British children). The one sample t test compares the mean (196 minutes) with the known population.
2.
Null hypothesis: British children are not statistically different from American children regarding TV watching.
Alternative Hypothesis: British children are statistically different from American children regarding TV watching.
3.
OneSample Statistics  
N  Mean  Std. Deviation  Std. Error Mean  
TV watched per night in minutes  20  165.85  29.290  6.550 
OneSample Test  
Test Value = 196  
t  df  Sig. (2tailed)  Mean Difference  95% Confidence Interval of the Difference  
Lower  Upper  
TV watched per night in minutes  4.603  19  .000  30.150  43.86  16.44 
4.
165.85 is the Mean number of minutes watched per night.
5.
The tstatistics is 4.603
6.
The Pvalue is .000
7.
Given that t (20) = 4.603 with a p < 0001, Signify that the level of significance is less than .05. This denotes a failure to accept or reject the null hypothesis. In the case of determining statistically if there is a difference between the amount of TV watched per day by American children and British children, the results of a single t test suggests that there were significant differences.
Fail to accept the null hypothesis t=4.603 all. The alternative hypothesis is t>4.603. The only difference between the null and alternative hypothesis is that the possible value of the population parameter included within Ha will always differ from Ho. What we really think will occur.
1.
There are two normally distributed but independent populations with separate samples containing different sets of individual subjects and this test compares the groups to infer some type of differences between the groups. The 20 American children and 20 British children are the two groups under study. The Independent Means application is the appropriate type of t test to get the scores.
2.
Null There is no significant difference between the amount of time American children and British children watch T.V. The Means are equal.
AlternativeThere is significant difference between the amount of time American children and British children watch T.V. The Means are not equal.
3.
Group Statistics  
Country of residence  N  Mean  Std. Deviation  Std. Error Mean  
Minutes of TV watched per night  America  20  165.85  29.290  6.550 
Britain  20  186.75  29.563  6.611 
4.
186.75 is the mean number of minutes of British children who watched TV.
5.
165.85 is the mean number of minutes of American children who watched TV.
6.
The t statistic is 2.246.
7.
The p value of significance is .031
8.
A twosample t test was conducted. The null hypothesis is false therefore the researcher would reject. This means the Alternative hypothesis is true because the means are different. This test suggests that with a t of 2.246 and a df of 38 and level of significance of .031 which is less than .05 the alpha level of significance, we conclude that there was a significant difference between the children in America and Britain with regards to watching TV, so we will reject the null hypothesis.
9.
There are always observable differences between the means of any two groups; this is due in part to chance random events or just dumb luck.
Social competency in pair 1 indicates that the participants did not make sufficient gains, t (45) = 3.229, p < .002. The null hypothesis is false therefore the researcher would reject. This means that Alternative hypothesis is true because the means are different. d= (100.487113.87)/18.695 SD
d=.4814. In this case, the effect size is small. The effect size (d) for these gains was small.
Assignment 5: ANOVA
Open the divorce.sav data file and run the oneway ANOVA to answer the following question:
What are the effects of marital status on life satisfaction?
 State the independent and dependent variables.
 State the null and alternate hypotheses.
 Run the appropriate analysis using SPSS (Hint: Use the General Linear Models and Univariate Procedure. Select Estimates of Effect Size under the Options tab).
 What are the mean and standard deviation for each of the levels of the IV?
 Report the appropriate F statistic, degrees of freedom, p value, and eta squared (η^{2}).
 What is your decision regarding the null hypothesis (i.e., did you reject or fail to reject the null)? Explain your decision (1 sentence)
 Using Morgan et al text. write up a sample results section.
Use the goggles.sav data to run a twoway ANOVA. Consider the following example and then answer the subsequent questions:
Derived from Field (2005), an anthropologist was interested in the effects of alcohol on mate selection at nightclubs. Her rationale was that after alcohol had been consumed, subjective perception of physical attractiveness would become more inaccurate. She was also interested in whether this effect was different for men and women. She picked 48 students: 24 male and 24 female. She then took groups of eight participants to a nightclub and gave them either a nonalcoholic lager, 2 pints of strong lager, or 4 pints of strong lager. At the evening she took a photograph of the person that the participant was chatting up. She then got a pool of independent judges to assess the attractiveness of the person in each photograph (out of 100).
