Humble Independent School District / Homepage



Chapter 10 Quiz Review 16-17Name: _________________________________ Period: ____________1.A state policeman has a pet theory that people who drive red cars are more likely to drive too fast. On his day off, he borrows one of the department’s radar guns, parks his car in a rest area, and measures the proportion of red cars and non-red cars that are driving too fast. (He decides ahead of time to define “driving too fast” as exceeding the speed limit by more than 5 miles per hour). To produce a random sample, he rolls a die and only includes a car in his sample if he rolls a 5 or a 6. He finds that 18 of 28 red cars are driving too fast, and 75 of 205 other cars are driving too fast. (a) Is this convincing evidence that people who drive red cars are more likely to drive too fast, as the policemen has defined it? Support your conclusion with a test of significance, using a = 0.05.(b) Construct and interpret a 95% confidence interval for the difference in proportion of red cars that drove too fast and other cars that drive too fast.2.A study of “adverse symptoms” in users of over-the-counter pain relief medications assigned subjects at random to one of two common pain relievers: acetaminophen and ibuprofen. In all, 650 subjects took acetaminophen, and 44 experienced some adverse symptom. Of the 347 subjects who took ibuprofen, 49 had an adverse symptom. (a) Does the data provide convincing evidence that the two pain relievers differ in the proportion of people who experience an adverse symptom? Support your conclusion with a test of significance. Use a = 0.05.3.Jordan’s cat “Fern” is a finicky eater. Jordan is trying to determine which of two brands of canned cat food Fern prefers, Tab-a-Cat or Chow Lion. For two months, she flips a coin each day to decide which of the two foods to feed Fern, and weighs how much Fern eats in grams. Here is the data:nsTab-a-Cat3185.23.45Chow Lion3082.14.62(a) Find the standard error for the difference in the mean amount of Tab-a-Cat that Fern eats and the mean amount of Chow Lion she eats.(b) Construct and interpret a 99% confidence interval for the difference in mean amount of food Fern eats when she is offered Tab-a-Cat and when she is offered Chow Lion.(c) Complete a significance test to test the hypothesis that the mean amount of Tab-a-Cat Fern eats is higher than the mean amount of Chow Lion she eats. 4.As a non-native English speaker, Sanda is convinced that people find more grammar and spelling mistakes in essays when they think the writer is a non-native English speaker. To test this, she randomly sorts a group of 40 volunteers into two groups of 20. Both groups are given the same paragraph to read. One group is told that the author of the paragraph is someone whose native language is not English. The other group is told nothing about the author. The subjects are asked to count the number of spelling and grammar mistakes in the paragraph. While the two groups found about the same number of real mistakes in the passage, the number of things that were incorrectly identified as mistakes was more interesting. Here are the results:Number of “mistakes” found“Native English Speaker”01300100021200320020“Non-native English speaker”21501487601014742145Do these data provide convincing evidence that readers are more likely to incorrectly identify errors in writing if they think the author’s native language is not English? Support your conclusions with an appropriate statistical test.Chapter 10 Quiz Review 16-17Answer SectionOTHER1.ANS:A. State: We wish to test , where and are the proportion of red cars and other car, respectively, who are driving too fast. We will use a significance level of a = 0.05. Plan: The procedure is a two-sample z-test for the difference of proportions. Conditions: Random: The policemen chose cars randomly by rolling a die. 10%: We can safely assume that the number of cars driving past the rest area is essentially infinite, so the 10% restriction does not apply. Large counts: The number of successes and failures in the two groups are 18, 10, 75, and 130—all of which are at least 10. Do: , and , so ; One-tailed P-value = 1 – 0.9968 = 0.0032. [A 2-proportion z-test on the calculator yields z = 2.807 and a P-value of 0.0025] Conclude: A P-value of 0.0032 is less than a = 0.05, so we reject H0. We have sufficient evidence to conclude that the proportion of red cars that drive too fast on this highway is greater than the proportion of non-red cars that drive too fast. B. State: We wish to estimate, with 95% confidence, the difference , as defined in part A. .Plan: We should use a 