a. The null hypothesis is

H0: The mean difference of alcohol rating is equal zero

b. The appropriate statistical test to use is the paired sample t test.

c. The analysis

Paired Samples Statistics

Mean

N

Std. Deviation

Std. Error Mean

Pair 1

rating before alcohol

6.99

10

1.08

.34

rating after 2units alcohol

5.27

10

1.86

.59

Paired Samples Correlations

N

Correlation

Sig.

Pair 1

rating before alcohol & rating after 2units alcohol

10

-.001

.997

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

rating before alcohol – rating after 2units alcohol

1.72

2.16

.68

.17

3.26

2.51

9

.03

Interpretation: it can be seen that the p-value = 0.03 which is less than the 0.05level of significance. Since the p-value is less than 0.05level of significance we have statistical reason to reject the null hypothesis and conclude that the true mean difference

between the ratings of alcohol is not equal zero. It can also be seen on the correlation table that the correlation coefficient is -0.001 which implies a weak negative correlation.

d. Supposing the analyzer uses ANOVA, we have

The planned contrast name to be used here is the pos hoc test.

e. Result of ANOVA

rating state

Mean

N

Std. Deviation

rating before alcohol

6.990

10

1.0806

rating after 2units alcohol

5.270

10

1.8691

rating after 4units alcohol

4.540

10

.7351

Total

5.600

30

1.6448

It can be seen that the means of each alcohol rating are different

Is there evidence of a significant overall effect of alcohol consumption on perceptions of fearfulness?

ANOVA

output

Sum of Squares

df

Mean Square

F

Sig.

Between Groups

31.646

2

15.823

9.126

.001

Within Groups

46.814

27

1.734

Total

78.460

29

Here, it can be seen on the table above that the p-value = 0.001 which is less than 0.05level of significance. Since the p-value is less than 0.05level of significance we can conclude that the alcohol consumption on perceptions of fearfulness is significantly different for at least one of the alcohol rating state.

It shows us of what we see in the compare means output. The points show the average of each group. It’s much easier to see from this graph that the rating before alcohol had the slowest , while the rating after 4 units alcohol had the fastest mean rating state.

Multiple Comparisons

output

LSD

(I) rating state

(J) rating state

Mean Difference (I-J)

Std. Error

Sig.

95% Confidence Interval

Lower Bound

Upper Bound

rating before alcohol

rating after 2units alcohol

1.7200*

.5889

.007

.512

2.928

rating after 4units alcohol

2.4500*

.5889

.000

1.242

3.658

rating after 2units alcohol

rating before alcohol

-1.7200*

.5889

.007

-2.928

-.512

rating after 4units alcohol

.7300

.5889

.226

-.478

1.938

rating after 4units alcohol

rating before alcohol

-2.4500*

.5889

.000

-3.658

-1.242

rating after 2units alcohol

-.7300

.5889

.226

-1.938

.478

*. The mean difference is significant at the 0.05 level.

Is there evidence of a significant change in perceptions of fearfulness after drinking the first 2 units of alcohol?

Answer: from the table above it can be seen that there is significant change after drinking the first 2units of alcohol since the p-value = 0.007 and the mean difference is 1.72.

Is there evidence of a significant change in perceptions of fearfulness between drinking the first 2 units of alcohol and a further 2 units of alcohol?

Answer: from the table above it can be seen that there is not a significant change between drinking the first 2units of alcohol and a further 2units of alcohol since the p-value = 0.22.

f. Is there evidence of a violation of the assumption of sphericity? State the appropriate statistics to justify your answer.

Answer: since the p-value = 0.445 which is greater than 0.05level of significance, we have statistical reason to conclude that there is no sphericity issue here. The test statistics used is the Mauchly’s test of sphericity

Mauchly’s Test of Sphericityb

Measure:rating

Within Subjects Effect

Mauchly’s W

Approx. Chi-Square

df

Sig.

Epsilona

Greenhouse-Geisser

Huynh-Feldt

Lower-bound

factor1

.817

1.620

2

.445

.845

1.000

.500

g. Create a bar graph to display this data, including all the required elements for publication in a scientific report.

