a. The null hypothesis isH0: The mean difference of alcohol rating is equal zerob. 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 haveThe 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 TWOa. 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 2dpAnswer: 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 THREEa. 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