Introduction to Statistics

Name _____________________ I.D. Number ___________Unit 5 Evaluation 419 MTHH 041Unit 5 EvaluationEvaluation 05Introduction to Statistics(MTHH 041 055)This evaluation will cover the lessons in this unit. It is open book, meaning you can use yourtextbook, syllabus, and other course materials. You will need to understand, analyze, and apply theinformation you have learned in order to answer the questions correctly. To submit the evaluation,follow the directions provided.Multiple-ChoiceSelect the response that best completes the statement or answers the question.Use this information to answer questions 1 – 5.Oceanographers believe that 28% of coral in a large reef have a fungus. A survey team willinspect 142 different locations on the reef._ 1. Find p ˆ . a. 0.72 b. 0.20 c. 0.28 d. 0.80 2. Findqˆ .a. 0.72b. 0.20c. 0.28d. 0.803. Find the standard deviation of the proportion.a. 0.280b. 0.0377c. 0.720d. 0.80 4. What would be the 95% confidence interval be if we use 2 standard deviations from themean?a. (0.28, 0.3554)b. (0.2046, 0.3177)c. (0.2423, 0.3177)d. (0.2046, 0.3554)Unit 5 Evaluation 420 MTHH 0415. Would a proportion of 37% lead you to question the parameter of 28%? Explain.a. Yes, since it is inside the confidence interval.b. Yes, since it is outside the confidence interval.c. No, since it is inside the confidence interval.d. No, since it is outside the confidence interval.Use this information to answer questions 6 – 8A chicken hatchery claims that 80% of their eggs successfully hatch. A farmer purchases 60eggs from the hatchery to start his own flock. 6. What proportion of these eggs should the farmer expect to hatch?a. 0.20b. 0.95c. 0.80d. 0.607. How many chicks should the farmer expect?a. 60 chicksb. 80 chicksc. 48 chicksd. 75 chicks 8. What’s the standard deviation of hatching rates for samples of eggs this size?a. 0.0516b. 0.1032c. 0.80d. 0.0067Unit 5 Evaluation 421 MTHH 0419. A CNN/Opinion Research poll in August 2010 asked Americans if they believed PresidentBarack Obama was born outside the country. Based on those results, we’vesimulated several samples of 50 people. This histogram displays the proportion ofpeople who answered yes to this question in each of those samples. What does thestriped bar represent?a. The striped bar represents a 95% confidence interval between 0.20 and 0.25 thatthe President was born outside the country.b. The striped bar represents a Normal model, N(0.20, 0.25).c. The striped bar represents 47 samples where between 20 and 25 percent believedthe President was born outside the country.d. The striped bar represents between 0.20 and 0.25 of the 50 people believe that thePresident was born outside the country.Use this information to answer questions 10 – 11.A regular die is tossed 60 times and the proportion of rolls that are 1 or 2 is recorded. This isrepeated many times. 10. Find the mean and standard deviation of the sampling distribution of sample proportions.a.13; 0.0609b.13; 0.0430c.23; 0.0609d.23; 0.0430Unit 5 Evaluation 422 MTHH 04111. Describe how the mean and standard deviation of the sampling distribution model wouldchange if we were to roll the die fewer times.a. Mean would stay the same; standard deviation would decrease.b. Mean would stay the same; standard deviation would increase.c. Mean would decrease; standard deviation would increase.d. Mean would decrease; standard deviation would stay the same. 12. A national telephone survey by Rasmussen Reports in early August 2010 found that35% of voters surveyed believe increasing taxes for those making more than $250,000would be good for the economy. The margin of error was ±3%. What can you concludeabout all American voters?a. We can be reasonably certain that the true percentage of American voters whobelieve increasing taxes for those making more than $250,000 is between 29%and 41% (±2 standard deviations).b. We can be reasonably certain that the true percentage of American voters whobelieve increasing taxes for those making more than $250,000 is between 32%and 38%.c. We can be reasonably certain that 35% of those surveyed make between $80,000and $95,000.d. We can be reasonably certain that all American voters believe that taxes should beincreased for those making more than $250,000.13. In another telephone survey of 1,000 Americans chosen randomly by RasmussenReports, 40% say they have chosen not to fill a prescription because it cost too much.Find the 95% confidence interval.a. (0.3845, 0.4155)b. (0.5845, 0.6155)c. (0.3696, 0.4304)d. (0.559, 0.631) 14. 