SPSS Data Analysis: Descriptive Statistics

Descriptive Statistics

Part 1

Descriptive Statistics

Q1) Provide the means for the following variables:

Age:  28.2349 

HHSize: 3.09 

Descriptive Statistics
	N	Minimum	Maximum	Mean	Std. Deviation
Age	149	2.00	56.00	28.2349	10.68302
How many people are living or staying at your address, including yourself?	150	0	9	3.09	1.573
Valid N (listwise)	149

 Q2) Next, in the menu bar, click on ANALYZE/DESCRIPTIVE STATISTICS/FREQUENCIES

What percentage of the sample is:
Married:  45.6% 

Has less education than a college degree?

32.0%

 Worked for pay?  98.0%  

Q3)  For this assignment, you also need to generate a depression score for each participant. There are 20 items from the CES-D scale (variables CES_0001 to CES_0020).

I felt I was just as good as other people.
	Frequency	Percent	Valid Percent	Cumulative Percent
Valid	Rarely or none of the time (less than 1 day )	12	8.0	8.1	8.1
Some or a little of the time (1-2 days)	13	8.7	8.7	16.8
Occasionally or a moderate amount of time (3-4 days)	40	26.7	26.8	43.6
Most or all of the time (5-7 days)	84	56.0	56.4	100.0
Total	149	99.3	100.0
Missing	System	1	.7
Total	150	100.0

Reverse score of CES4 Categories
	Frequency	Percent	Valid Percent	Cumulative Percent
Valid	Rarely or none of the time (less than 1 day )	84	56.0	56.4	56.4
Some or a little of the time (1-2 days)	40	26.7	26.8	83.2
Occasionally or a moderate amount of time (3-4 days)	13	8.7	8.7	91.9
Most or all of the time (5-7 days)	12	8.0	8.1	100.0
Total	149	99.3	100.0
Missing	System	1	.7
Total	150	100.0

Yes, the above results do match what I would expect to find.

Q4) Transform->Compute Variable

There are times when we will want to compute a new variable based on the data we have. Create a new variable summing the 20 items from the CES-D scale.

The mean for Depression sum score is 13.57 while the mean for the Media use Score is 17.45

Descriptives
	Statistic	Std. Error
Depression sum score	Mean	13.5683	.89572
95% Confidence Interval for Mean	Lower Bound	11.7972
Upper Bound	15.3395
5% Trimmed Mean	12.7966
Median	11.0000
Variance	111.522
Std. Deviation	10.56042
Minimum	.00
Maximum	49.00
Range	49.00
Interquartile Range	13.00
Skewness	1.080	.206
Kurtosis	.706	.408
Media Use Score	Mean	16.3014	.58018
95% Confidence Interval for Mean	Lower Bound	15.1542
Upper Bound	17.4486
5% Trimmed Mean	16.0958
Median	16.0000
Variance	46.789
Std. Deviation	6.84027
Minimum	3.00
Maximum	37.00
Range	34.00
Interquartile Range	9.00
Skewness	.458	.206
Kurtosis	-.066	.408

Q5) Some of the data is missing for individual CES-D items. The important thing in dealing with missing data is to figure out if the data is missing randomly or if there is some pattern (reason) to why the data points are missing. Does there appear to be a pattern to the missing data?

How might one deal with the missing data? (Do not do this, simply report what you think based on our discussion this week).

Answer

There is no pattern to the missing data but rather they appear to be missing at random. Missing data might be dealt with by removing the missing cases or doing imputation for the missing cases.

Q6)  Examine the Descriptive Statistics output you generated for CESDTOT and Media Use for outliers. Remember that univariate outliers are those with very large standardized scores (z scores greater than 3.3) and that are disconnected from the distribution.  SPSS DESCRIPTIVES will give you the z scores for every case if you select save standardized values as variables and SPSS FREQUENCIES will give you histograms (use SPLIT FILE/ Compare Groups under DATA for grouped data).

