Students Again

Political Research Methods Homework

1.

a. Compute the mean, median, and mode for the following three sets of scores (3x 3 points each = 9 points).

Set 1 = 3,7,5,4,5,6,7,8,6,5

Set 2 = 34,54,17,26,34,25,14,24,25,23

Set 3 = 154,167,132,145,154,145,113,156,154,123

A)

Set #1

MEAN =  Xbar = 3+7+5+4+5+6+7+8+6+5 / 10 = 5.6

MEDIAN = 10 + 1 / 2 = 5.5 = 5 + 6 / 2 = 5.5

MODE (occurs Most often) = 5

Set #2

MEAN = Xbar = 34+54+17+26+34+25+14+24+25+23 / 10 = 27.6

MEDIAN = 10 + 1 / 2 = 5.5 = 25 + 25 / 2 = 25

MODE (occurs most often)= 25, 34 (bimodal)

Set #3

MEAN = Xbar = 154+167+132+145+154+145+113+156+154+123 / 10 = 144.3

MEDIAN = 10 + 1 / 2 = 5.5 = 145 + 154 / 2 = 149.5

MODE (occurs most often)= 154

b. Explain the differences between the mean, median, and mode, as well as when each measure should be used (3 points).

B)

The mean is the sum of a set of scores divided by the total number of scores in the set.  This is calculated by adding up a set of scores and dividing by the number of scores. The mean is the most commonly used measure of central tendency. The mean is the point in a distribution around which the scores above it balance with those below it.

The median is the measure of central tendency that cuts the distribution into two equal parts. In other words, it is the middlemost point in a distribution. In an odd number of cases the median will be the case that falls exactly in the middle of the distribution. In an even number of cases the median is always the point above which 50% of the cases fall and below which 50% of the cases fall.

The mode is the most frequent, most typical, or most common value in a distribution. Some frequency distributions contain two or more modes. These distributions are referred to as bimodal as opposed to unimodal (single mode distributions).

When deciding to use the mean, median, or mode several factors are involved three of which include: 1) level of measurement, 2) Shape or form of the distribution of data, and 3) research objective. In evaluating proper use in relation to the level of measurement the following should be considered. The mode requires only a frequency count and can be applied to an set of data at the nominal, ordinal, or interval level of measurement. The median requires an ordering of categories from highest to lowest. For this reason, it can only be obtained from ordinal or interval data, not from nominal data. The use of the mean is exclusively restricted to interval data. Applying the mean to ordinal or nominal data will produce a meaningless result.

c. What are the differences between variability and central tendency? Why do we need to know both? (2 points)

C)

Central Tendency is a social researcher’s more precise version of what might be called an average by others. It is called central tendency because it is generally located toward the middle or center of a distribution where most of the data tend to be concentrated. Central tendency is needed because it is a useful way to describe a group as a whole by finding a single number that represents what is average or typical of that set of data.

Variability shows how scores are scattered around the center of distribution. It is needed because it shows this spread, width, or dispersion of scores.

d. For the following set of scores, compute the standard deviation (show your work): 31, 42, 25, 55, 34, 25, 44, 35 (4 points).

D)

Step #1 (Find the mean) – 31+42+25+55+34+25+44+35 = 291 / 8 = 36.375

Step# 2 (Subtract the mean from each raw score)

55 – 36.375 = 18.625

44 - 36.375 = 7.625

42 - 36.375 = 5.625

35 - 36.375 = -1.375

34 - 36.375 = -2.375

31 - 36.375 = -5.375

25 - 36.375 = -11.375

25 - 36.375 = -11.375

Step# 3 (Square each deviation then add together the squared deviations)

55 – 36.375 = 18.625 =  346.891

44 - 36.375 = 7.625 =  58.1406

42 - 36.375 = 5.625 = 31.6406

35 - 36.375 = -1.375 = 1.89063

34 - 36.375 = -2.375 = 5.64063

31 - 36.375 = -5.375 = 28.8906

25 - 36.375 = -11.375 = 129.391

25 - 36.375 = -11.375 = 129.391

TOTAL = 731.876

Step# 4 (Divide by N to get the square root of the result)

S = (square root of) 731.875 / 8

= (square root of) 91.4845

= 9.56475

9.56475 is the standard deviation.

