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Name_________________________________ Date______________
AP Statistics
Zielenkievicz
Chapter 10 Take Home
1. Just before the presidential election in November 2008, a local newspaper conducted a poll of residents of a medium-sized city and found that 120 out of a simple random sample of 250 men intended to vote for Barack Obama and 132 out of an SRS of 240 women intended to vote for Obama.
(a) Is this convincing evidence that there was a gender difference in Obama’s support in this city? Support your conclusion with a test of significance, using α = 0.05.
(b) Construct and interpret a 95% confidence interval for the difference in proportion of women and men who supported Obama in this city.
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 α = 0.05.
(b) Find the margin of error for a 95% confidence interval for the difference in the proportions of people who experience adverse reactions to these two mediations. You need not carry out all the steps in constructing a confidence interval.
(c) In parts (a) and (b) above, you should have used two different formulas for estimating the standard deviation of the randomization distribution of [pic]. Explain why it makes sense to do this.
3. You have two large bins of several thousand plastic beads. Bin A is 60% red beads; Bin B is 48% red beads. Suppose you take a random sample of 80 beads from each bin and calculate [pic]= the proportion of red beads in the sample from Bin A and [pic]= the proportion of red beads in the sample from Bin B.
(a) Describe the sampling distribution of [pic].
(b) What is the probability that the proportion of red beads you select from Bin B is higher than the proportion of red beads you select from Bin A?
4. In a study of heart surgery, one issue was the effect of drugs called beta-blockers on the pulse rate of patients during surgery. The available subjects were divided at random into two groups of 30 patients each. One group received a beta-blocker; the other group received a placebo. The pulse rate of each patient at a critical point during the operation was recorded. The treatment group had a mean pulse rate of 65.2 and standard deviation 7.8. For the control group, the mean pulse rate was 70.3 and the standard deviation was 8.3.
(a) Find the standard error for the difference in mean pulse rate between the two groups.
(b) Construct and interpret a 99% confidence interval for the difference in mean pulse rates.
(c) Suppose we want to test the hypothesis that beta-blockers reduce mean pulse rate. State the null and alternative hypotheses for this test.
(d) The test statistic is t = –2.453. Determine the P-value and draw an appropriate conclusion, using α = 0.05.
5. Nicotine patches are often used to help smokers quit. Does giving medicine to fight depression also help? A randomized double-blind experiment assigned 244 smokers to receive nicotine patches and another 245 to receive both a patch and the antidepressant drug bupropion. After a year, 40 subjects in the nicotine patch group had abstained from smoking, as had 87 in the patch-plus-drug group.
(a) Construct and interpret a 99% confidence interval for the difference in the proportion of smokers who abstain when using buproprion and a nicotine patch and the proportion who abstain when using only a patch.
(b) Based only on this interval, do you think that the difference in proportion of abstaining smokers is significant? Justify your answer.
6. Researchers studying the acquisition of pronunciation often compare measurements made on the recorded speech of adults and children. One variable of interest is called "voice onset time" (VOT), the length of time between the release of a consonant sound (such as "p") and the beginning of an immediately following vowel (such as the "a" in "pat"). For speakers of English, this short time lag can be heard as a period of breathiness between the consonant and the vowel.
Here are the results for some randomly-selected 4-year-old children and adults asked to pronounce the word “pat.” VOT is measured in milliseconds and can be either positive or negative.
| |n |[pic] |s |
|Children |10 |60.67 |39.89 |
|Adults |20 |88.17 |24.74 |
You are interested in whether there is a difference in the VOT of adults and children, so you plan to test [pic] against [pic], where [pic] and [pic] are the population mean VOT for adults and children, respectively.
(a) What additional information would you need to confirm that the conditions for this test have been met?
(b) Assuming the conditions have been met, calculate the test statistic and P-value for this test.
(c) Interpret the P-value in the context of this study, and draw the appropriate conclusion at the α = 0.05 level.