Did you find any univariate outliers? Briefly write up your conclusion about univariate outliers, using data to back up your report.

Answer

Extreme Values
	Case Number	Value
Depression sum score	Highest	1	105	49.00
2	87	45.00
3	64	41.00
4	134	40.00
5	31	37.00a
Lowest	1	140	.00
2	129	.00
3	111	.00
4	66	.00
5	19	.00
Media Use Score	Highest	1	87	37.00
2	105	35.00
3	64	34.00
4	115	29.00
5	137	29.00
Lowest	1	102	3.00
2	70	3.00
3	100	5.00
4	35	5.00
5	111	6.00b
a. Only a partial list of cases with the value 37.00 are shown in the table of upper extremes.
b. Only a partial list of cases with the value 6.00 are shown in the table of lower extremes.

Yes there were cases of univariate outliers since we observed z score values greater than 3. 

Q7)  Finally, write up the results of your descriptive statistics analysis (Q1-6) in APA format as if you were describing the analysis for your dissertation (it will probably be only a paragraph). Make sure to include figures (e.g., a box plot).  The APA formatting may be difficult, but it will be helpful in the long run to spend some time learning it properly now.  

Answer

A descriptive analysis was performed to understand the distribution of the datasets. The mean age was found to be 28.23 (SD = 10.68) while the average household size (HHSize) was found to be 3.09 (SD = 1.57). This can be seen in the table presented below;

Descriptive Statistics
	N	Min.	Max.	M	SD
Age	149	2.00	56.00	28.23	10.68
Household size (HHSize)	150	0	9	3.09	1.57
Valid N (listwise)	149

In terms of the marital status, 45.6% (n = 68) of the participants were married and 98% (n = 147) worked for pay.

From the boxplots constructed, the plots revealed that outliers were present in the Media use score as well as the depression sum scores

Summary of Key Variables

There was however no pattern for the missing data but rather the data seemed to be missing at random. The missing data were random for the various variables and not associated with say a particular subject or particular item.

Part 2

Inferential Statistics

Paired T-test

The hypothesis of the test is given below

Results are presented below;

Paired Samples Statistics
	Mean	N	Std. Deviation	Std. Error Mean
Pair 1	Pre	20.81	45	7.159	1.067
Post	16.24	45	7.218	1.076

Paired Samples Correlations
	N	Correlation	Sig.
Pair 1	Pre & Post	45	.729	.000

Paired Samples Test
	Paired Differences	t	df	Sig. (2-tailed)
Mean	Std. Deviation	Std. Error Mean	95% Confidence Interval of the Difference
Lower	Upper
Pair 1	Pre – Post	4.565	5.291	.789	2.975	6.154	5.788	44	.000

A paired-samples t-test was conducted to compare pre-treatment scores to post-treatment scores. There was significant difference in the treatment scores for pre-treatment (M = 20.81, SD = 7.16) and post-treatment (M = 16.24, SD = 7.22) conditions; t(44) = 5.788, p = 0.000. These results suggest that there is a significant overall change between pre and post PTSD symptoms. The overall treatment effect was quite significant in the sense that people get better over time.

ANOVA

The first ANOVA we conducted was to compare the 4 groups on the first time point.  We sought to investigate whether the groups have a different amount of PTSD before they start treatment. The hypothesis tested is as follows;

Results are given below

ANOVA
Pre
	Sum of Squares	df	Mean Square	F	Sig.
Between Groups	18.566	3	6.189	.113	.952
Within Groups	2236.252	41	54.543
Total	2254.818	44

A one-way between subjects ANOVA was conducted to compare the PTSD before for four different independent groups. There was no significant effect of groups on PSTD scores at the 5% level of significance for the four conditions [F(3, 41) = 0.113, p = 0.952].

The above results clearly shows that the mean scores for the different groups are the same at the start. There is thee =fore no need to have a post-hoc test since there are no differences in the mean scores.