2-sample z-interval for . The conditions were addressed in part A. .Do: The critical z for 95% confidence is 1.96, so the interval is , or .Conclude: We are 95% confident that the interval from 0.070 to 0.470 captures the true difference in the proportion of red cars and non-red cars that drive too fast on this highway.PTS:12.ANS:A. State: We are testing the hypotheses versus where and are the proportions of people who experience adverse reactions to acetaminophen and ibuprofen, respectively, among all patients (similar to those in this experiment) who might receive these drugs. We will use a significance level of a = 0.05. Plan: The procedure is a two-sample z-test for the difference of proportions. Conditions: Random: Problem states that subjects were assigned at random. 10%: Since no sampling took place, the 10% condition does not apply. Large counts: The number of successes (adverse reactions) and failures in the two groups are 44, 606, 49, 298—all of which are at least 10. Independent: Random assignment means we can view these groups as independent, and adverse reactions in one person should not influence reactions in another person. Do: , , so . This test statistic is so large that by Table A the P-value is very close to 0 (by calculator, P-value = 0.00014). Conclude: A P-value of 0.00014 is much less than _ = 0.05, so we reject H0. We have sufficient evidence to conclude that the proportions of people who experience adverse reactions to acetaminophen and ibuprofen are different. B. The critical z for 95% confidence is 1.96, so the margin of error is . C. When we conduct a significance test, we assume that the null hypothesis is true. In this case, , so we are assuming that . This means we are assuming the two samples are independent samples that estimate the same quantity. It thus makes sense to combine this information into single pooled estimate when estimating . When constructing a confidence interval, we are not making the assumption that , so we should not pool the data.PTS:13.ANS:A. Let = mean amount of Tab-a-Cat Fern eats and = mean amount of Chow Lion she eats. Standard error of is . B. State: We wish to estimate, with 99% confidence, the difference , as defined in part A. . Plan: We should use a 2-sample t-interval for . Conditions: Random: Feedings of each type of food were randomly determined by coin flip. 10%: We can view this study as randomly assigning 61 feedings to one of two groups, thus sampling did not take place, so the 10% condition does not apply. Normal/Large Sample: since both samples are at least 30, the Normal condition is satisfied. Do: Using the conservative degrees of freedom of 29, the critical t-value for 99% confidence is 2.756, so the interval is , or . [Using a calculator and 53.64 degrees of freedom, the interval is ]. Conclude: We are 99% confident that the interval from 0.22 to 5.98 captures the true difference in the mean amount of Tab-a-Cat Fern eats and the mean amount of Chow Lion she eats. C. . D. Using Table A and df?=?29, 0.0025< P-value < 0.005. Using the calculator and df = 53.64, P-value = 0.0023. Since in both cases the P-value is less than a =?0.01, we reject H0. We have convincing evidence that Fern eats more, on average, when offered Tab-a-Cat than when offered Chow Lion.PTS:14.ANS:1. State: We are testing the hypotheses versus , where is the mean number of mistakes someone finds if they think the writer is a non-native English speaker and is the mean number of mistakes someone finds if they think the writer is a native English speaker. We will use a significance level of a = 0.05. Plan: The procedure is a two-sample t-test for the difference of means. Conditions: Random: The problem states that subjects were assigned at random. 10%: Since no sampling took place, the 10% condition does not apply. Normal/Large Sample: Since n?=?20 for each group, we need to examine the distributions of sample data for strong skew or outliers. The dot plots at right indicate that both distributions are moderately skewed right, but the sample sizes are probably large enough to compensate for this. Do: Summary statistics at right. Using Table?A and df?=?19, 0.0005< P-value < 0.001. Using the calculator and df = 25.56, P-value?=?0.00055. Conclude: Since in both cases the P-value is less than _?=?0.05, we reject H0. We have convincing evidence that the mean number of non-existent “mistakes” people find when reading something they think has been written by a non-Native English speaker is greater than the mean number they find if they think the writer’s native language is English.PTS:1 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download