# Descriptive and Inferential Statistics Analysis for Desklib

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This analysis provides a descriptive and inferential statistics analysis for Desklib, an online library for study material with solved assignments, essays, dissertation etc. It includes means, frequencies, depression score, outliers, and ANOVA results. The analysis is presented in APA format with figures and tables.

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Statistics

Student Name:

Instructor Name:

Course Number:

2 September 2018

Student Name:

Instructor Name:

Course Number:

2 September 2018

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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.

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.

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?

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?

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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

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.

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.

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.

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From the boxplots constructed, the plots revealed that outliers were present in the Media use

score as well as the depression sum scores

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

score as well as the depression sum scores

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

H0 : μPre=μPost

μPre=mean for the pre

μPre=mean for the post

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

Paired T-test

The hypothesis of the test is given below

H0 : μPre=μPost

μPre=mean for the pre

μPre=mean for the post

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;

H0 : μ1=μ2=μ3=μ4

H A : At least one of the meansis different

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.

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;

H0 : μ1=μ2=μ3=μ4

H A : At least one of the meansis different

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.

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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

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,

χ2 (2)=20.01, 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.

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,

χ2 (2)=20.01, 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.

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.

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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

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

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

2. 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

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

2. 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

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*. Correlation is significant at the 0.05 level (2-tailed).

3. 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

4. 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.

5. 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

3. 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

4. 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.

5. 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

[AFTER ANSWERING Q5, CONTINUE ON TO NEXT PAGE FOR Q’s 6 – 16]

6. [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).

decrease in Ebola search volume index would result to a subsequent decrease in voter

intention index

[AFTER ANSWERING Q5, CONTINUE ON TO NEXT PAGE FOR Q’s 6 – 16]

6. [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.

7. 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.

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.

7. 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.

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8. 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.

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.

9. 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.

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.

10. 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.

11. 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

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.

11. 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

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A positive relationship was observed between voter intention index for the first week of

October and the daily Ebola search volume.

12. 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.

13. 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).

October and the daily Ebola search volume.

12. 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.

13. 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.outbrea

k.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

Differen

ce

Std.

Error

Differen

ce

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

14. 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

Independent sample t-test

Group Statistics

Two.weeks.prior.to.outbrea

k.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

Differen

ce

Std.

Error

Differen

ce

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

14. 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

15. 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

16. 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.

announced

15. 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

16. 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.

1 out of 25

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