QUESTION TWO

a. Using the “Question 2 data.sav”, create a scatter graph of the relationship between defendants’ physical attractiveness and sentence length, making sure to include all the elements appropriate to a scientific publication.

b. From looking at this graph, what would you hypothesis to the relationship between attractiveness and length of sentence?

Answer: it can be seen on the chart above that there is relationship between attractiveness and length of sentence but it seems to be a negative relationship.

c. What is the statistical correlation between attractiveness and sentence length in this sample? How much variation length is predicted by variation in attractiveness? Report your analysis, with statistics, in a way that is suitable for a scientific report.

Correlations

attractiveness

sentence

attractiveness

Pearson Correlation

1

-.528**

Sig. (2-tailed)

.000

N

400

400

sentence

Pearson Correlation

-.528**

1

Sig. (2-tailed)

.000

N

400

400

**. Correlation is significant at the 0.01 level (2-tailed).

Answer: for the first question, the correlation between attractiveness and sentence length is significant at 0.01level of significance since p-value = 0.000. This implies that there is relationship between the two variables but a weak negative relationship because the correlation coefficient is -0.528.

Secondly, the amount of variation length predicted by variation in attractiveness is 27.1 since the value for R-squared = 0.271.

d. The researchers are also interested in the relationship between defendants’ age and sentence length. They perform a linear regression, to explore this relationship. Perform a linear regression yourself, and report the results in full. Include a sentence, in plain English to describe the relationship between age and sentence length. Report statistics to 2dp

Answer:

Descriptive Statistics

Mean

Std. Deviation

N

sentence

3.003

1.2295

400

Age

23.35

4.021

400

Correlations

sentence

Age

Pearson Correlation

sentence

1.000

-.413

Age

-.413

1.000

Sig. (1-tailed)

sentence

.

.000

Age

.000

.

N

sentence

400

400

Age

400

400

It can be seen that Age and Sentence length are weak negative correlated since the correlation coefficient is -0.413.

Model Summary

Model

R

R Square

Adjusted R Square

Std. Error of the Estimate

Change Statistics

R Square Change

F Change

df1

df2

Sig. F Change

1

.413a

.171

.169

1.1210

.171

81.929

1

398

.000

a. Predictors: (Constant), Age

ANOVAb

Model

Sum of Squares

df

Mean Square

F

Sig.

1

Regression

102.959

1

102.959

81.929

.000a

Residual

500.160

398

1.257

Total

603.118

399

a. Predictors: (Constant), Age

b. Dependent Variable: sentence

Coefficientsa

Model

Unstandardized Coefficients

Standardized Coefficients

t

Sig.

B

Std. Error

Beta

1

(Constant)

5.953

.331

18.003

.000

Age

-.126

.014

-.413

-9.051

.000

a. Dependent Variable: sentence

It can also be seen on the table above that after formulating the regression model. That is

Sentence length = 5.95 – 0.13age. This implies that a unit increase in age will decrease the sentence length by 0.13 which implies that there is relationship between age and sentence length.

e. The defendants in the sample are all aged between 18 and 40. Using your output from part d, estimate what the average sentence would be for a 47 year old.

Answer: since the regression model is given as Sentence length = 5.95 – 0.13age, substituting 47 into the model we have sentence length = -0.16. This implies that the average sentence will be -0.16 for age 47.

f. The researchers are interested in whether attractiveness still predicts sentence length after controlling for age. Perform a hierarchical (two-step) linear regression to test this. Is attractiveness still a predictor of sentence length when age is included in the model, and is the direction of the effect still the same? How much variation in sentence length does attractiveness explain after controlling for age? Write your results in full.

The output of the analysis

Descriptive Statistics

Mean

Std. Deviation

N

sentence

3.003

1.2295

400

Age

23.35

4.021

400

attractiveness

3.615

1.4641

400

Model Summary

Model

R

R Square

Adjusted R Square

Std. Error of the Estimate

Change Statistics

R Square Change

F Change

df1

df2

Sig. F Change

1

.845a

.714

.712

.6592

.714

4.954E2

2

397

.000

a. Predictors: (Constant), attractiveness, Age

ANOVAb

Model

Sum of Squares

df

Mean Square

F

Sig.