 State the independent variables and the dependent variable.
 State the null and alternate hypotheses.
 Run the appropriate analysis and include the Levene’s (Homogeneity of Variance) test and the test of BetweenSubjects Effects (Cut and Paste Levene’s Test Output Below – Explain the meaning of this test).
Tests of BetweenSubjects Effects (Include Data from Output in the Figure Provided Below)
Source  Df  Mean Square  F  Sig. 
Gender  
Alcohol


Gender*alcohol 
 Report the mean and standard deviation for each level of the IV (Cut and Paste Output).
 What is your decision concerning the null hypothesis (i.e., did you reject or fail to reject the null)? Explain your response.
 Post the Estimated Marginal Means of Attractiveness of Data .
 Using Morgan et al, write up a sample results section.
Solution
Assignment 5
What are the effects of marital status on life satisfaction?
a.
Marital status is Independent variable.
Life satisfaction is Dependent variable.
b.
The null hypothesis is that there is no significant difference in life satisfaction based on marital status.
The alternate hypothesis is that there is significant difference in life satisfaction based on marital status.
c.
BetweenSubjects Factors  
Value Label  N  
current marital status  1  marred  55 
2  separated  34  
3  divorced or DTS  112  
4  widowed  3  
5  cohabit  25 
Descriptive Statistics  
Dependent Variable: life satisfaction  
current marital status  Mean  Std. Deviation  N 
marred  4.9362  .79650  55 
separated  4.5335  1.03186  34 
divorced or DTS  4.8241  .85747  112 
widowed  4.6133  2.12660  3 
cohabit  4.9152  .88940  25 
Total  4.8151  .89571  229 
d.
Married – The mean is 4.936. The standard deviation is .121
Separated – The mean is 4.534. The standard deviation is .153
Divorced or DTS – The mean is 4.824. The standard deviation is .084
Widowed – The mean 4.613. The standard deviation is .516
Cohabit – The mean is 4.915. The standard deviation is .179
e.
Tests of BetweenSubjects Effects  
Dependent Variable: life satisfaction  
Source  Type III Sum of Squares  df  Mean Square  F  Sig.  Partial Eta Squared 
Corrected Model  3.884^{a}  4  .971  1.215  .305  .021 
Intercept  1320.218  1  1320.218  1651.762  .000  .881 
status  3.884  4  .971  1.215  .305  .021 
Error  179.038  224  .799  
Total  5492.225  229  
Corrected Total  182.922  228  
a. R Squared = .021 (Adjusted R Squared = .004) 
Fstatistics = 1.22
DF = 4
Pvalue = 0.305
Eta Squared = .021
f.
The researcher would fail to accept the null hypothesis (reject the null hypothesis) because the p value is more than .05.
g.
Subjects who were married, separated or cohabitating have little to no difference (12, 15, and 17). Whereas those who are widowed and those who are life satisfaction is significant don’t have a consistent sample size to make a determination. There are only 3 samples for the widowed category.
a.
The independent variables are gender (male/female) and the effects of alcohol.
The dependent variable is perception of physical attractiveness.
b.
The null hypothesis –
 Perception based on gender is that males have wide ranging effects over females.
The first null hypothesis is that there is no significant difference between the male and female perception of physical attractiveness and the means are equal, so we accept the null hypothesis.
 Perception based on alcohol.
The second hypothesis is that there is an interaction between males and females and the effects of alcohol so we would reject the null hypothesis.
 There is an interaction between gender and alcohol and the perception.
 The alternative hypothesis is that there is a significant difference in the perception of attractiveness by males compared to female participants after alcohol consumption. The means are significantly different so we reject the null hypothesis.
c.
Levene’s Test of Equality of Error Variances^{a}  
Dependent Variable: Attractiveness of Date  
F  df1  df2  Sig. 
1.527  5  42  .202 
Tests the null hypothesis that the error variance of the dependent variable is equal across groups.  
a. Design: Intercept + alcohol + gender + alcohol * gender 
WE can proceed with the ANOVA since the Levene’s test has a significance of .20 which is greater than .05. This indicates that the researcher meets the assumption of the test.
Source  DF  Mean square  F  Sig 
Gender  1  168.750  2.032  .161 
Alcohol  2  1666.146  20.065  .000 
Gender*Alcohol  2  989.062  11.911  .000 
d.
