Exercises 1 – 5: Election Results<\/p>\n
The following is part of an article that appeared in a newspaper:<\/p>\n
With the election for governor still more than a year away, a new poll shows the race is already close. The Republican governor had 47%, and the Democratic challenger had 45% in a poll released Tuesday of 800 registered voters.<\/p>\n
\u201cThat\u2019s within the poll\u2019s margin of error of 3.5 percentage points, making it essentially a toss-up,\u201d said the poll\u2019s director.<\/p>\n
Exercise 1.
\nWhy don\u2019t the two percentages add up to 100%?
\nAnswer:
\nThe other 8% might be undecided voters or voters who want to vote for a third-party candidate.<\/p>\n
Exercise 2.
\nWhat is meant by the margin of error of 3.5 percentage points?
\nAnswer:
\nIt is unlikely that the estimate of 47% of the proportion of all voters who would vote for the Republican candidate will be farther from the actual population proportion than the margin of error of 3.5%, or the proportion of all voters who would vote for the Republican candidate is likely between 43.5% and 50.5%.<\/p>\n
Exercise 3.
\nUsing the sample size of 800 and the proportion 0.47, calculate the margin of error associated with the estimate of the proportion of all registered voters who would vote for the Republican governor.
\nAnswer:
\n0.035<\/p>\n
Exercise 4.
\nWhy did the poll director say that the election is \u201cessentially a toss-up\u201d?
\nAnswer:
\nEven though the sample proportion who preferred the Republican governor is 0.47, this is just an estimate, and it might be too high by as much as 0.035. If this was the case, the Democratic challenger would win.<\/p>\n
Exercise 5.
\nIf the sample size had been 2, 500 registered voters, and the results stated 47% would vote for the Republican governor, and 45% said they would vote for the Democratic challenger, what would the margin of error have been?
\nCould the director still say that the election was a toss-up?
\nAnswer:
\nThe margin of error would be 0.02. The director could still say the election was a toss-up.<\/p>\n
Exercises 6 – 8: Chocolate Chip Claim<\/p>\n
The Nabisco Company claims that there are at least 1,000 chocolate chips in every 18-ounce bag of their Chips Ahoy! cookies. An article in a local newspaper reported the efforts of a group of students in their attempt to validate the Nabisco claim.<\/p>\n
The article reported that the students randomly selected 42 bags of cookies from local grocery stores and counted the number of chocolate chips in the cookies in each bag. The students found the sample mean was 1,261.6 chips, and the sample standard deviation was 117.6 chips. The article stated that the students\u2019 data supported the Nabisco Company claim.<\/p>\n
Exercise 6.
\nUsing the students\u2019 results, calculate the margin of error associated with the estimate of the mean number of chocolate chips in an 18-ounce bag of Chips Ahoy! chocolate chip cookies. Write a sentence interpreting the margin of error.
\nAnswer:
\nThe margin of error is 36.3. The population mean the number of chocolate chips in an 18-ounce bag is likely between 1,225.3 and 1,297.9.<\/p>\n
Exercise 7.
\nDo you agree that the student data supported the Nabisco Company claim? Explain.
\nAnswer:
\nStudents should agree with the claim. The population mean number of chocolate chips is likely in the range 1261.6 \u00b1 36.3, or between 1,225.3 and 1,297.9 This range is consistent with the claim of at least 1,000 chocolate chips in each 18-ounce bag.<\/p>\n
Exercise 8.
\nComment on the procedure that the students used to collect their data.
\nAnswer:
\nIt was important that the 42 bags were randomly selected. Since they were all collected from local grocery stores, it might not be reasonable to extend the conclusion to beyond the local grocery stores.<\/p>\n
Exercises 9 – 15: Understanding a Poll<\/p>\n
George Gallup founded the American Institute of Public Opinion (Gallup Poll) in 1935. The translated company is famous for its public opinion polls, which are conducted in the United States and other countries.<\/p>\n
Gallup published a graph in May 2013 titled Percent in U.S. Who Exercise for at Least 30 Minutes Three or More Days per Week. Use the graph found on the Gallup website http:\/\/www.gallup.com\/poll\/162194\/americans-exercise-habits- worsen-slightly-2013.aspx) to answer the following questions:<\/p>\n
Exercise 9.
\nWhat percent of those surveyed said that they exercise at least 30 minutes three or more days a week at the start of 2013?
\nAnswer:
\nApproximately 48%<\/p>\n
Exercise 10.
\nDescribe the patterns that you observe in the graph.
\nAnswer:
\nThe percent of adults who say they exercise at least 30 minutes three or more days a week follows an up-and-down cycle; at the beginning of the year, the percent is low; by midyear, the percent rises to its highest; by the end of the year, the percent goes back down.<\/p>\n
Exercise 11.
