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Chapter 8 Test ReviewName: _____________________________________Period: ____________________Multiple ChoiceIdentify the choice that best completes the statement or answers the question.____1.A 95% confidence interval for the mean ? of a population is computed from a random sample and found to be 9 ? 3. We may conclude thata.there is a 95% probability that ? is between 6 and 12.b.95% of values sampled are between 6 and 12.c.if we took many, many additional random samples and from each computed a 95% confidence interval for ?, approximately 95% of these intervals would contain ?.d.there is a 95% probability that the true mean is 9 and a 95% chance that the true margin of error is 3.e.all of the above are true.____2.A polling organization announces that the proportion of American voters who favor congressional term limits is 64 percent, with 95% confidence and margin of error of 3 percent. This means thata.if the poll were conducted again in the same way, there is a 95% chance that the fraction of voters favoring term limits in the second poll would be between 61 percent and 67 percent.b.there is a 95% probability that the true percentage of voters favoring term limits is between 61 and 67 percent.c.if the poll were conducted again the same way, there is a 95% probability that the percentage of voters favoring term limits in the second poll would be within 3 percent of the percentage favoring term limits in the first poll.d.among 95% of the voters, between 61 percent and 67 percent favor term limits.e.none of the above.____3.A political candidate is told by his polling organization that a 90% confidence interval for the proportion of voters who support his candidacy is 0.45 to 0.53. What are the point estimate and margin of error for this interval?a.Point estimate = 0.50; margin of error = 0.08.b.Point estimate = 0.49; margin of error 90%c.Point estimate = 0.49; margin of error 0.08.d.Point estimate = 0.49; margin of error = 0.04.e.Point estimate = 0.49; margin or error cannot be determined without sample size.____4.Other things being equal, the margin of error of a confidence interval increases asa.the sample size increases.b.the sample mean increases.c.the population standard deviation increases.d.the confidence level decreases.e.none of the above.____5.A polling organization announces that the proportion of American voters who favor congressional term limits is 64 percent, with a 95% confidence margin of error of 3 percent. If the opinion poll had announced the margin of error for 80% confidence rather than 95% confidence, this margin of error would bea.3%, because the same sample is used.b.less than 3%, because we require less confidence.c.less than 3%, because the sample size is smaller.d.greater than 3%, because we require less confidence.e.greater than 3%, because the sample size is smaller.____6.A quality control inspector is testing microprocessor chips made during a single day by a new machine to determine the proportion of defective chips. She selects an SRS of 80 chips from the 3000 chips produced by the machine on that day. It turns out that six of the chips are defective. Which of the following conditions for constructing a confidence interval for the proportion of defective chips has been violated?a.b.An SRS has been taken from the population of interest.c.The population is at least 10 times the size of the sample.d.The population is approximately Normally distributed.e.There appear to be no violations.____7.Eighty rats whose mothers were exposed to high levels of tobacco smoke during pregnancy were put through a simple maze. The maze required the rats to make a choice between going left or going right at the outset. Sixty of the rats went right when running the maze for the first time. Assume that the eighty rats can be considered an SRS from the population of all rats born to mothers exposed to high levels of tobacco smoke during pregnancy. (Note that this assumption may or may not be reasonable, but researchers often assume lab rats are representative of such larger populations because lab rats are often bred to have very uniform characteristics.) The standard error for the proportion p of those who went right the first time when running the maze isa.0.0023.b.0.0072.c.0.0484.d.0.0548.e.0.0559.____8.A noted psychic was tested for ESP. The psychic was presented with 200 cards face down and asked to determine if the card featured one of five symbols: star, cross, circle, square, or three wavy lines. The psychic was correct in 48 cases. If p represents the proportion of correct guesses the psychic made, what is the standard error of p?a.0.0009b.0.0283c.0.0302d.0.0354e.0.4330Scenario 8-4A sociologist is studying the effect of having children within the first two years of marriage on the divorce rate. Using hospital birth records, she selects a random sample of 200 couples that had a child within the first two years of marriage. Following up on these couples, she finds that 80 couples are divorced within five years.____9.Use Scenario 8-4. A 90% confidence interval for the proportion p of all couples that had a child within the first two years of marriage and are divorced within five years isa.0.40 ± 0.004.b.0.40 ± 0.035.c.0.40 ± 0.044.d.0.40 ± 0.057.e.0.40 ± 0.068.