(d) Given your conclusion in part (c), which type of error, Type I or Type II, is it possible to make? Describe that error in the context of this study.
7. An association of Christmas tree growers in Indiana sponsored a sample survey of 500 randomly selected Indiana households to help improve the marketing of Christmas trees. One question the researchers asked was, “Did you have a Christmas tree this year?” Respondents who had a tree during the holiday season were asked whether the tree was natural or artificial. Respondents were also asked if they lived in an urban area or in a rural area. The tree growers want to know if there is a difference in preference for natural trees versus artificial trees between urban and rural households. Among the 160 who lived in rural areas, 64 had a natural tree. Among the 261 who lived in an urban area, 89 had a natural tree.
(a) Construct and interpret a 95% confidence interval for the difference in the proportion of rural and urban Indiana residents who had a natural Christmas tree this year.
(b) Suppose the tree growers asked 10 different questions about Christmas tree preferences in an effort to find differences between urban and rural residents. Explain why it would be unwise to perform 10 different 2-sample significance tests (α = 0.05) on their data and conclude that any significant result establishes a difference between the two groups.
Chapter 10 Take Home
Answer Section
SHORT ANSWER
1. ANS:
(a) State: We wish to test[pic], where [pic] and[pic] are the proportion of male and female voters in this city, respectively, who supported Obama. We will use a significance level of α = 0.05. Plan: The procedure is a two-sample z-test for the difference of proportions. Conditions: Random: The problem states that two SRSs were taken. Normal: The number of successes and failures in the two groups are 120, 130, 132, and 108—all of which are at least 10. Independent: We have two independent samples of men and women. It’s reasonable to assume that individual responses are independent as well, and that there are at least [pic]male and [pic] female voters in the city. Do: [pic], and [pic], so [pic]; Two-tailed P-value = (2)(0.0606) = 0.1212. Conclude: A P-value of 0.1212 is greater than α = 0.05, so we cannot reject H0. We do not have sufficient evidence to conclude that the there is a difference in the proportion of males and females who supported Obama.
(b) State: We wish to estimate, with 95% confidence, the difference [pic], as defined in part (a).
Plan: We should use a 2-sample z-interval for [pic]. The conditions were addressed in part (a).
Do: The critical z for 95% confidence is 1.96, so the interval is [pic], or [pic].
Conclude: We are 95% confident that the interval from – 0.158 to 0.018 captures the true difference in proportion of male and female voters supporting Obama in this city.
PTS: 1 REF: Quiz 10.1A TOP: Comparing Two Proportions
2. ANS:
(a) State: We are testing the hypotheses [pic], where [pic] and [pic]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 α = 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. Normal: 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: [pic], [pic], so
[pic]. This test statistic is so large that by Table A the P-value is very close to 0 (by calculator, P-value = 0.00012). Conclude: A P-value of 0.00012 is much less than α = 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 [pic].
(c) when we conduct a significance test, we assume that the null hypothesis is true. In this case, [pic], so we are assuming that [pic]. 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 [pic] when estimating [pic]. When constructing a confidence interval, we are not making the assumption that [pic], so we should not pool the data.
PTS: 1 REF: Quiz 10.1C TOP: Comparing Two Proportions
3. ANS:
(a) [pic]; [pic];
Since [pic]all of which are at least 10, the shape of the distribution is approximately Normal.
(b) [pic].
PTS: 1 REF: Quiz 10.1A TOP: Comparing Two Proportions
4. ANS:
(a) Let [pic]= true mean pulse rate of patients similar to those in the experiment who take beta-blockers and [pic]= mean pulse rate of similar patients who do not take beta blockers. Standard error of is [pic].
