SPSS Final. Please read the instructions. Seriously. Read the instructions.

Please turn in your SPSS final by December 5th by 7:00 pm. You will need to attach:

1. Your answers in Word or Rich text format INCLUDING necessary SPSS tables

2. Your SPSS output3. Your SPSS data4. Honor Pledge with your typed signature (read before you start)

This combines many of the different things that we have done over the semester.

Part of the final is new material, equivalent to one SPSS homework, covering correlation & regression. Please remember that this part is worth 50 points.

1. (16 pts) Using the World95 data: a. Producing scatter-plots for these relationships:

i. (a) percentage of females who read (literacy) and average female life expectancy;

ii. (b) fertility and average female life expectancy;

This shows a positive linear relationship

This shows a negative linear relationship

iii. (c) GDP and female life expectancy; and

iv. (d) aids cases and average female life expectancy

v. Interpret whether the results show linear or curvilinear relationships. (stated in the textboxes)

vi. Which of these relationships should you NOT use a Pearson correlation for? Explain.-You would not use a Pearson Correlation for the curvilinear relationship between GDP and average female life expectancy because Pearson’s correlation only measures linear relationships. The same applies to the no relationship between Aids cases and average female life expectancy.

b. Reporting and interpreting the correlations, p-values, and effect sizes for (indicate whether they significant correlations (alpha= .05) and include the appropriate table):

This shows a curvilinear relationship

This shows no relationship

i. (a) percentage of females who read (literacy) and female life

expectancy; -There was a significant positive correlation between percentage of females who read (M = 67.26, SD = 28.607) and average female life expectancy (M = 70.16, SD = 10.572), r (83) = .819, p < .001.- r2 = .6708, 67.08% of the variance in percentage of average female life expectancy can be accounted for by the percentage of females who read.

ii. (b) infant mortality and female life expectancy; and

-There was a significant negative correlation between infant mortality (M = 42.313, SD = 38.0792) and average female life expectancy (M = 70.16, SD = 10.572), r (107) = -.962, p < .001.-r2 = .925, 92.5% of the variance in percentage of average female life expectancy can be accounted for by the infant mortality rate.

iii. (c) fertility and female life expectancy.

-There was a significant negative correlation between fertility (M= 3.563, SD = 1.9025) and average female life expectancy (M = 70.16, SD = 10.572), r = -.838, p < .001.-r2 = .702, 70.2% of the variance in average female life expectancy can be accounted for by the fertility.

2. (22 pts) Complete a regression analysis to predict female life expectancy (dependent) using percentage of females who read as the predictor (independent). When you are doing the regression, click on “Statistics” and select “estimates”, “Model fit”, and “R-square change”. Click on Save and select “Unstandardized predicted values” and “Unstandardized residual values.”

a. Look at the data view. You now have two additional columns. i. Explain what these two variables are.

-The first column is the unstandardized predicated value for each data score (using the regression model) and the second column is the unstandardized residual for each data score.

ii. How do these columns related to the standard error of the estimate?-The second column shows how much error there is in each estimate from the first column from the actual data.

b. Look at the output view.

i. Explain R, R-square, and standard error of the estimate in relation to the predictor and criterion variables.- There was a significant positive correlation between percentage of females who read (M = 67.26, SD = 28.607) and average female life expectancy (M = 70.16, SD = 10.572), r (83) = .819, p < .001.- r2 = .670, 67.0% of the variance in percentage of average female life expectancy can be accounted for by the percentage of females who read.-On average, there is an error of 6.195 when predicting the average female life expectancy

ii. Explain whether the regression equation is “significant”. What does

this mean?-yes, the regression equation is significant (p < .001). This means that using the predictor (% of females who read) is better than using the mean to predict average female life expectancy.

c. Look at the coefficient table. i. Give the unstandardized regression equation

Y-hat = 47.170 + .307(females who read)

ii. Give the standardized regression equation. Zy-hat = .819(Zfemales who read)

iii. Using the unstandardized equation, predict the life expectancy for women in a country with:

1. a 20% literacy rateY-hat = 47.170 + .307(20) = 53.31

2. a 95% literacy rate (You will use the numbers “20” and “95” not .20 and .95)Y-hat = 47.170 + .307(95) = 76.335

d. Based on this information, a researcher concluded: “There is a serious problem with literacy worldwide. Based on these data, we strongly encourage governments around the world to put their resources into improving literacy, especially among women. This will result in an increase in the life expectancy of women, an important goal for all governments.” Is the researcher correct in her recommendation? Explain.-The researcher cannot assume this because he does not have causation. While there is a strong correlation between these two variables, literacy is not the only variable involved in female life expectancy.