2.

a. What are the characteristics of the normal curve? What human behavior, trait, or characteristic can you think of that is distributed normally? (2 points)

A)

A normal curve has symmetry in that if the curve were folded at its highest point at the center it would create two equal halves (each a mirror image). Another characteristic is that a normal curve is unimodal. This means that there is only one point of maximum frequency in the middle of the curve at which the mean, median, and mode coincide. In addition, from the central peak of the normal distribution the curve gradually falls of at both tails extending indefinitely in either direction while getting closer and closer to the base line without actually reaching it. Height and weight are two examples of human characteristics that may be distributed normally.

b. What is the central limit theorem and what implications does it have for our understanding of the representativeness of sample distributions? (2 points)

B)

The central limit theorem states that if repeated samples are taken from a population and each sample mean is calculated and plotted, the distribution eventually would resemble the normal curve. This provides several characteristics of a sampling distribution of means: 1) the sampling distribution of means approximates the normal curve, 2) the mean of a sampling distribution of means is equal to the true population mean, and 3) the standard deviation of sampling distribution of means is smaller than the standard deviation of the population. With the central limit theorem most of the outcomes occur around the population mean because these events have a high probability of occurring. Thus, the farther removed from the mean, the less likely an outcome will occur. However, in reality, researchers often draw a single sample. This does not provide information about the population mean or standard deviation. As a result, the standard error of means (which can be calculated at different confidence levels) provides an estimate of the variation around the population mean.

c. Explain the difference between a Type 1 and Type 2 error, as well as the implications that statistical significance has on our likelihood of making each type of error (2 points).

C)

Type I and Type II errors hinge on understanding the null and research hypotheses. The research hypothesis states the causal relationship between the independent and dependent variables. The null hypothesis states that there is no relationship between the independent and dependent variables. A Type I error is a false positive. In other words, an error was made by rejecting a hypothesis that should have been accepted. A type II error, or false negative, is the error of accepting a hypothesis that should have been rejected. Sample size matters. Large samples are more likely to be representative and are less likely to result in sample specific results. Small samples have fewer combinations of the values of the variables. This increases the likelihood of obtaining a combination of values suggesting a strong relationship by mere chance. The smaller the relationship the larger the sample size needs to be in order to detect it. The larger the relationship the small the sample size needs to be to detect the relationship. However, since a researcher will not know the relationship beforehand so it is wise to err on the side of caution and use large sample sizes. Testing significance is important in making decisions on determining levels of significance and avoiding error.

d. Explain the difference between statistical and substantive significance (2 points).

D)

Statistical significance is a measure of technical success. Substantive significance focuses on the interpretation of the statistical results in the contexts of the researcher’s theory. The most common mistake is when a researcher finds statistics that are significant and assumes that there is a substantive significance. Statistical significance only tells the likelihood of occurrence, but not the substantive importance. In other words, what do the numbers mean?

3.

What is the standard error of the mean, what is its importance to statistical inference, and what limitations hinder the ability to calculate it? (4 points).

3)

The standard error of the mean provides an estimate of the variation around the population mean. The standard error of the mean allows a researcher to generate confidence intervals at 68%, 95%, or 99% levels (the higher the level the more restrictive). Confidence intervals are a range in which the researcher might expect to find the true mean of the population. These intervals allow the researcher to asses the confidence that the sample mean is an accurate estimate of the population mean. The standard error of the mean is estimated from sampling data. This adds additional uncertainty beyond that of sample variability. As a consequence, researchers want a wider range of estimating this value. Therefore, a t distribution is used instead of using the normal curve. Estimating the standard error of the mean with sample data is very important because it is the basis for all interval statistics.

4.

Compute the z-score for the following raw scores where the mean is 33 and the standard deviation is 4.3. (show your work) (1 point each).

a. 43

b. 21

A)

z = 43 – 33 / 4.3 = 2.32558

B)
z = 21 – 33 / 4.3 = - 2.7907

Using z scores and a distribution of scores with a mean of 55 and a standard deviation of 4.38 determine the following (show your work) (2 points each).

c. What is the probability of a score falling between 50 and 60?

d. What is the probability of a score falling between a score of 64 and 67?

e. What is the probability of a score falling below a score of 57?

C)

z = 50 – 55 / 4.38 = - 1.14155

z = 60 – 55 / 4.38 = 1.14155

Area from the mean to a z-score of 1.14115 according to appendix = 37.29

= 37.29 x 2 = 74.58%

= The probability of a score falling between 50 and 60 is 74.58%

P = 74.58%

D)

z = 64 – 55 / 4.38 = 2.05479

Area from the mean to a z-score of 2.05479 according to appendix = 47.98

z = 67 – 55 / 4.38 = 2.73973

Area from the mean to a z-score of 2.73973 according to appendix = 49.69

= 49.69 – 47.98 = 1.71

= The probability of a score falling between 64 and 67 is 1.71%

P = 1.71%

 

E)

z = 57 – 55 / 4.38 = .456621

Area beyond a z-score of .456621 according to appendix = 32.64

= 100 – 32.64= 67.36

= The probability of a score falling below 57 is 67.36%

P = 67.36%

 

1.

Suppose you are a researcher interested in studying gender differences in citizens’ attitudes towards former president Bill Clinton. Using the NES2000 dataset, you conduct a difference of means test where the dependent variable is Bill Clinton’s thermometer rating (clintpre) and the independent variable is sex (gender). You generate the following output. Gender N Mean Std. Deviation Std. Error Mean Pre:Thermometer Bill Clinton 1. Male 787 54.15 29.558 1.054 2. Female 1007 56.52 29.772 .938 Begin your interpretation by providing a brief interpretation of the group means and standard deviations for males and females Next, using the estimates of the standard error of the mean for males and females, develop a 95% confidence interval around each group mean and interpret what this means. Last, using the value of 1.412 for your estimate of the standard error of the difference of means, calculate a t-test for your analysis and interpret the t value and its associated significance level in the context of the hypothesis under investigation.

 

1)

In this difference of means test a sample size of 787 males and 1007 female respondents were collected. Of that collection 54.15 for males and 56.52 for females is the point in the distribution at which the scores above balance with the scores below. The standard deviation (male is 29.558 and female is 29.772) measures the average variability in the distribution. In other words, it is the average of deviations from the mean. The standard error of the difference between the mean (male is 1.054 and female is .938) represents the estimate of how different the mean of the two independent variables can be because of sampling error.

 

Standard Error of Mean:

Male = 1.054

Female = .938

Independent = 1.412

 

Xbar 1 = 54.15

Xbar 2 = 56.52

 

t = Xbar1 – Xbar2 / S Xbar1 – Xbar2

 

Male = 54.15-56.52 / 1.054 = -2.25 = |2.25|

Female = 54.15 – 56.52 / .938 = -2.53 = |2.53|

Independent = 54.15 – 56.52 / 1.412 = -1.68 = |1.68|

 

Degrees of Freedom for t test:

N1 + N2 – 2

781 + 1007 – 2 = 1786

Critical value according to table in the book = 1.960

 

Therefore, with the male and female independent variable we cannot reject the null hypothesis because the obtained values exceed the critical value. However, with the independent we must reject the null because the obtained value does not exceed the critical value.

 

2.

You are interested in knowing if ideological differences exist in citizens' attitudes towards big business. Using data taken from the 2000 ANES, you wish to calculate an ANOVA to determine if these differences exist and if they are statistically significant. For the analysis, the dependent variable is the thermometer ratings of big business and the independent variable is a three level, measure of ideology (liberal, moderate, conservative). Specifically, using the output below, calculate the within groups sum of squares, the within groups and between groups degrees of freedom, the mean square between and the mean square within, the F ratio, and its statistical significance. Next, provide a brief interpretation of the various sums of squares and interpret the F value and its accompanying significance level in the context of the theory under investigation. ANOVA D2k. Thermometer big business 2) Sum of Squares df Mean Square F Sig. Between Groups 6353.209 2 3176.6 8.06 .000 Within Groups 537455 1364 394.029 Total 543808.433 1366

SS Within:

543808.433 = SS Within + 6353.209

-6353.209                          -6353.209

SS Within = 537455

 

Degrees of Freedom:

df  = K – 1

= 3 – 1

= 2 (Between Groups)

 

df  Within

1366 -2 = 1364 (Within Groups)

 

Mean Square:

MS Within = SS Within / df Within

= 537455 / 1364

= 394.029

 

MS Between = SS Between / df Between

= 6353.209 / 2

= 3176.6

 

F Value:

F = MS Between / MS Within

= 3176.6 / 394.029

= 8.06

 

Significance:

.000

 

This significance value is because for the level of significance = .05 the F value exceeds the critical value of 2.99 obtained from the table in the book. Therefore we reject the null hypothesis and accept the research hypothesis that ideological differences exist in citizens’ attitudes towards big business. Also, the Square of Sums Mean within the group is large enough to be significant.

 

 

3.

Suppose you are a researcher interested in examining the relationship between education and campaign involvement. Specifically, you want to test the hypothesis that higher levels of education positively correlate with campaign activity. Using SPSS and the NES2000 dataset, you conduct a Spearman’s correlation between the variables educ3 (independent variable) and campacts (dependent variable). You generate the following output: Correlations Number of R's campaign acts (1225-1229, 1231, 1233) Education: 3 categories Spearman's rho Number of R's campaign acts (1225-1229, 1231, 1233) Correlation Coefficient 1.000 .131(**) Sig. (1-tailed) . .000 N 1552 1548 Education: 3 categories Correlation Coefficient .131(**) 1.000 Sig. (1-tailed) .000 . N 1548 1800 ** Correlation is significant at the 0.01 level (1-tailed). Interpret the output in terms of its statistical and substantive significance.

 

3)

The output of Spearman's rho correlation shows that the higher levels of education positively correlate with campaign involvement activity at significant level of 0.01 (1-tailed) . Since the value of the Spearman's rho correlation is .131 for campaign acts and education, the more specific correlation level is a moderate positive correlation. Since the one-tailed test of significance value is .000, it can be determined that the critical value is more than the correlation value of .131. If this is the case then the null hypothesis is to be rejected. Since the null hypothesis is rejected, this particular study is relevant and accepts the research hypothesis. Results suggest that there is a positive correlation of higher levels of education positively correlates with campaign involvement in the population from which the sample was taken.

 

4.

Using the “Computational Formula for Correlation,” calculate the correlation coefficient for the following data. After you calculate the value of r, determine if it is statistically significant using a two tailed test and provide an interpretation of the statistical and substantive significance of the analysis. The easiest way to complete this question is to generate a five columned table in Word and then copy and paste the table into the response box. 4) X Y X2 Y2 XY 9 16 81 256 144 6 11 36 121 66 7 10 49 100 70 20 17 400 289 340 24 5 576 25 120 17 12 289 144 204 2 8 4 64 16 15 15 225 225 225 11 14 121 196 154 8 22 64 484 176 ?       119 130 1845 1904 1515

 

N = 10

SumX = 119

SumY = 130

Xbar = 119 / 10 = 11.9

Ybar = 130 / 10 = 13

SumX Squared = 1845

SumY Squared = 1904

SumXY = 1515

 

r = SumXY – NxbarYBar / Square Root of (SumX Squared – Nxbar Squared)(SumY Squared – Nxbar Squared)

= 1515 – 10 (11.9)(13) / Square Root of [1845 – 10 (11.9)Squared] [1904 – 10(13)Squared]

= -32 / Square Root of 91784.6

= -32 / 302.96

= - .11

 

t = r Square Root of N – 2 / Square Root of 1 – r Squared

= - .11 Square Root of 10 – 2 / Square Root of 1 – (- .11) Squared

= -.31 / .993932

= - .31

= |.31|

 

With the level of significance of .05 and 8 degrees of freedom a critical value of 2.306 (according to the back of book) or higher would be required to reject the null hypothesis of p = 0 in a two-tailed test. Therefore, we cannot reject the null hypothesis with our calculated absolute t value of .31. Our correlation value of - .11 can be considered a weak negative correlation and the measure of that correlation has moderate positive significance.

 

5.

You are interested in examining the relationship between voters’ attitudes towards Bill Clinton and George W. Bush. Specifically, you hypothesize that as a voters’ attitudes towards Bill Clinton become more positive, their attitudes towards George Bush will become more negative and vice versa. To evaluate this hypothesis, you use the NES2000 dataset and run a Pearson’s bivariate correlation between the variables gbushpre and clintpre and generate the following output: Correlations Pre:Thermometer Bill Clinton Pre:Thermometer George W Bush Pre:Thermometer Bill Clinton Pearson Correlation 1 -.405(**) Sig. (2-tailed) .000 N 1794 1758 Pre:Thermometer George W Bush Pearson Correlation -.405(**) 1 Sig. (2-tailed) .000 N 1758 1761 ** Correlation is significant at the 0.01 level (2-tailed). Interpret the output in terms of its statistical and substantive significance.

 

5)

The output of Pearson’s bivariate correlation shows that the relationship between voter’s attitudes towards Bill Clinton and G.W. Bush is a negative correlation. Since the value of the Pearson’s correlation is -.405 for G.W. Bush and Bill Clinton the more specific correlation is a moderate negative correlation. Since the two-tailed test of significance value is .000 it can be determined that the critical value is more than the correlation value of absolute value of -.405. If this is the case then the null hypothesis is to be rejected. Since the null hypothesis is rejected, this particular study is relevant and accepts the research hypothesis. Results suggest that there is a negative correlation between the voter’s attitude between G.W. Bush and Bill Clinton in the population from which the sample was taken.

 

1.
	In a regression model (e.g., Y = a + bx), what specifically do a and b represent?

 

The term ‘a’ in the regression model (Y = a + bX + e) is called the Y-intercept. This refers to the expected level of Y. This can be seen as a base-line amount because it is what Y should be before the level of X is taken into account. The term ‘b’ is called the slope (a.k.a. regression coefficient) for X. This represents the amount that Y changes (increase or decrease) for each change of one unit in X. Regression involves placing a line through scatter points. With an accurately drawn line, the value of ‘a’ (the Y-intercept) would be where the line crosses the Y axis. The value of ‘b’ (the slope) matches to the incline or rise of the line for a unit increase in X. When ‘b’ (the slope) and ‘a’ (the Y-intercept) are computed they form the placement for the regression line.

 

2.

Below is the output for a regression model examining the relationship between a state's divorce rate (the dependent variable, which is measured as the number of divorces per 1,000 citizens) and the percent of a state's residents who are Christian adherents (the independent variable, which is measured as the percent of the population that are Christian adherents). Provide a substantive interpretation of the R-square value, the coefficient for the constant (e.g., the intercept), and the slope estimate for the independent variable. Do not worry about interpreting the standard errors and the t values.  However, for those of you who are interested, the standard errors in a regression analysis are akin to those used in the t-test in that they represent sample variability in the estimate of the intercept and slope.  The t ratios, which are used to determine statistical significance of the intercept and slope, are obtained by dividing the intercept and slope by their standard error and then using the t table to determine if the obtained values exceed the critical values for the appropriate degrees of freedom. Model Summary R Square .120 Coefficients Coefficients t Sig. B Std. Error (Constant) 6.435 .802 8.023 .000 Percent of pop who are Christian adherents -.035 .015 -2.420 .020

R-square = amount of variance in Y that is explained by X.

Y = percent residence whom are Christian adherents (Independent Variable)

X = divorce rate (Dependent variable)

The R-square value (.120) shows that the divorce rate only explains 12% of the variance percent residents that are Christian adherents. Therefore, of the percent residents whom are Christian adherents, 88% can be explained by other factors. The coefficient for the constant of 6.435 is what the expected value of percent residents whom are Christian adherents without taking the divorce rate into account. The slope estimate for the independent variable of -.035 is the amount that the percent residence whom are Christian adherents changes for each unit (divorce).

 

 

3.
	Using the output in the previous question, generate predicted values for the dependent variable (divorces per 1,000 population) for states that have 27%, 66%, and 85% Christian adherents (show your work).

 

a = 6.435

b = -.035

Y = 6.435 + (-.035)x

- Predicted value for divorce per 1,000 population for a state with 27% Christian adherents:

.27 -6.435 = 6.435 + (-.035)x -6.435

-6.165 = -.035x

-6.165 / -.035 = -.035x /.-.035

176.143 = x

- Predicted value for divorce per 1,000 population for a state with 66% Christian adherents:

.66 -6.435 = 6.435 + (-.035)x -6.435

-5.775 = -.035x

-5.775 / -.035 = -.035x /.-.035

165 = x

- Predicted value for divorce per 1,000 population for a state with 85% Christian adherents:

.85 -6.435 = 6.435 + (-.035)x -6.435

-5.585 = -.035x

-5.585 / -.035 = -.035x /.-.035

159.57 = x

*numbers can be rounded because there cannot be fractional divorces (i.e. 27% and 85% results).