Repeated Measures ANOVA

Multivariate Testsa
Effect	Value	F	Hypothesis df	Error df	Sig.
Time	Pillai’s Trace	.451	17.677b	2.000	43.000	.000
Wilks’ Lambda	.549	17.677b	2.000	43.000	.000
Hotelling’s Trace	.822	17.677b	2.000	43.000	.000
Roy’s Largest Root	.822	17.677b	2.000	43.000	.000
a. Design: Intercept  Within Subjects Design: Time
b. Exact statistic

Mauchly’s Test of Sphericitya
Measure:   MEASURE_1
Within Subjects Effect	Mauchly’s W	Approx. Chi-Square	df	Sig.	Epsilonb
Greenhouse-Geisser	Huynh-Feldt	Lower-bound
Time	.628	20.010	2	.000	.729	.747	.500
Tests the null hypothesis that the error covariance matrix of the orthonormalized transformed dependent variables is proportional to an identity matrix.
a. Design: Intercept  Within Subjects Design: Time
b. May be used to adjust the degrees of freedom for the averaged tests of significance. Corrected tests are displayed in the Tests of Within-Subjects Effects table

Tests of Within-Subjects Effects
Measure:   MEASURE_1
Source	Type III Sum of Squares	df	Mean Square	F	Sig.
Time	Sphericity Assumed	833.372	2	416.686	28.070	.000
Greenhouse-Geisser	833.372	1.458	571.726	28.070	.000
Huynh-Feldt	833.372	1.495	557.500	28.070	.000
Lower-bound	833.372	1.000	833.372	28.070	.000
Error(Time)	Sphericity Assumed	1306.337	88	14.845
Greenhouse-Geisser	1306.337	64.136	20.368
Huynh-Feldt	1306.337	65.773	19.861
Lower-bound	1306.337	44.000	29.689
Tests of Within-Subjects Contrasts
Measure:   MEASURE_1
Source	Time	Type III Sum of Squares	df	Mean Square	F	Sig.
Time	Linear	748.638	1	748.638	32.442	.000
Quadratic	84.734	1	84.734	12.813	.001
Error(Time)	Linear	1015.361	44	23.076
Quadratic	290.976	44	6.613

Tests of Between-Subjects Effects
Measure:   MEASURE_1
Transformed Variable:   Average
Source	Type III Sum of Squares	df	Mean Square	F	Sig.
Intercept	40707.654	1	40707.654	322.283	.000
Error	5557.659	44	126.310

Mauchly’s Test of Sphericity indicated that the assumption of sphericity had been violated, , p  = .000, and therefore, a Greenhouse-Geisser correction was used. A repeated measures ANOVA with a Greenhouse-Geisser correction determined that mean PSTD scores differed statistically significantly between time points (F(1.458, 64.136) = 28.07, P = 0.000). Therefore, we can conclude that a long-term intervention elicits a statistically significant reduction in PSTD scores.

Repeated measures ANOVA

Multivariate Testsa
Effect	Value	F	Hypothesis df	Error df	Sig.
Time	Pillai’s Trace	.525	22.088b	2.000	40.000	.000
Wilks’ Lambda	.475	22.088b	2.000	40.000	.000
Hotelling’s Trace	1.104	22.088b	2.000	40.000	.000
Roy’s Largest Root	1.104	22.088b	2.000	40.000	.000
Time * Group	Pillai’s Trace	.375	3.151	6.000	82.000	.008
Wilks’ Lambda	.647	3.247b	6.000	80.000	.007
Hotelling’s Trace	.513	3.335	6.000	78.000	.006
Roy’s Largest Root	.437	5.976c	3.000	41.000	.002
a. Design: Intercept + Group  Within Subjects Design: Time
b. Exact statistic
c. The statistic is an upper bound on F that yields a lower bound on the significance level.