3,500 randomly chosen voters are asked in a national poll if they approve of the job thepresident is doing. Which best describes a sampling distribution of the sampleproportion in this situation?a. A confidence interval containing the true proportion of voters who approve of thepresident.b. The proportion of these 3,500 voters who approve the president.c. The proportion of all voters who approve the president.d. The proportions who approve the president within all possible samples of this size.15. Which statement about a confidence interval is true?a. There is a 95% probability that the true proportion is this interval.b. The interval contains 95% of the true proportion.c. 95% of the time, the sample proportion and the true proportion will be the same.d. 95% of intervals constructed this way will contain the true proportion.Unit 5 Evaluation 423 MTHH 041 16. Which is a condition that must be met before using a Normal model for a samplingdistribution of proportions?a. We must expect more than 10 successes and more than 10 failures.b.  npqc.ˆ ˆ ( )ˆpq SE pnd. The outcomes within each sample are not independent.17. Which of the following statements is true?a. Using a non-random sample does not affect the accuracy of a confidence interval.b. Our method requires that the sample size must be less than 10% of the populationwe are investigating.c. The value of p depends on our sample.d. The data values can be dependent on each other. 18. The jobless rate for Nevada in July 2010 was 13%. In a random sample of 50 employableadults in one city, 8 were unemployed. Which is true?a. p = 0.16 andpˆ= 0.13b. The 95% confidence interval for this city’s jobless rate is 0.044 to 0.235.c. The sample size is not large enough to use a Normal model for the samplingdistribution.d. This situation does not meet the 10% Condition for using a Normal model to modelthe sampling distribution.19. A governor is concerned about his “negatives” – the percentage of state residents whoexpress disapproval of his job performance. His political committee pays for a seriesof TV ads, hoping that they can lower negatives to below 30%. The pollsters test thehypothesis that the ads produced no change against the alternative that the negativesare now below 30% and find a P-value of 0.22. Which conclusion is correct?a. There’s a 22% chance that the ads worked.b. There’s a 78% chance that the ads worked.c. There’s a 22% chance that their poll is correct.d. There’s a 22% chance that natural sampling variation could produce poll results likethese if there is really no change in public opinion.Unit 5 Evaluation 424 MTHH 041 20. Nationally, 29% of high school students report riding in a car driven by someone who hadbeen drinking during the 30 days preceding their survey. A state youth councilbelieves this proportion is lower in their state where there had been an intense mediacampaign against riding with drivers who have been drinking. A survey of randomlychose students from across the state shows 304 of 1214 students answering yes tothis question. What are the hypotheses in this survey?a. H0: p = 0.29; HA: p < 0.29 b. H0: p = 0.29; HA: p > 0.29c. H0: p = 0.71; HA: p > 0.71d. H0: p = 0.71; HA: p < 0.71 21. A large company randomly samples 5% of its customers and compares their rate of satisfaction to last year’s satisfaction rate. The z-score for the observed satisfaction rate in this sample is z = 2.12. What should the company conclude? a. The satisfaction rate has only increased. b. The satisfaction rate has only decreased. c. The satisfaction rate has changed. d. There has been no change in the satisfaction rate. 22. An engineer tests air conditioner compressors made with a new design to see if fewer have operational defects than those made with the previous design. The z-score for the observed success rate in this sample is z = –1.82. What should the researcher conclude? a. The old design has fewer operational defects. b. The new design has fewer operational defects. c. There is no difference in the designs. d. There is not enough information to draw a conclusion. 