____10.Use Scenario 8-4. Which of the following is closest to the sample size you would need to estimate p with a margin of error of 0.02 with 90% confidence? Use 0.4 from the first sample as an approximation of p.a.24b.600c.1624d.2305e.3842Other1.A survey of a random sample of 1280 student loan borrowers found that 218 had loans totaling more than $40,000 for their undergraduate education. Construct and interpret a 95% confidence interval to estimate the population proportion of student loan borrowers who have loans totaling more than $40,000.2.A cereal maker’s container machine is designed to fill boxes so that the mean weight of cereal in the boxes is 18 ounces. A simple random sample of 30 boxes produced by the machine yields a mean weight of 17.92 ounces and a standard deviation of 0.2 ounces. The distribution of box weights is summarized in the Minitab output below: MEAN STDEV SEMEAN MIN Q1 MEDIAN Q3 MAX 17.920 0.200 0.0365 17.55 17.76 17.90 18.00 18.25(a) Construct and interpret a 90% confidence interval to estimate the true mean weight of cereal in the boxes. (b) Does the interval in A. give you reason to suspect that the machine is not filling boxes with the correct amount of cereal? Explain your reasoning.3.The level of dissolved oxygen in a river is an important indicator of the water’s ability to support aquatic life. You collect water samples at 15 randomly chosen locations along a stream and measure the dissolved oxygen. Here are your results in milligrams per liter:4.535.043.295.234.135.504.834.405.426.384.014.662.875.735.55Construct and interpret a 95% confidence interval to estimate the mean dissolved oxygen level in this stream.4.Political parties rely heavily upon polling to measure their support in the electorate. In a country with four major political parties, a poll is conducted and the Coffee Party is supported by 435 of the 1183 randomly-selected voters who were polled. (a) Construct and interpret a 98% confidence interval to estimate the proportion of voters who support the Coffee Party.(b) The Coffee Party leaders claim that they have the support of 42% of the electorate. Does your interval in part A. suggest that this claim is plausible? Explain your reasoning.(c) The random sample was actually 1200 people, but 17 people never responded. Does this change your answer to part B.? Explain why or why not.Chapter 8 Test ReviewAnswer SectionMULTIPLE CHOICE1.ANS:CPTS:1TOP:Interpret confidence level2.ANS:EPTS:1TOP:Interpret confidence level3.ANS:DPTS:1TOP:Margin of error and point estimate4.ANS:CPTS:1TOP:Factors influencing width of confidence interval5.ANS:BPTS:1TOP:Factors influencing width of confidence interval6.ANS:APTS:1TOP:Conditions for confidence interval for p7.ANS:CPTS:1TOP:Standard error of p-hat8.ANS:CPTS:1TOP:Standard error of p-hat9.ANS:DPTS:1TOP:Calculate confidence interval for p10.ANS:CPTS:1TOP:Choosing sample size (proportions)OTHER1.ANS:State: We wish to estimate, with 95% confidence, the true proportion of student loan borrowers who have loans totaling more than $40,000 for their undergraduate education. Plan: We will use a one-sample z-interval for a population proportion. Conditions: Random: The problem states that the survey involved a “random sample.” We must assume this was an SRS. There are millions of student loan borrowers, so the sample is less than 10% of the population. Normal/Large Counts: Do: , so the 95% confidence interval is . Conclude (i.e., interpret the interval): We are 95% confident that the interval 0.149 to 0.191 contains the true proportion of student loan borrowers who have loans totaling more than $40,000 for their undergraduate education.PTS:12.ANS:A. State: We wish to estimate, with 90% confidence, the true mean weight of cereal in boxes filled by this machine. Plan: We will use a one-sample t-interval for a population mean. Conditions: Random: The question states that an SRS was taken. While the population of all boxes filled by the machine is not infinite, it is very large, satisfying the 10% condition. NormalLarge Sample: Since and the five-number summary shows no outliers, the t-interval is appropriate. Do: Critical t* for 90% confidence and 29 df is 1.699, so the 90% confidence interval is . Conclude (i.e., interpret the interval): We are 90% confident that the interval 17.858 to 17.982 ounces contains the true mean weight of cereal boxes filled by this machine. B. Since the claimed mean of 18 ounces is not in this confidence interval, we have evidence that the mean is not 18PTS:13.ANS:State: We wish to estimate, with 95% confidence, the true mean level of dissolved oxygen in the stream. Plan: We will use a one-sample t-interval for a population mean. Conditions: Random: We will assume that taking samples at randomly-chosen locations in the stream constitutes a simple random sample. Water samples from streams are essentially taken from an infinite population, so the 10% condition does not apply. Normal/Large sample: The sample size is small, but a dotplot of the sample data (right) shows little skew and no outliers. Do: ; critical t* for 95% confidence and 14 df is 2.145, so the 95% confidence interval is Conclude (i.e., interpret the interval): We are 95% confident that the interval 4.250 to 5.292 mg per liter contains the true mean level of dissolved oxygen in the stream. PTS:14.ANS:A. State: We wish to estimate, with 98% confidence, the true proportion of voters who support the Coffee Party. Plan: We will use a one-sample z-interval for a population proportion. Conditions: Random: The problem states that voters were “randomly selected.” We must assume this was an SRS and that the country is large enough so that 1183 is less than 10% of the population. Normal/Large counts: ; Do: so the 98% confidence interval is . Conclude (i.e., interpret the interval): We are 98% confident that the interval 0.3351 to 0.4003 captures the true proportion of voters who support the Coffee Party. B. Since the entire interval is below 0.42, we are 98% confident that the true proportion of Coffee Party supports is different from 42%. C. Consider the most extreme scenario, in which all 17 individuals who did not respond actually support the Coffee Party. Then and the 98% confidence interval is or 0.3442 to 0.4092, and 0.42 is still (just barely!) above the interval. We still can still be confident that the true proportion of voters supporting the Coffee Party is different from 0.42. But it’s close! [Note: how and when a student rounds results may change their answer to this question. It should be considered correct if the conclusion is consistent with calculated results.]PTS:1 ................
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