Mauchly’s Test of Sphericitya
Measure:   MEASURE_1
Within Subjects Effect	Mauchly’s W	Approx. Chi-Square	df	Sig.	Epsilonb
Greenhouse-Geisser	Huynh-Feldt	Lower-bound
Time	.714	13.488	2	.001	.777	.862	.500
Tests the null hypothesis that the error covariance matrix of the orthonormalized transformed dependent variables is proportional to an identity matrix.
a. Design: Intercept + Group  Within Subjects Design: Time
b. May be used to adjust the degrees of freedom for the averaged tests of significance. Corrected tests are displayed in the Tests of Within-Subjects Effects table.

Tests of Within-Subjects Effects
Measure:   MEASURE_1
Source	Type III Sum of Squares	df	Mean Square	F	Sig.
Time	Sphericity Assumed	778.949	2	389.474	32.412	.000
Greenhouse-Geisser	778.949	1.555	500.953	32.412	.000
Huynh-Feldt	778.949	1.723	452.034	32.412	.000
Lower-bound	778.949	1.000	778.949	32.412	.000
Time * Group	Sphericity Assumed	321.009	6	53.501	4.452	.001
Greenhouse-Geisser	321.009	4.665	68.815	4.452	.002
Huynh-Feldt	321.009	5.170	62.095	4.452	.001
Lower-bound	321.009	3.000	107.003	4.452	.008
Error(Time)	Sphericity Assumed	985.328	82	12.016
Greenhouse-Geisser	985.328	63.752	15.456
Huynh-Feldt	985.328	70.651	13.946
Lower-bound	985.328	41.000	24.032

Tests of Within-Subjects Contrasts
Measure:   MEASURE_1
Source	Time	Type III Sum of Squares	df	Mean Square	F	Sig.
Time	Linear	698.646	1	698.646	38.467	.000
Quadratic	80.303	1	80.303	13.680	.001
Time * Group	Linear	270.703	3	90.234	4.968	.005
Quadratic	50.306	3	16.769	2.857	.049
Error(Time)	Linear	744.658	41	18.162
Quadratic	240.670	41	5.870

Tests of Between-Subjects Effects
Measure:   MEASURE_1
Transformed Variable:   Average
Source	Type III Sum of Squares	df	Mean Square	F	Sig.
Intercept	40618.546	1	40618.546	331.760	.000
Group	537.877	3	179.292	1.464	.238
Error	5019.782	41	122.434

Part 3

1. Open the data file (called RSM801Week3.sav). Explore the data file. Note, you will not analyze all of these variables. Try to find the variables that are relevant to the study description above. Write the names of the two variables here:

Answer

The variables are voter intention index and Ebola search volume index

1. Run a correlation analysis to test if there is an association between the Ebola search volume index and the voter intention index. A correlation coefficient indicates the strength and direction of a relationship between two variables.  (select Analyze – correlate – bivariate).  Be sure to check off Options – Means and standard deviations. In SPSS, be sure to check the box of the correct correlation type (Pearson, Spearman or Kendall’s tau-b) for this data.  Indicate the correlation type and why this is the right choice for these variables:

Answer

Correlation Type: Pearson Correlation  

Why: Because the data is an interval scale

Correlations
	Voter Intention Index	Ebola Search Volume Index
Voter Intention Index	Pearson Correlation	1	.505*
Sig. (2-tailed)		.012
N	24	24
Ebola Search Volume Index	Pearson Correlation	.505*	1
Sig. (2-tailed)	.012
N	24	65
*. Correlation is significant at the 0.05 level (2-tailed).

1. Write an APA formatted sentence describing the mean and standard deviation for these two variables.   

Answer

The average voter intention index was 1.12 (SD = 0.89) while the average Ebola search volume index was 24.17 (SD = 22.85).

Descriptive Statistics
	N	Minimum	Maximum	Mean	Std. Deviation
Voter Intention Index	24	-.40	2.40	1.1167	.88596
Ebola Search Volume Index	65	2.86	70.86	24.1712	22.84665
Valid N (listwise)	24

1. Report the results of the correlation in an APA statement (e.g., r (N-2) = .xx, p = .yyy. Be sure to include degrees of freedom and statistical significance.  

