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SOC 353 1 Homework #3: Confidence Intervals and Hypothesis Testing with Means and Proportions Due Date: 11:59 p.m. on October 3, 2021 (via Canvas) Directions: Answer each question below in its entirety. Submit your answers through Canvas by the due date. Your submission should be a Word document and named with the following structure: LASTNAME_hw3.doc. Be sure to show your work and to list the names of any students with whom you worked on this assignment. Please round to the nearest hundredth in your calculations. Part I: Confidence Intervals 1.) Below are some descriptive statistics from a subset of the Congressional Candidate Websites Study, specifically a random sample (stratified by state) of candidates running for the U.S. House of Representatives in 2002. The variable, “TVADS,” is a count of the number of ads sponsored by each candidate during their campaign. Calculate the standard error and interpret it. Please show your work. Variable | Obs Mean Std. Dev. Min Max ————-+——————————————————-TVADS | 116 2.767241 3.669754 0 20 2.) Now calculate both a 95% and 99% confidence interval for the “TVADS” variable above. Interpret the results and compare the two confidence intervals. If you were a researcher having to make a tradeoff between precision and confidence, which confidence interval would you report? Why did you make the decision that you did? 3.) Below are more descriptive statistics for this same random sample of candidates. The variable, “ENDORSE,” is a dichotomous variable indicating whether or not the candidate received non-partisan official endorsements. For this variable, 0 = no endorsements and 1 = received at least one endorsement. Calculate and interpret a 99.9% confidence interval for the 1 category. Please show your work. . tab ENDORSE2 if Y2002==1 & SENATE==0 ENDORSE2 | Freq. ————+———–0 | 45 1 | 71 ————+———–Total | 116 SOC 353 2 Part II: One-Sample Hypothesis Tests 4.) The Congressional Candidate Websites Study contains data for the entire population of major party Senate candidates in the 2002, 2004, and 2006 campaign cycles and a random sample of House candidates. The variable of interest here, “FSPENT,” is the amount of money (in real dollars) spent by the candidate on their campaign. The mean amount of money spent in the population of Senate candidates was $5,759,084. A researcher wants to assess whether or not House candidates during this same time span—as reflected by a random sample of 530 House candidates—are any different from their Senate colleagues in terms of their propensity to spend money on the campaign. What sort of hypothesis test would be appropriate here? Explain. (Hint: You’ll want to think about the probability distribution that would be appropriate given the information provided.) 5.) Below are some descriptive statistics on the “FSPENT” variable from the random sample of House candidates. Using this information, carry out the hypothesis test you proposed in your answer to Question 1 using 𝛼 = .001. Be sure to follow the “5 step” system outlined in class and in your textbook. Be sure to interpret your results and show your work. Variable | Obs Mean Std. Dev. Min Max ————-+——————————————————-FSPENT | 530 943002.9 927594.1 0 5040897 Part III: Two-Sample Hypothesis Tests 6.) An Amazon warehouse manager wants to know if their day crew is more time efficient with their work than the night crew. They draw a random sample of 23 day crew workers and 24 night crew workers. Using these samples, they find that the day crew manages to box an average of 132 boxes per hour while the night crew manages 122 boxes per hour. The standard deviation for the day crew is 20 boxes per hour; the standard deviation for the night crew is 16 boxes per hour. Using this information and an a = .05, decide on the appropriate hypothesis test, calculate the test statistic, and interpret the results. Be sure to follow the “five step” system for hypothesis testing outlined in class and in the textbook. Show all of your work. 7.) A researcher studying perceptions of neighborhood safety wants to know if there is a difference between men and women when it comes to fear of walking in their neighborhood at night. They run a hypothesis test in R using the “fear” measure in the 2014 General Social Survey (GSS)—a dichotomous variable where 0 indicates they are not afraid to walk in their neighborhood at night and 1 indicates they are afraid. Using the output below, answer the following questions: a. What is the dependent variable? What is the independent variable? b. What test do you think they ran, and why did they run it? c. What is the null hypothesis? What is the alternative hypothesis? d. The z-statistic is -8.5867. What is this number telling us? Be sure to interpret it in relation to the null hypothesis. SOC 353 3 e. How would you interpret the outcome of this test? What is the p-value telling you? PAfraid 0.224 0.421 -0.197 Male Female Difference z-statistic p-value (twotailed) -8.587 0.000 Part IV: ANOVA 0 2 y 4 6 8 8.) Which of the following partitioned distributions has the statistically significant F-ratio? How do you know? Explain. 0 .2 .4 .6 .8 1 .8 1 x Group 1 Group 2 Group 3 Group 4 0 .5 y 1 1.5 Group 5 0 .2 .4 .6 x Group 1 Group 2 Group 3 Group 4 Group 5 9.) A researcher is examining whether or not systolic blood pressure readings vary across age groups in the U.S. adult population. They ran an ANOVA test and presented the output below. They also provided some descriptive statistics for the systolic blood pressure variable, computed separately for each age group. Using this information, answer the following questions: a. What is your null hypothesis? What is your alternative hypothesis? b. What is the F-ratio (i.e., the F-statistic), and what is its interpretation? c. What is the critical F-value, assuming an a = .01 (hint: you’ll need your textbook— or some copy of the F-distribution—for this one). SOC 353 4 d. Do you reject or fail to reject your null hypothesis? What is your interpretation of the results? ANOVA Results: > summary(aov(nhanes$bpsystol ~ nhanes$agegrp)) Df Sum Sq Mean Sq F value Pr(>F) nhanes$agegrp 5 1276984 255397 605.9

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