3. (12 pts) Now, use a regression analysis to predict female life expectancy using both percentage of females who read and fertility rate.

a. Provide the SPSS regression tables.

b. Look at the coefficient table:

i. Give the unstandardized regression equationY-hat = 66.212 + .169(females who read) - .2488(fertility)

ii. Give the standardized regression equationZy-hat = .452(Zfemales who read) - .437(Zfertility)

iii. Explain R & R-square in relation to the question.-R = .853, R is the Predicted female life expectancy – the actual female life expectancy based upon the two predictors.-R2 = .727, .72.7% of the variance in life expectancy can be explained by the % of females who read and the fertility rate.

iv. Explain the standard error of the estimate. Compare this to the standard error for the previous regression. Which regression has a better standard error of the estimate? Explain.-On average, there is a 5.673 error when predicting female life expectancy. This is a smaller error than the previous regression. This means that this multiple regression has a better standard error of the estimate, meaning that it is more accurate in its estimate of female life expectancy.

v. Is the regression equation significant? Explain what this means and how you know.-yes, the regression equation is significant (p < .001). Using the two predictors is better than using the mean to predict the average female life expectancy.

vi. Is either a significant predictor? Explain your answers. -Females who read is a significant predictor because the p-value is less than .001 as well as the predictor of fertility.

vii. Should you use both of these predictors? Explain why or why not.-Yes you should use both predictors. Because both predictors are significant and have a lower standard error of the estimate, both would be more accurate in predicting the average female life expectancy.

SPSS Output:

SPSS Data

Cumulative Portion of final: 80 pts

Make sure you read the description of the data and the instructions.

Use the Psyc 210 final data set (link is online) to answer the questions. This is part of a national data set called the General Social Survey. It is given to a representative sample of the population in the United States. The variables are described in the data set as well as the ranges of those values. Make sure you look at the values so that you understand what a higher/lower number means.

For each of the following questions, (a) indicate the statistic you chose, (b) explain why you used that statistic, and (c) state your conclusion with regard to the original variables, including a statistical sentence. Make sure to include the necessary (d) tables from SPSS, and (e) graphs where indicated. As you can see, you only need 2 sentences in most cases to describe your results.

Report means and standard deviations when appropriate. Also, report confidence intervals if appropriate.

EXAMPLE QUESTION with APPROPRIATE ANSWERS:

Question: Is there a difference between the average income of this sample and the population (μ= 30)?

I chose a single sample t test because we are comparing a mean from our sample to the mean of the population. (Here I indicated what statistic and why; parts a & b)

“Our sample (M=25, SD=3) had a lower average income than the general population (μ=30), t(32)= 3.21, p = .02. (Here I indicated the groups, the means & SD where appropriate, the direction of the difference, the statistic, the df, the test statistic, and the p-value; part c).

[tables] (Here I included the appropriate tables; parts d and/or e here)

Formatting: 5 pts. Including all materials: 10 pts

Questions:

1. Is there a difference between men & women (gender) on their incomes (rincom06)? On their ideal number of children (chldideal)? Include bar graphs for each analysis. 10 pts

-I chose an independent samples t test because we are comparing the difference in men and women (two separate factors) on income and their ideal number of children (separately)

-Males (M = 16.63, SD = 5.460) had statistically significant higher average reported income than females (M = 14.39, SD = 6.201), t(911.201) = 5.850, p < .001, d =0.39, 95% CI [ 1.489, 2.992]

-There is not a statistically significant difference between males (M = 3.04, SD = 1.914) and females (M = 3.16, SD = 1.920) in reported ideal number of children, t(909) = -.919, p = .359, d =-0.061, 95% CI [-.367, .133].

Tables for Income:

Tables for Ideal Number of Children:

2. Do people’s marital satisfaction ratings change after they have children? There are two measures of marital satisfaction, where higher scores indicate higher satisfaction. Show a graph. 8 pts

-I chose a dependent samples t test because we are finding the difference in marital satisfaction before and after having children.