Answer

The two variables had a moderate positive relation, r(24) = .49, p = 0.012.

1. Write a sentence interpreting what these findings mean.  That is, following periods characterized by especially heavy periods of Ebola related Internet search activity, was this search activity related or unrelated to voting intentions? If related, were U.S. voters more likely to vote for a Republican or Democrat candidate? Does the relationship appear to be direct (i.e., positive) or inverse (i.e, negative)?

Answer

Results showed that there is a significant positive relationship between voter intention index and Ebola search volume index. This means that an increase in the Ebola search volume index would result to an increase in voter intention index. On the other hand a decrease in Ebola search volume index would result to a subsequent decrease in voter intention index

1. [This question is optional extra credit] Next, to test whether the association between these variables is stronger during the period just prior to and after the Ebola outbreak, select only the scores from the two-week period including the last week of September and the first week of October (use Data – Select Cases – If Condition is Satisfied, specifying the condition (1) that would meet this criteria). Re-run the correlation analyses for the association between Ebola search volume index and voter intention index. Also, compute the correlation analysis between Daily Ebola search volume and voter intention index. Which correlation value was stronger? Write an APA statement summarizing these results.  

Answer

Correlations
	Voter Intention Index	Ebola Search Volume Index	Daily Ebola Search Volume
Voter Intention Index	Pearson Correlation	1	.988**	.607
Sig. (2-tailed)		.000	.111
N	8	8	8
Ebola Search Volume Index	Pearson Correlation	.988**	1	.693**
Sig. (2-tailed)	.000		.006
N	8	14	14
Daily Ebola Search Volume	Pearson Correlation	.607	.693**	1
Sig. (2-tailed)	.111	.006
N	8	14	14
**. Correlation is significant at the 0.01 level (2-tailed).

A Pearson correlation test was performed to check the relationship between Ebola search volume index and Daily Ebola search volume with the Voter intention index during the period just prior to and after the Ebola outbreak. Results showed that a very strong positive relationship between Voter intention index and Ebola search volume index, r(8) = 0.988, p = 0.000. A strong positive but insignificant relationship was observed between Voter intention index and Daily Ebola search volume, r(8) = 0.607, p = 0.111.

Correlation value was stronger between voter intention index and Ebola search volume index.

1. Make sure that all the data is selected (i.e., that the Select cases is set to “All Cases”). Run a correlational analysis between ALL of the scale variables.

Which one is the strongest pair?

Which pair has the weakest relationship?

Correlations
	Voter Intention Index	Ebola Search Volume Index	Daily Ebola Search Volume
Voter Intention Index	Pearson Correlation	1	.505*	.169
Sig. (2-tailed)		.012	.430
N	24	24	24
Ebola Search Volume Index	Pearson Correlation	.505*	1	.831**
Sig. (2-tailed)	.012		.000
N	24	65	65
Daily Ebola Search Volume	Pearson Correlation	.169	.831**	1
Sig. (2-tailed)	.430	.000
N	24	65	65
*. Correlation is significant at the 0.05 level (2-tailed).
**. Correlation is significant at the 0.01 level (2-tailed).

The stronger pair is between Voter intention index and Ebola Search volume index

The pair with weak relationship is between Voter intention index and Daily Ebola Search volume.

1. Prepare a series of scatterplots (making sure to follow APA-style guidelines). Select Graphs – Chart Builder and then choose “Scatter/Dot” as the chart type. First, depict the relationship between day and the voter intention index for the month of September (demonstrating the relationship for voter intention index for the month prior to the Ebola outbreak was announced). Please note that you will need to change the select function as you did in Question 6 (but this time using the variable month, selecting September as the month). Include the figure (formatting with an APA style title) and write a sentence describing the relationship.

Answer

A negative relationship was observed between voter intention index for the month prior to the Ebola outbreak was announced and the daily Ebola search volume.