23. National data in 2004 showed that about 55% of the adult population had never smoked cigarettes. Public health officials wonder if this rate has now increased. What type of test should be done? a. Two-tailed test b. Lower-tailed test c. Upper-tailed test d. There are no tails for this test. Unit 5 Evaluation 425 MTHH 041 24. National data in 2004 showed that about 55% of the adult population had never smoked cigarettes. Public health officials wonder if this rate has now increased. What would be a Type I Error? a. I conclude that 55% of adults have never smoked cigarettes. However, more than 55% have never smoked cigarettes. b. I conclude that 55% of adults have never smoked cigarettes. However, it is much less than that. c. I conclude that more than 55% of adults have never smoked cigarettes when actually fewer have never smoked. d. I conclude that 55% of adults have never smoked cigarettes when actually it is more than that. 25. Based on feedback from customers, a company has reduced the amount of sodium in its organic canned soups. Data from extensive consumer testing on the previous recipe showed 73% rating the old soup as satisfactory or better. The company fears that fewer people will be satisfied with the new recipe with less sodium. What would be Type II error? a. I conclude that the proportion of people who like the soup has not changed when actually fewer people do like the soup. b. I conclude that the proportion of people who like the soup has not changed and it, in fact, has not. c. I conclude that fewer people are satisfied with the new recipe when actually the proportion has not changed. d. I conclude that more people are satisfied with the new recipe when actually the proportion has not changed. 26. What is the first step in a hypothesis test? a. Form a null hypothesis and an alternative hypothesis. b. Check assumptions and conditions to use a model for the sampling distribution. c. State a conclusion in the context of a problem. d. Compute the mechanics (find standard deviation, name the test, draw curve, find z). 27. In a hypothesis test, what should we conclude if the data would be very unusual if our original assumption about the parameter were correct? a. The null hypothesis must be false. b. The null hypothesis must be true. c. There’s evidence that the alternative hypothesis is true. d. There’s evidence that the null hypothesis is false. Unit 5 Evaluation 426 MTHH 041 28. An investigative reporter is conducting a poll to test the city council’s claim that they have an 85% approval rating among voters. The reporter finds 60 of 84 people polled approve the city council. Which is true? a. p ˆ = 0.85 b. p = 0.71 c. H0: p < 0.71 d. HA: p ≠ 0.85 29. A statistics professor wants to see if more than 80% of her students enjoyed taking her class. At the end of the term, she takes a random sample of students from her large class and asks, in an anonymous survey, if the students enjoyed taking her class. Which set of hypotheses should she test? a. H0: p < 0.80; HA: p > 0.80b. H0: p = 0.80; HA: p > 0.80c. H0: p > 0.80; HA: p = 0.80d. H0: p = 0.80; HA: p < 0.80 30. Suppose that a manufacturer is testing one of its machines to make sure that the machine is producing more than 97% good parts(H0: p = 0.97 and HA: p > 0.97). Thetest results in a P-value of 0.122. Unknown to the manufacturer, the machine isactually producing 99% good parts. What probably happens as a result of the testing?a. They correctly fail to reject H0.b. They fail to reject H0, making a Type I error.c. They fail to reject H0, making a Type II error.d. They reject H0, making a Type I error.31. A report on health care in the US said that 28% of Americans have experienced timeswhen they haven’t been able to afford medical care. A news organization randomlysampled 801 black Americans, of whom 304 reported that there had been times in thelast year when they had not been able to afford medical care. Assuming that this testmeets the criteria for assumptions and conditions, what would be the z-score?a. 6.27b. 0.3795c. 0.0995d. 1.9632. The step of hypothesis testing that tests the assumptions and conditions is called what?a. Conclusionb. Hypothesesc. Mechanicsd. ModelUnit 5 Evaluation 427 MTHH 04133. In 2005 the U.S. Census Bureau reported that 68.9% of American families owned theirhomes. Census data reveal that the ownership rate in one small city is much lower.The city council is debating a plan to offer tax breaks to first-time home buyers in orderto encourage people to become homeowners. They decide to adopt the plan on a 2-year trial basis and use the data they collect to make a decision regarding tax breaks.Since this plan costs the city tax revenue, they will continue to us it only if there isstrong evidence that the rate of home ownership is increasing. What would be theeffect of a Type II error?a. The city concludes that home ownership is on the rise.b. The city foregoes tax revenue.c. The city withdraws help from those who might otherwise have been able to buy ahouse.d. The city continues to lose money.34. The seller of a loaded die claims that it will favor the outcome 6. We don’t believe thatclaim, and roll the die 200 times to test an appropriate hypothesis. Our P-value turnsout to be 0.03. Which conclusion is appropriate? Explaina. There’s a 3% chance that a loaded die could randomly produce the results weobserved so it’s reasonable to conclude that the die is fair.b. There’s a 3% chance that a fair die could randomly produce the results weobserved, so it’s reasonable to conclude that the die is loaded.c. There’s a 3% chance that the die is fair.d. There’s a 97% chance that the die is fair.35. A company with a fleet of 150 cars found that the emissions systems of 7 out of 22 theytested failed to meet pollution control guidelines. Is this strong evidence that morethan 20% of the fleet may be out of compliance? Which conditions/assumptions arenot met in order to use the Normal model?i. Independence assumptionii. Randomization Conditioniii. 10% Conditioniv. Success/Failure Conditiona. ii and ivb. i and iiic. ii and iiid. iii and iv36. As the value of α gets smaller, how does this affect the z-score?a. The z-score of a one-tailed test is higher.b. The z-score for a one-tailed test is lower.c. The z-score for a two-tailed test is lower.d. There is no change in the z-score.Unit 5 Evaluation 428 MTHH 041__ 37. Here is a histogram of waiting times of the 1,159 calls to a customer service center lastmonth. The mean wait was 47.7 minutes with a standard deviation of 33.14 minutes.Describe the center, and spread in context of the sampling distribution of means ofrandom samples of 75 calls taken from this population.

a. mean: 47.7 minutes; standard deviation: 33.14 minutesb. mean: 74 df; standard deviation: 3.83 minutesc. mean: 47.7 minutes; standard deviation: 3.83 minutesd. mean: 74 df; standard deviation: 33.14 minutes_ 38. A government report on housing costs says that single-family home prices nationwide are skewed to the right, with a mean of $235,700. We collect price data from a random sample of 50 homes in Orange County, California. Why is it okay to use these data for inference even though the population is skewed? a. We can then say the degrees of freedom are 49. b. As the sample size increases, the distribution of sample means will be a t-model. c. We are talking about the sampling distribution of prices which always follows a Normal model. d. We can find the mean and the standard deviation. 39. A hypothesis test of whether the mean number of hours adults spend on their cellphones is more than 30 minutes per day produces a P-value of 0.112. Explain whatthis means in context.a. We would expect about 11.2% of samples of the same size to have a distribution ashigh or higher than the one we saw in our sample.b. We would expect about 11.2% of samples of the same size to have a sample meanas high or higher than the one we saw in our sample.c. 11.2% of our sample population spent more than 30 minutes per day on their cellphones.d. The mean length of time that adults spent on their cell phones was more than 30minutes 11.2% of the time.Unit 5 Evaluation 429 MTHH 04140. How does the t-model compare to the Normal model for various sample sizes?a. As the sample size decreases, the t-model comes closer to approximating a Normalmodel.b. The larger