1

Regression

430.580

2

215.290

495.370

.000a

Residual

172.538

397

.435

Total

603.118

399

a. Predictors: (Constant), attractiveness, Age

b. Dependent Variable: sentence

Coefficientsa

Model

Unstandardized Coefficients

Standardized Coefficients

t

Sig.

B

Std. Error

Beta

1

(Constant)

10.503

.255

41.108

.000

Age

-.218

.009

-.712

-24.583

.000

attractiveness

-.668

.024

-.795

-27.456

.000

a. Dependent Variable: sentence

Is attractiveness still a predictor of sentence length when age is included in the model, and is the direction of the effect still the same?

Answer: Yes, attractiveness is a predictor of sentence length when age is included in the model and also the direction is still the same because it also has negative effect on the dependent variable sentence length.

How much variation in sentence length does attractiveness explain after controlling for age?

Answer: The amount of variation explained in sentence length by attractiveness is 71.4%.

QUESTION THREE

a. Perform an analysis to test the researchers’ hypothesis, including all the required follow up tests and descriptive that you believe are necessary. Write up this analysis in full, describing any statistical assumptions that you think are necessary, or explaining why you think that such tests are not necessary. You should report effect sizes as well as statistical significance.

Answer: the analysis

Between-Subjects Factors

Value Label

N

Thirst

0

Not thirsty

80

1

Thirsty

80

Charity

0

No drought

80

1

Drought

80

Descriptive Statistics

Dependent Variable:Donation

Thirst

Charity

Mean

Std. Deviation

N

Not thirsty

No drought

3.3441

1.21184

40

Drought

4.5197

1.07498

40

Total

3.9319

1.28270

80

Thirsty

No drought

5.1382

1.40957

40

Drought

7.1109

1.13906

40

Total

6.1245

1.61451

80

Total

No drought

4.2411

1.58767

80

Drought

5.8153

1.70613

80

Total

5.0282

1.82266

160

Levene’s Test of Equality of Error Variancesa

Dependent Variable:Donation

F

df1

df2

Sig.

.593

3

156

.620

Tests the null hypothesis that the error variance of the dependent variable is equal across groups.

a. Design: Intercept + Thirst + Charity + Thirst * Charity

It can be seen on the table above that p-value = 0.620 which is greater than 0.05level of significance. We have statistical reason to fail to reject the null hypothesis and conclude that the variance are equal.

Tests of Between-Subjects Effects

Dependent Variable:Donation

Source

Type III Sum of Squares

df

Mean Square

F

Sig.

Partial Eta Squared

Intercept

Hypothesis

4045.260

1

4045.260

40.813

.099

.976

Error

99.118

1

99.118a

Thirst

Hypothesis

192.308

1

192.308

30.261

.114

.968

Error

6.355

1

6.355b

Charity

Hypothesis

99.118

1

99.118

15.597

.158

.940

Error

6.355

1

6.355b

Thirst * Charity

Hypothesis

6.355

1

6.355

4.302

.040

.027

Error

230.431

156

1.477c

a. MS(Charity)

b. MS(Thirst * Charity)

c. MS(Error)

From the table above it can be seen that we have a statistically significant interaction at the p = 0.040level. We can see from the table above that there was no statistically significant difference in mean differences Thirst (p = .114), there is also no statistically significant differences between charity (p =0.158).

Lack of Fit Tests

Dependent Variable:Donation

Source

Sum of Squares

df

Mean Square

F

Sig.

Partial Eta Squared

Lack of Fit

.000

0

.

.

.

.000

Pure Error

230.431

156

1.477

§ If the P-value is smaller than the significance level ?, we reject the null hypothesis in favor of the alternative. We state “there is sufficient evidence at the ? level to conclude that there is lack of fit in the model. Here it can be seen that p-value is less than 0.05level of significance we can conclude that there is lack of fit in the model.

b. Fill in the following table with the statistics you produced in your analysis. Pay attention to the number of decimal places required.

Effect:

F-value (2 decimal places)

Partial eta squared (2 decimal places)

Thirst condition

30.26

0.97

Campaign type

15.60

0.94

Thirst condition*Campaign type

4.30

0.03

c. Produce an appropriate graph to show your findings