1. Second, depict the relationship between day and the voter intention index for the last week of September (i.e., the week prior to the outbreak was announced, Sept 24-30). Include the figure (formatting with an APA style title) and write a sentence describing the relationship.

Answer

A positive relationship was observed between voter intention index for the last week of September and the daily Ebola search volume.

1. Third, depict the relationship between day and the voter intention index for the month of October (i.e., the month after the outbreak was announced). Include the figure (formatting with an APA style title) and write a sentence describing the relationship.

Answer

A negative relationship was observed between voter intention index for index for the month of October and the daily Ebola search volume.

1. Finally, depict the relationship between day and the voter intention index for the first week of October (i.e., 10/1-10/7, the week after the Ebola outbreak was announced). Include the figure (formatting with an APA style title) and write a sentence describing the relationship.

Answer

A positive relationship was observed between voter intention index for the first week of October and the daily Ebola search volume.

1. Does viewing these graphs influence your interpretation of the correlation analyses above? How so?

Answer

Yes viewing these graphs influence my interpretation of the correlation analyses above. This is because daily Ebola search volume influences voter intention index differently depending on the period when the Ebola was announced.

1. Conduct a t-test on voter intention index to assess whether voter intentions differed whether they were expressed before (group = 1) or after (group = 2) the initial Ebola outbreak was announced (Variable: newmonth).  Based on this data, put a check in front of the type of t-test you should carry out (and carry it out, using the Pace book and/or notes from RSM701).

Answer

Independent sample t-test 

Group Statistics
	Two.weeks.prior.to.outbreak.only	N	Mean	Std. Deviation	Std. Error Mean
Voter Intention Index	Not in the 2 week window	16	1.5750	.62450	.15612
Within 2 week window	8	.2000	.55032	.19457

Independent Samples Test
	Levene’s Test for Equality of Variances	t-test for Equality of Means
F	Sig.	t	df	Sig. (2-tailed)	Mean Difference	Std. Error Difference	95% Confidence Interval of the Difference
Lower	Upper
Voter Intention Index	Equal variances assumed	.167	.687	5.276	22	.000	1.37500	.26063	.83449	1.91551
Equal variances not assumed			5.512	15.850	.000	1.37500	.24946	.84575	1.90425

1. Examining your output, write a sentence reporting the mean support for Republican (relative to democratic) candidates for the Month prior to the outbreak as well as proceeding the outbreak.   Please be sure to use terms such as more, less, or about the same to indicate whether support was greater prior to or after the outbreak was announced.

Answer

The mean support for Republican (relative to democratic) candidates for the Month prior to the outbreak as well as proceeding the outbreak was more than during the outbreak period. His shows that support was greater prior to or after the outbreak was announced

1. Assess whether this change was statistically significant.  Write an APA statement with the t-test findings (e.g., t (df) = xx.xx, p = .yyy).  Don’t forget to use the Levene’s test to determine whether you should be reporting the row of findings that consider “equal variances assumed” (when Levene’s test p > .05) or equal variances not assumed (when Levene’s test p < .05). Interpret this finding.

Answer

An independent samples t-test was performed to compare the average voter intention index. The Levene’s test showed that we assume equal variances (p-value < 0.05). Results showed that the average voter intention index Not in the 2 week window (M = 1.58, SD = 0.62, N = 12) was significant different with the average voter intention index within 2 week window (M = 0.20, SD = 0.55, N = 8), t (22) = 5.276, p < .05, two-tailed. The difference of 1.375 showed a significant difference. Essentially results showed that Ebola outbreak did significantly reduce the voter intention index

1. What does the t-test tell you? What does the correlation tell you? How are each useful to understanding this data?

Answer

T-test tells us the difference in the average voter intention index for the two time points while correlation tells us the relationship that exists between the voter intention index and daily Ebola search volume. The two tests are important since they are able to tell us the relationship that the different factors have on the voter intention index.