Tests of BetweenSubjects Effects  
Dependent Variable: Attractiveness of Date  
Source  Type III Sum of Squares  df  Mean Square  F  Sig.  Partial Eta Squared 
Corrected Model  5479.167^{a}  5  1095.833  13.197  .000  .611 
Intercept  163333.333  1  163333.333  1967.025  .000  .979 
alcohol  3332.292  2  1666.146  20.065  .000  .489 
gender  168.750  1  168.750  2.032  .161  .046 
alcohol * gender  1978.125  2  989.063  11.911  .000  .362 
Error  3487.500  42  83.036  
Total  172300.000  48  
Corrected Total  8966.667  47  
a. R Squared = .611 (Adjusted R Squared = .565) 
e.
We would reject the null hypothesis because p is more than .05 since the means are not equal.
f.
1. Grand Mean  
Dependent Variable: Attractiveness of Date  
Mean  Std. Error  95% Confidence Interval  
Lower Bound  Upper Bound  
58.333  1.315  55.679  60.988 
Gender  
Dependent Variable: Attractiveness of Date  
Gender  Mean  Std. Error  95% Confidence Interval  
Lower Bound  Upper Bound  
Male  56.458  1.860  52.705  60.212 
Female  60.208  1.860  56.455  63.962 
Alcohol Consumption  
Dependent Variable: Attractiveness of Date  
Alcohol Consumption  Mean  Std. Error  95% Confidence Interval  
Lower Bound  Upper Bound  
None  63.750  2.278  59.153  68.347 
2 Pints  64.688  2.278  60.090  69.285 
4 Pints  46.562  2.278  41.965  51.160 
g.
The attractiveness data (score) is lower when the subject has consumed 4 pints of alcohol. This scoring is increases with men than with women. The interaction effect shows that men are more influence by alcohol than women. The increase in consumption by males alcohol drinking causes women to be less attractive. Thus, plays a role in the companion that they choose. Men will chose an unattractive companion whereas, women won’t. Even when they are drinking, women are selective about the companion they choose.
Assignment 6: NonParametrics
ChiSquare Test of Association (Independence)
A school system is concerned about the low graduation rate among their high school students. The superintendent assigned a task force to research possible reasons that could explain the low graduation rate. The task force decided to conduct a preliminary literature review about current graduation rate research. The literature review signaled that among several key factors that are related to completion of high school is the development and execution of a school guidance intervention plan. The task force decided to investigate if such is the case in the high school with the highest dropout rate in their district. The following data was collected:
Valenti High School Data
Guidance Intervention Plan  
Students obtaining a high school diploma  No  Yes 
No  577  46 
Yes  381  492 
In order to run the chisquare test of association (aka, chisquare test of independence), open the file named:
assign 6 prob 1 data.sav
Under View > Value labels you can toggle between the variable category label (no, yes) or the dummy codes (numeric representation) for the category label (0 = no, 1 = yes)
Make sure you review both the “Data view” and “Variable view” so you understand the variable types and other properties – and how the variables were created and entered.
Go to Analyze > Descriptive statistics > Frequencies, and move the two variables to the “Variables” box, then click “ok”
This gives you the frequencies and percents of each of the variables
To run the chisquare of association test, go to Analyze > Descriptive statistics > Crosstabs, then move High School Diploma (HSD) to the “Row” box, and Guidance Intervention Plan (GIP) to the “Column” box
Select “Statistics” then select the “Chi square” and “Phi and Cramer’s V” boxes
Click “Continue”
Select “Cells” then select “Observed” and “Expected” under “Counts” (“Round cell counts” is selected by default)
Click “Continue” then “ok”
You will use the output for questions 13.
You were asked to analyze the data and present the findings in the upcoming school board meeting. Your report must include the following:
 A short explanation about the
 ChiSquare Test of Association
 The Case Processing Summary table
 Cross tabulation table
 Chisquare Tests table
 Clustered Bar Chart
 Write down the value of the Pearson chisquare and its associated tail probability (pvalue). Is it significant? (Use the Morgan text pp. 3638, for the APA style writeup)
 In terms of the experimental hypothesis, what has this test shown?