\nGive some reasons why you think the graph follows the pattern that you described.
\nAnswer:
\nThe pattern seems to be following the seasons of the year. During the winter, the percent is low; during the summer, the percent reaches its peak.<\/p>\n
Following are the survey methods that Gallup used to collect the data:<\/p>\n
Results are based on telephone interviews conducted as part of the Gallup-Healthways Well-Being Index survey June 1 – 30, 2013, with a random sample of 15,235 adults, aged 18 and older, living in all 50 u.s. states and the District of Columbia.<\/p>\n
For results based on the total sample of national adults, one can say with 95% confidence that the maximum margin of sampling error is \u00b11 percentage point.<\/p>\n
Exercise 12.
\nUsing the value of 0.538 for the proportion of those surveyed who said they exercise at least 30 minutes three or more days a week in the most recent poll, calculate the margin of error. How does your margin of error compare to the value reported by Gallup?
\nAnswer:
\nThe margin of error is 0.008, which rounds up to 0.01.<\/p>\n
Exercise 13.
\nInterpret the phrase \u201cmargin of sampling error is \u00b11 percentage point.\u201d
\nAnswer:
\nIt is very likely that the proportion of all adults who said they exercise at least 30 minutes three or more days a week is within 0.01 of the sample proportion 0.538.<\/p>\n
Exercise 14.
\nWhy is it important that Gallup selects a random sample of adults?
\nAnswer:
\nIt helps to ensure that the sample taken is representative of the population.<\/p>\n
Exercise 15.
\nIf Gallup had used a random sample of 1,500, what would happen to the margin of error? Explain your answer.
\nAnswer:
\nThe margin of error would increase to 0.026.<\/p>\n
ME = 2\\(\\sqrt{\\frac{0.538(1-0.538)}{1500}}\\)<\/p>\n
Question 1.
\nThe British Medical Journal published a study whose objective was to investigate estimation of calorie content of meals from fast-food restaurants. Below are the published results.<\/p>\n
Participants: 1,877 adults and 330 school-age children visiting restaurants at dinnertime (evening meal) in 2010 and 2011; 1, 178 adolescents visiting restaurants after school or at lunchtime in 2010 and 2011<\/p>\n
Results: Among adults, adolescents, and school-age children, the mean actual calorie content of meals was 836 calories (SD 465), 756 calories (SD 455), and 733 calories (SD 359), respectively. Compared with the actual figures, participants underestimated calorie content by means of 175 calories, 259 calories, and 175 calories, respectively.<\/p>\n
a. Calculate the margin of error associated with the estimate of the mean number of actual calories in the meals eaten by each of the groups: adults, adolescents, and school-age children.
\nAnswer:
\nMargin of error for adults: 21.47
\nMargin of error for adolescents: 26.51
\nMargin of error for school-age children: 39.52<\/p>\n
b. Write a sentence interpreting the margin of error for the adult group.
\nAnswer:
\nIt is very likely that the sample mean of 836 calories is within 21.47 of the actual mean number of calories in meals eaten by all adults.<\/p>\n
c. Explain why the margin of error for the estimate of the mean number of actual calories in meals eaten by adults is smaller than the margin of error of the mean number of actual calories in meals eaten by school-age children.
\nAnswer:
\nThe size of the sample for school-age children is considerably smaller than the sample size for the adults.<\/p>\n
d. Write a conclusion that the researchers could draw from this study.
\nAnswer:
\nThe mean number of calories in a meal consumed by adults is greater than the mean number of calories in a meal consumed by adolescents.<\/p>\n
Question 2.
\nThe Gallup organization published the following results from a poll that it conducted:<\/p>\n
By their own admission, many young Americans, aged 18 to 29, say they spend too much time using the Internet (59%), their cell phones or smartphones (58%), and social media sites such as Facebook (48%). Americans\u2019 perceptions that they spend \u201ctoo much\u201d time using each of these technologies decline with age.<\/p>\n
Conversely, older Americans are most likely to say they spend too much time watching television, and among all Americans, television is the most overused technology tested.<\/p>\n
Results are based on telephone interviews conducted as part of Gallup Daily tracking April 9 – 10, 2012, with a
\nrandom sample of 1,051 adults, aged 18 and older, living in all 50 u.s. states and the District of Columbia.<\/p>\n
For results based on the total sample of national adults, one can say with 95% confidence that the maximum margin of sampling error is \u00b14 percentage points.<\/p>\n
a. Write a newspaper headline that would capture the main idea from the poll.