- People’s marital satisfaction before children (M = 4.00, SD = .97) is statistically significantly higher than it is after children (M = 3.70, SD = 1.00), t(99) = 2.346, p = .021, d =.235, 95% CI: [ .047, .563 ].

3. Do people’s become less religious after attending college? There are two measures of religious belief, where higher scores indicate more religious. Show a graph. 8 pts.

-I chose a dependent samples t test because we are finding the difference in religious belief before and after college.

-On average, people are more religious before college (M = 5.01, SD = .977) than they are after college (M = 4.05, SD = 1.03), t(98) = 6.76, p < .001, d = .689, 95% CI: [.681, 1.247].

4. You are interested in how region of the country people are from (East/NE, South, & West) and political party orientation (Democrat, Independent, Republican) are related to how often people attend religious services. Describe your results thoroughly (HINT HINT) and include a graph (include an explanation of the graph). 24 pts

-I chose a 3x3 between groups factorial ANOVA because we are testing two factors with 3 levels on how often people attend religious services.

-There was a statistically significant main effect for region, F (2, 1387) = 18.569, p < .001, R2 = .026. From the results, we can see that there is a significant difference between East/ North East and South (p < .001) as well as a significant difference between West and South (p < .001), but there is not a statistically significant difference between West and East/ North East ( p = .366).

-There was a statistically significant main effect for political party orientation, F (2, 1387) = 7.876, p = .001, R2 = .010. From the results, we can see that there is a significantly significant difference between Democrat and Republican (p = .012) as well as between independent and republican ( p <.001), but there is not a statistically significant difference between Independent and Democrat (p = .099).

-There was a significant interaction between the region of the country people are from and political party orientation on reported religious service attendance, F(4, 1387) = 3.856, p = .004, R2 = .011, 1.1% of the variance in attending religious service is explained by the interaction between region and political party orientation.

Tests of Between-Subjects Effects

Dependent Variable: HOW OFTEN R ATTENDS RELIGIOUS SERVICES

Source

Type III Sum of

Squares df Mean Square F Sig.

Partial Eta

Squared

Corrected Model 528.906a 8 66.113 9.164 .000 .050

Intercept 12189.850 1 12189.850 1689.621 .000 .549

CountryRegion 267.939 2 133.970 18.569 .000 .026

PoliticalParty 99.207 2 49.604 6.876 .001 .010

CountryRegion *

PoliticalParty111.266 4 27.816 3.856 .004 .011

Error 10006.579 1387 7.215

Total 27316.000 1396

Corrected Total 10535.484 1395

a. R Squared = .050 (Adjusted R Squared = .045)

Multiple Comparisons

Dependent Variable: HOW OFTEN R ATTENDS RELIGIOUS SERVICES

Tukey HSD

(I) Region of Country (J) Region of Country

Mean Difference

(I-J) Std. Error Sig.

95% Confidence Interval

Lower Bound Upper Bound

East & North east South -.85* .163 .000 -1.24 -.47

West .26 .190 .366 -.19 .70

South East & North east .85* .163 .000 .47 1.24

West 1.11* .194 .000 .66 1.57

West East & North east -.26 .190 .366 -.70 .19

South -1.11* .194 .000 -1.57 -.66

Based on observed means.

The error term is Mean Square(Error) = 7.215.

*. The mean difference is significant at the .05 level.

-This graph shows the interaction between the region and the political party affiliation. In general, those in the

south typically have higher attendance than the other two regions. Republicans have the highest over all

attendance, regardless of their region.

5. Is there a difference in how often people attend religious services across different racial groups? Show the appropriate statistics (with the means and standard deviations) and explain your answer. Include a graph. Make sure to do all the appropriate analyses (HINT, HINT). 15 pts

-I chose a one-way between groups ANOVA because we are testing one factor (racial group) with 3 levels on how often people attend religious services.

-The mean religious service attendance differed significantly between white (M = 3.32, SD = 2.729), black (M = 4.20, SD = 2.69), and other (M = 3.45, 2.828), F(2, 1395) = 9.288, p <.001, R2 =.013. From the results we can tell that there is a statistically significant difference between white and black attendance (p < .001) as well as black and other (p = .043), but there is not a significant difference between white and other (p = .860).

SPSS Output:

SPSS Data: