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Chapter 12: Problem 16

Religion in Congress Is the religious make-up of the United States Congress reflective of that in the general population? The following table shows the religious affiliation of the 535 members of the 114 th Congress along with the religious affiliation of a random sample of 1200 adult Americans. $$ \begin{array}{lcc} \text { Religion } & \begin{array}{c} \text { Number of } \\ \text { Members } \end{array} & \begin{array}{c} \text { Sample of } \\ \text { Residents } \end{array} \\ \hline \text { Protestant } & 306 & 616 \\ \hline \text { Catholic } & 164 & 287 \\ \hline \text { Mormon } & 16 & 20 \\ \hline \text { Orthodox Christian } & 5 & 7 \\ \hline \text { Jewish } & 28 & 20 \\ \hline \text { Buddhist/Muslim/Hindu/Other } & 6 & 57 \\ \hline \text { Unaffiliated/Don't Know/Refused } & 10 & 193 \\ \hline \end{array} $$ (a) Determine the probability distribution for the religious affiliation of the members of the 114 th Congress. (b) Assuming the distribution of the religious affiliation of the adult American population is the same as that of the Congress, determine the number of adult Americans we would expect for each religion from a random sample of 1200 individuals. (c) The data in the third column represent the declared religion of a random sample of 1200 adult Americans (based on data obtained from Pew Research). Do the sample data suggest that the American population has the same distribution of religious affiliation as the 114 th Congress? (d) Explain what the results of your analysis suggest.

Short Answer

Expert verified
Congress's religious make-up significantly differs from the general U.S. population based on the chi-square test.

Step by step solution

01

- Determine Relative Frequencies for Congress Members

Calculate the relative frequency for each religious affiliation by dividing the number of members by the total number of members (535).
02

- Calculate Relative Frequencies

Protestant = \(\frac{306}{535} = 0.572\) Catholic = \(\frac{164}{535} = 0.307\) Mormon = \(\frac{16}{535} = 0.03\) Orthodox Christian = \(\frac{5}{535} = 0.009\) Jewish = \(\frac{28}{535} = 0.052\) Buddhist/Muslim/Hindu/Other = \(\frac{6}{535} = 0.011\) Unaffiliated/Don't Know/Refused = \(\frac{10}{535} = 0.019\)
03

- Expected Number of Americans for Each Religion

Multiply the relative frequencies from Step 2 by 1200 to find the expected values. Protestant = \(0.572 \times 1200 = 686.4\) Catholic = \(0.307 \times 1200 = 368.4\) Mormon = \(0.03 \times 1200 = 36\) Orthodox Christian = \(0.009 \times 1200 = 10.8\) Jewish = \(0.052 \times 1200 = 62.4\) Buddhist/Muslim/Hindu/Other = \(0.011 \times 1200 = 13.2\) Unaffiliated/Don't Know/Refused = \(0.019 \times 1200 = 22.8\)
04

- Chi-Square Test for Goodness of Fit

Use the chi-square test to compare the observed and expected frequencies. Calculate the chi-square statistic using the formula: \[\chi^2 = \sum \frac{(O-E)^2}{E}\] where O represents the observed frequency and E represents the expected frequency.
05

- Calculate the Chi-Square Statistic

Protestant: \(\frac{(616 - 686.4)^2}{686.4} = 6.00\) Catholic: \(\frac{(287 - 368.4)^2}{368.4} = 19.32\) Mormon: \(\frac{(20 - 36)^2}{36} = 7.11\) Orthodox Christian: \(\frac{(7 - 10.8)^2}{10.8} = 1.34\) Jewish: \(\frac{(20 - 62.4)^2}{62.4} = 28.24\) Buddhist/Muslim/Hindu/Other: \(\frac{(57 - 13.2)^2}{13.2} = 117.78\) Unaffiliated/Don't Know/Refused: \(\frac{(193 - 22.8)^2}{22.8} = 1261.58\) Total Chi-Square Statistic: \(\chi^2 = 1441.37\)
06

- Compare with Critical Value

With 6 degrees of freedom (7 - 1 categories), and typically at a significance level of 0.05, the critical value is approximately 12.59. Since the calculated chi-square statistic (1441.37) is much greater than 12.59, we reject the null hypothesis.
07

- Interpretation of Results

The sample data suggest that the American population does not have the same distribution of religious affiliation as the 114th Congress. The significant difference indicates that Congress is not reflective of the general population in terms of religious distribution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Distribution
Probability distribution is a statistical function that describes the likelihood of obtaining the possible values that a random variable can take. In the context of this exercise, we create a probability distribution for the religious affiliations of Congress members.
We calculate this by dividing the number of members of each religion by the total number of Congress members.
For example, the relative frequency for Protestant is obtained by \(\frac{306}{535} = 0.572\). Similar calculations are done for other religions to derive their respective probability distributions.
This helps us understand how the Congress members are distributed across different religions.
Without a proper understanding of probability distributions, it would be challenging to conduct further analysis like the Chi-Square Test of Goodness of Fit.

Key points to remember:
  • Probability distribution provides a map of how frequently values occur.
  • It is foundational for statistical analysis and hypothesis testing.
Relative Frequency
Relative frequency is a measure used to describe how often a certain event occurs compared to the total number of events observed.
It is an essential concept in probability and statistics as it allows us to convert raw data into a form that is easier to interpret and compare.
In this exercise, we calculate the relative frequency to understand the proportion of Congress members that belong to each religion.
For instance, the relative frequency of Protestant members is \( 306/535 鈮� 0.572 \) or roughly 57.2%.
These relative frequencies help us envisage how common each religion is among Congress members compared to the general population.

Key takeaways:
  • Relative frequency provides a normalized frequency of occurrences.
  • It helps simplify comparisons across different groups or samples.
Expected Frequency
Expected frequency is a crucial concept in statistical tests like the Chi-Square Test of Goodness of Fit.
It represents the number of observations we would expect to see in a specific category if the null hypothesis were true.
To find the expected frequency for this exercise, we use the relative frequencies from Congress and multiply them by the sample size (1200 adult Americans).
For instance, the expected frequency of Protestants would be: \(0.572 \times 1200 = 686.4 \).
Calculating expected frequencies allows us to create a benchmark against which we can compare the observed data.

Key points:
  • Expected frequency is used to determine what we 'expect' to occur under certain assumptions.
  • It forms the basis for comparing observed data in hypothesis testing.
Statistical Significance
Statistical significance helps us determine whether the result of an analysis is likely due to chance or reflects a true effect.
In the Chi-Square Test of Goodness of Fit, we compare the chi-square statistic to a critical value based on the significance level (commonly 0.05) and degrees of freedom.
In this exercise, our calculated chi-square statistic is 1441.37, which is much greater than the critical value of 12.59.
This large difference indicates that the observed distribution of religious affiliations in the population is significantly different from what we would expect if it mirrored Congress.

Key takeaways:
  • Statistical significance indicates whether an observed effect is likely to be real or due to random chance.
  • It is determined by comparing a test statistic to a critical value.
Null Hypothesis
The null hypothesis is a statement that there is no effect or no difference, and it is the assumption we test against alternative hypotheses.
In the context of this Chi-Square Test of Goodness of Fit, the null hypothesis claims that the distribution of religious affiliations in Congress is reflective of the general American population.
Based on our chi-square statistic, we reject the null hypothesis since there is a significant difference between the observed and expected frequencies.

Key insights:
  • The null hypothesis provides a baseline assumption for statistical testing.
  • Rejection of the null hypothesis suggests that there is a significant effect or difference.

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Most popular questions from this chapter

Social Well-Being and Obesity The Gallup Organization conducted a survey in 2014 asking individuals questions pertaining to social well-being such as strength of relationship with spouse, partner, or closest friend, making time for trips or vacations, and having someone who encourages them to be healthy. Social well-being scores were determined based on answers to these questions and used to categorize individuals as thriving, struggling, or suffering in their social wellbeing. In addition, body mass index (BMI) was determined based on height and weight of the individual. This allowed for classification as obese, overweight, normal weight, or underweight. The data in the following contingency table are based on the results of this survey. $$ \begin{array}{lccc} & \text { Thriving } & \text { Struggling } & \text { Suffering } \\ \hline \text { Obese } & 202 & 250 & 102 \\ \hline \text { Overweight } & 294 & 302 & 110 \\ \hline \text { Normal Weight } & 300 & 295 & 103 \\ \hline \text { Underweight } & 17 & 17 & 8 \\ \hline \end{array} $$ (a) Researchers wanted to determine whether the sample data suggest there is an association between weight classification and social well-being. Explain why this data should be analyzed using a chi-square test for independence. (b) Do the sample data suggest that weight classification and social well- being are related? (c) Draw a conditional bar graph of the data by weight classification. (d) Write some general conclusions based on the results from parts (b) and (c).

Determine \((a)\) the \(\chi^{2}\) test statistic, \((b)\) the degrees of freedom, (c) the critical value using \(\alpha=0.05,\) and (d) test the hypothesis at the \(\alpha=0.05\) level of significance. \(H_{0}:\) The random variable \(X\) is binomial with \(n=4, p=0.3\) \(H_{1}:\) The random variable \(X\) is not binomial with \(n=4\) \(p=0.3\) $$ \begin{array}{llllll} \boldsymbol{X} & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} \\ \hline \text { Observed } & 260 & 400 & 280 & 50 & 10 \\ \hline \text { Expected } & 240.1 & 411.6 & 264.6 & 75.6 & 8.1 \\ \hline \end{array} $$

Does the location of your seat in a classroom play a role in attendance or grade? To answer this question, professors randomly assigned 400 students a general education physics course to one of four groups. The 100 students in group 1 sat 0 to 4 meters from the front of the class, the 100 students in group 2 sat 4 to 6.5 meters from the front, the 100 students in group 3 sat 6.5 to 9 meters from the front, and the 100 students in group 4 sat 9 to 12 meters from the front. (a) For the first half of the semester, the attendance for the whole class averaged \(83 \% .\) So, if there is no effect due to seat location, we would expect \(83 \%\) of students in each group to attend. The data show the attendance history for each group. How many students in each group attended, on average? Is there a significant difference among the groups in attendance patterns? Use the \(\alpha=0.05\) level of significance. (b) For the second half of the semester, the groups were rotated so that group 1 students moved to the back of class and group 4 students moved to the front. The same switch took place between groups 2 and \(3 .\) The attendance for the second half of the semester averaged \(80 \% .\) The data show the attendance records for the original groups (group 1 is now in back, group 2 is 6.5 to 9 meters from the front, and so on ). How many students in each group attended, on average? Is there a significant difference in attendance patterns? Use the \(\alpha=0.05\) level of significance. Do you find anything curious about these data? $$ \begin{array}{lllll} \hline \text { Group } & 1 & 2 & 3 & 4 \\ \hline \text { Attendance } & 0.84 & 0.81 & 0.78 & 0.76\\\ \hline \end{array} $$ (c) At the end of the semester, the proportion of students in the top \(20 \%\) of the class was determined. Of the students in group \(1,25 \%\) were in the top \(20 \%\); of the students in group \(2,21 \%\) were in the top \(20 \%\); of the students in group \(3,15 \%\) were in the top \(20 \%\); of the students in group \(4,19 \%\) were in the top \(20 \% .\) How many students would we expect to be in the top \(20 \%\) of the class if seat location plays no role in grades? Is there a significant difference in the number of students in the top \(20 \%\) of the class by group? (d) In earlier sections, we discussed results that were statistically significant, but did not have any practical significance. Discuss the practical significance of these results. In other words, given the choice, would you prefer sitting in the front or back?

The following table contains observed values and expected values in parentheses for two categorical variables, \(X\) and \(Y\), where variable \(X\) has three categories and variable \(Y\) has two categories: $$ \begin{array}{cccc} & \boldsymbol{X}_{\mathbf{1}} & \boldsymbol{X}_{\mathbf{2}} & \boldsymbol{X}_{3} \\ \hline \boldsymbol{Y}_{\mathbf{1}} & 34(36.26) & 43(44.63) & 52(48.11) \\ \hline \boldsymbol{Y}_{\mathbf{2}} & 18(15.74) & 21(19.37) & 17(20.89) \\ \hline \end{array} $$ (a) Compute the value of the chi-square test statistic. (b) Test the hypothesis that \(X\) and \(Y\) are independent at the \(\alpha=0.05\) level of significance.

At Joliet Junior College, the mathematics department decided to offer a redesigned course in Intermediate Algebra, called the Math Redesign Program (MRP). Laura Egner, the coordinator of the program, wanted to determine if the grade distribution in the course differed from that of traditional courses. The following shows the grade distribution of traditional courses based on historical records and the observed grades in three pilot classes in which the MRP program was utilized. $$ \begin{array}{lcccccc} & \mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D} & \mathbf{F} & \mathbf{W} \\\ \hline \begin{array}{l} \text { Traditional } \\ \text { Distribution } \end{array} & 0.133 & 0.191 & 0.246 & 0.104 & 0.114 & 0.212 \\ \hline \begin{array}{l} \text { Observed Counts } \\ \text { in MRP Program } \end{array} & 7 & 16 & 10 & 13 & 6 & 12 \\ \hline \end{array} $$ (a) How many students were enrolled in the MRP program for the three pilot courses? Based on this result, determine the expected number of students for each grade assuming there is no difference in the distribution of MRP student grades and traditional grades. (b) Does the sample evidence suggest that the distribution of grades is different from the traditional classes at the \(\alpha=0.01\) level of significance? (c) Explain why it makes sense to use 0.01 as the level of significance. (d) Suppose the MRP pilot program continues in three more classes with the grades earned for all six pilot courses shown below. Notice that the sample size was simply doubled with the grade distribution remaining unchanged. Does this sample evidence suggest that the distribution of grades is different from the traditional classes at the \(\alpha=0.01\) level of significance? What does this result suggest about the role of sample size in the ability to reject a statement in the null hypothesis? $$ \begin{array}{lcccccc} & \mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D} & \mathbf{F} & \mathbf{W} \\\ \hline \begin{array}{l} \text { Observed Counts } \\ \text { in MRP Program } \end{array} & 14 & 32 & 20 & 26 & 12 & 24 \end{array} $$
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