\nAnswer:
\nAnswers will vary. An example is \u201cU.S. Young Adults Admit Too Much Time on Cell Phones, Web.\u201d<\/p>\n
b. Use the phrase from the article, \u201ctheir cell phones or smartphones (58%)\u201d to calculate the margin of error. Show your work.
\nAnswer:
\nME = 2\\(\\sqrt{\\frac{(0.58)(0.42)}{1051}}\\) = 0.03<\/p>\n
c. How do your results compare with the margin of error stated in the article?
\nAnswer:
\nThis value is less than the stated value of 0.04.<\/p>\n
d. Interpret the statement \u201cthe margin of sampling error is \u00b13 percentage points.\u201d
\nAnswer:
\nIt is very likely that the sample proportion 0. 581s within 0.04 of the proportion of all adults who would say that they use their cell phones too much.<\/p>\n
e. What would happen to the margin of error if Gallup had surveyed 100 people instead of the 1,051?
\nAnswer:
\nThe margin of error would increase to 0.10.<\/p>\n
Question 3.
\nThe Holiday Inn Resort Brand conducted the Kid Classified survey. 1,500 parents and children nationwide were interviewed via an online survey.<\/p>\n
The results of the survey state<\/p>\n
While many parents surveyed say they have some financial savings set aside specifically for vacation travel, more than half of parents in the survey (52%) noted that saving enough money was the biggest challenge to planning a family vacation, more so than coordination of family schedules (19%) or taking time off of work (12%).<\/p>\n
a. Calculate the margin of error associated with the estimate of the proportion of all parents who would say that saving enough money is the biggest challenge to planning a family vacation.
\nAnswer:
\nThe margin of error is 0.026.<\/p>\n
b. Write a sentence interpreting the margin of error.
\nAnswer:
\nIt is very likely that the sample proportion of 0.52 is within 0.026 of the proportion of all parents who would say that saving enough money is the biggest challenge to planning a family vacation.<\/p>\n
c. Comment on how the survey was conducted.
\nAnswer:
\nThis was an online survey. It is not known if this was a random sample. Only individuals who have Internet access could take this survey, which would eliminate anyone who did not have Internet access.<\/p>\n
The Gallup organization published the following results from a poll that is conducted.<\/p>\n
As health experts increasingly focus on the medical benefits of a healthy lifestyle and preventative healthcare, Americans say their doctor does commonly discuss the benefits of healthy habits with them. Specifically, 71% say their doctor usually discusses the benefits of engaging in regular physical exercise, and 66% say their doctor usually discusses the benefits of eating a healthy diet. Fewer Americans, 50%, say their doctor usually discusses the benefits of not smoking, although that number jumps to 79% among smokers.<\/p>\n
Survey Methods<\/p>\n
Results for this Gallup poll are based on telephone interviews conducted July 10 – 14, 2013, with a random sample of 2,027 adults, aged 18 and older, living in all 50 U.S. states and the District of Columbia.<\/p>\n
For results based on the total sample of national adults, one can say with 95% confidence that the margin of sampling error is \u00b13 percentage points.<\/p>\n
Question 1.
\nThe headline of the article is \u201cSmokers Much More Likely Than Nonsmokers to Say Doctor Discusses Not Smoking.\u201d Do you agree with this headline? Explain your answer.
\nAnswer:
\nYes, I agree with this headline because 79% of smokers say their doctor usually discusses the benefits of nonsmoking, while less than 50% of nonsmokers say their doctor discusses not smoking. The margin of error associated with these estimates is only 3%.<\/p>\n
Question 2.
\nUsing the data \u201c71% say their doctor usually discusses the benefits of engaging in regular physical exercise,\u201d calculate the margin of error. Show your work.
\nAnswer:
\nME = 2\\(\\sqrt{\\frac{(0.71)(0.29)}{2027}}\\) = 0.02<\/p>\n
Question 3.
\nHow do your results compare with the margin of error stated in the article?
\nAnswer:
\nThis value is lower than the stated value of 0.03.<\/p>\n
Question 4.
\nInterpret the margin of error in this context.
\nAnswer:
\nIt is very likely that the proportion of all adults who would say that their doctor usually discusses the benefits of engaging in physical exercise is within 3% of the sample proportion 0.71.<\/p>\n","protected":false},"excerpt":{"rendered":"
Engage NY Eureka Math Algebra 2 Module 4 Lesson 22 Answer Key Eureka Math Algebra 2 Module 4 Lesson 22 Exercise Answer Key Exercises 1 – 5: Election Results The following is part of an article that appeared in a newspaper: With the election for governor still more than a year away, a new poll … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/92781"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=92781"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/92781\/revisions"}],"predecessor-version":[{"id":136842,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/92781\/revisions\/136842"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=92781"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=92781"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=92781"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}