sample sizes increase the variation so the t-model and Normal modelbecome more similar.c. As the sample size increases, the t-model comes closer to approximating a Normalmodel.d. The difference between the two is the degrees of freedom.41. It is expected that a quarter of the responses to the question “Do you enjoy being aparent?” will be “unsure” and there will be twice as many “yes” as “no” responses. Findthe expected distribution for 86 responses.a. Yes: 0.5; No: 0.25; Not Sure: 0.25b. Yes: 42; No: 22; Not Sure: 22c. Yes: 50%; No: 25%; Not Sure: 25%d. Yes: 43; No: 21.5; Not Sure: 21.5Use the following information to answer questions 42 – 44.Here are the weights in ounces of a random sample of 25 oranges harvested at Sunny DayOrchards this season.12.6 10.0 11.4 13.6 12.88.0 9.7 10.1 9.5 7.19.0 9.3 6.5 15.9 7.414.1 5.1 8.8 4.9 8.110.8 7.5 4.6 10.1 10.242. Find the standard error for this sample.a. 0.5641b. 0.9843c. 0.5455d. 0.575843. What is the margin of error?a. 1.1885b. 2.0316c. 1.1643d. 1.126Unit 5 Evaluation 430 MTHH 04144. Find the 95% confidence interval.a. (8.2956, 10,673)b. (7.4524, 11.5156)c. (8.3197, 10.6483)d. (8.358, 10.61)45. Interpret the confidence interval (7.569, 9.325) for another sample from Blue StemOrchards, a neighboring orchard.a. 95% of all oranges from Blue Stem Orchards weigh between 7.569 ounces and9.325 ounces.b. I am 95% confident that the mean weight of an orange from Blue Stem Orchardsthis season is between 7.569 ounces and 9.325 ounces.c. I am 95% confident that a randomly selected orange from Blue Stem Orchards thisseason is between 7.569 ounces and 9.325 ounces.d. 95% of all samples from Blue Stem Orchards will have mean weight between 7.569ounces and 9.325 ounces.46. The distribution of salaries of television news broadcasters is skewed to the right. Whichis true according to the Central Limit Theorem?a. The distribution of all the salaries of the broadcasters can be described by a Normalmodel.b. The spread of the sampling distribution of sample means would be smaller for smallsamples than for larger samples.c. The sampling distribution of means for large random samples of these salaries canbe approximated by a Normal model.d. The standard deviation of the sampling distribution of means for large randomsamples of these salaries is equal topqˆˆn.Use this information to answer questions 47 – 49.A group of teenagers is chosen randomly and asked which of three indoor activities theypreferred. Of the 126 teenagers, 36 preferred listening to music, 16 preferred reading, andthe rest preferred going online or playing video games. You are wondering if theseresponses indicate a preference among teenagers in general.47. Find the actual count and the expected count for each activity.a. Music:36, 0.286; Reading: 16, 0.127; Online/Gaming: 74, 0.587b. Music:36, 42; Reading: 16, 42; Online/Gaming: 74, 42c. Music:42, 36; Reading: 42, 16; Online/Gaming: 42, 74d. Music:36, 126; Reading: 16, 126; Online/Gaming: 74, 126Unit 5 Evaluation 431 MTHH 04148. Find the2 statistic.a. 0.857b. 24.381c. 41.333d. 2 degrees of freedom49. Assuming a very small P-value for the2 statistic, what conclusions can be drawn?a. There is strong evidence that there is a difference in teenagers’ preference amongthese three indoor activities.b. There is strong evidence that there is no difference in teenagers’ preference amongthese three indoor activities.c. The probability that this is accurate is very small.d. A very small proportion of teenagers were sampled.__ 50. A2 statistic is calculated to be 37.12. What should you conclude?a. Reject the null hypothesis.b. Fail to reject the null hypothesis.c. The standard deviation is most cases will be less than 37.12.d. Not enough information is given.Carefully check your answers on this evaluation and make any corrections you feel arenecessary. When you are satisfied that you have answered the questions to the best of yourability, transfer your answers to an answer sheet. Please refer to the information sheet thatcame with your course materials.

Sample Solution
The post Introduction to Statistics

Image result for Order Now images

CLICK HERE TO ORDER NOW!!

Leave a Reply

Your email address will not be published. Required fields are marked *