ChiSquare Goodness of Fit Test
Following are the cumulative number of AIDS cases reported for Neptune County through December 31, 2008, broken down by ethnicity:
Ethnicity  Actual Number of AIDS Cases 
White  751 
Hispanic  225 
AfricanAmerican  100 
Asian, Pacific Islander  36 
Native American  5 
Total = 1117 
Ethnicity  Percentage of total county population  Expected cases by population 
1 = White  51  570 
2 = Hispanic  23  257 
3 = AfricanAmerican  4  45 
4 = Asian, Pacific Islander  21  235 
5 = Native American  1.0  10 
Total = 100%  Total = 1117 
In order to run the chisquare goodness of fit test, open the file named:
assign 6 prob 2 data.sav
Under View >Value labels you can toggle between the variable category label (White, Hispanic, African American, Asian Pacific Islander, Native American) or the dummy codes (numeric representation) for the category label (1 = White, 2 = Hispanic, 3 = African American, 4 = Asian Pacific Islander, 5 = Native American). For this test, we want to compares the observed (actual) AIDS cases to the expected cases (frequencies, based on county population).
Go to Analyze >Nonparametric tests >Chi square
Move the AIDS variable to the “Test variable list” box
Using the “Expected cases by population” numbers that correspond to each ethnic group in the table above, under “Expected values” select, “Values” then add the 5 population percentages that correspond with the frequency in the sample, using “Add” for each (570 > Add, 257 > Add, 45 > Add, 235 > Add, 10 > Add)
Select “Descriptive” > Continue under “Options”
Click “ok”
You will use the output for questions 13.
Perform a goodnessoffit test to determine whether the makeup of AIDS cases follows the ethnicity of the general population of Neptune County. State the following:
 Null Hypothesis
 Decision
 Reason for the Decision
 Does it appear that the pattern of AIDS cases in Neptune County corresponds to the distribution of ethnic groups in this county? Why or why not?
 You are asked to report your findings along with recommendations for prioritizing interventions by ethnic group to the County Health Commission. Write a one paragraph handout summarizing the findings and your recommendations. Also include the results, in APA style (Use the Morgan text pp. 3638, for the APA style writeup).
Solution
Assignment 6
1.
. ChiSquare Test of Association
The chisquare test for independence, also called Pearson’s chisquare test or the chisquare test of association, is used to discover if there is a relationship between two categorical variables.
. The Case Processing Summary table
Case processing summary table shows the total number of valid cases or entry in the given dataset and the total number of missing entry in the dataset.
. Cross tabulation table
Cross tabulation table is a tool that allow us to compare the relationship between two variable. It gives the observed and expected count of each variable under each category. The total of observed and expected count will always be the same.
. Chisquare Tests table
Chisquare test table provide us the test statistics, degree of freedom, and the significance value for each applied test. It shows how likely it is that an observed distribution is due to chance.
. Clustered Bar Chart
A clustered bar chart is a type of bar graph that allows for the display of two categorical variables. It consists of a grid and some vertical or horizontal columns that are arranged in groups or clusters.
2.
From the chi square test we have Pearson chi square value and p value are 378.628 and 0.000. Since the pvalue is less than the 5% level of significance, it can be said that the test is significant.
3.
Experimental hypothesis says that there is an association between Guidance intervention plan and High School Diploma. From the above results we have seen that the test has significant results and we are failed to accept the Null hypothesis.
Therefore we can conclude that there is an association between Guidance Intervention plan and the High School Diploma.
ChiSquare Goodness of Fit Test
 Null Hypothesis: Number of AIDS cases follow the given distribution of ethnicity in the country.
 Decision: Reject the Null hypothesis.
 Reason for the decision: Since Asymp.sig (pvalue) is 0.000 which is significantly less than the given level of significance.
 Since we have rejected the null hypothesis , we have strong evidence to conclude that it doesn’t appear that pattern of AIDS cases in Neptune County corresponds to distribution of ethnic group in this country.
 In the Null hypothesis we have assumed that the distribution number of AIDS cases is the same as distribution of ethnicity group in the county. From the chisquare goodness of fit test gives us χ^{2} test statistics = 299.697 with degree of freedom 4 and pvalue <0.001. If we look at the expected and observed count of the given number of cases of AIDS, we will observe that White and Asian Pacific Islanderhas very large differences. These all finding suggests and support to reject the claim that the number of AIDS cases follows the given distribution of ethnicity in the county.