(b) State: We wish to estimate, with 99% confidence, the difference [pic], as defined in part (a). Plan: We should use a 2-sample t-interval for [pic]. Conditions: Random: The subjects were randomly assigned to the experimental treatments. Normal: since both samples are at least 30, the Normal condition is satisfied. Independent: Random assignment means we can view these groups as independent, and pulse rates in one patient should not influence pulse rate in another patient. Do: Using the conservative degrees of freedom of 29, the critical t-value for 99% confidence is 2.756., so the interval is [pic], or [pic]. [Using a calculator and 57.78 degrees of freedom, the interval is [pic]]. Conclude: We are 99% confident that the interval from –10.83 to 0.63 captures the true difference in the mean pulse rate of patients receiving a beta-blocker during surgery and those who take a placebo.
(c) [pic]; [pic].
(d) Using Table A and df = 29, 0.01< P-value < 0.02. Using the calculator and df = 57.78, P-value = 0.0086. Since in both cases the P-value is less than α = 0.05, we reject H0. We have convincing evidence that the mean pulse rate of patients taking beta-blockers is lower than the mean pulse rate of patients taking a placebo.
PTS: 1 REF: Quiz 10.2A TOP: Comparing Two Means
5. ANS:
(a) State: We wish to estimate, with 99% confidence, the difference [pic], where [pic] and [pic] are, respectively, the proportion of subjects who abstain from smoking for a year while being treated with bupropion and nicotine patches or with patches alone (control). Plan: We should use a 2-sample z-interval for [pic]. Conditions: Random: Subjects were randomly assigned to one of the two treatment groups. Normal: Number of successes and failures in the two groups are 87, 158, 40, and 204, all of which are at least 10. Independent: Random assignment means we can view the two groups of subjects as independent, and it seem reasonable to assume that whether one subject abstains from smoking is independent of whether another subject does. Do: The critical z for 99% confidence is 2.576, and [pic]. The interval is [pic], or [pic]. Conclude: We are 99% confident that the interval from 0.091 to 0.291 captures the true difference in the proportion of smokers using bupropion and nicotine patches who abstain for a year versus those using only patches who abstain for a year.
(b) Since 0 is not within the 99% confidence interval of plausible values for the difference in proportion of patch-plus-buproprion abstainers and patch-only abstainers. this interval provides convincing evidence that there is a significant difference. A two-tailed test of significance would reject the null hypothesis that [pic] at the α = 0.01 level.
PTS: 1 REF: Test 10A
6. ANS:
(a) Since the subjects were randomly selected, the Normal condition is satisfied. Of greater concern is the small sample sizes: we need to know that there were no outliers in the samples, that the adult sample was not strongly skewed, and that the children sample is not skewed at all, since this sample is particularly small.
(b) [pic] Using Table B and conservative df = 9, the two tailed P-value is between 0.025 and 0.05. Using the calculator and df = 12.57, P-value = 0.034.
(c) If there is no difference between the mean VOT of adults and children, the probability of obtaining a sample mean difference this far or farther from 0 is 0.034. We reject H0. We have convincing evidence that there is a difference in the mean VOT of adults and children.
(d) It’s possible we made a Type I error, which would be to conclude that there is a difference in mean VOT between adults and children when there is none.
PTS: 1 REF: Test 10A
7. ANS:
State: We wish to estimate, with 95% confidence, the difference [pic], where[pic]and[pic]are, respectively, the proportion of rural and urban residents who had a natural Christmas tree this year. Plan: We should use a 2-sample z-interval for[pic]. Conditions: Random: The study used 500 randomly-selected Indiana households. We can view this as SRSs from rural and urban populations. Normal: Number of successes and failures in the two groups are 64, 96, 89, and 172, all of which are at least 10. Independent: It seem reasonable to assume that the samples are independent, and there are certainly more than 5000 households in Indiana. Do: The critical z for 95% confidence is 1.96 and [pic]. The interval is given by [pic], or [pic]. Conclude: We are 95% confident that the interval from –0.035 to 0.155 captures the true difference in the proportion of rural versus urban residents of Indiana who had a natural Christmas tree this year.
(b) At the α = 0.05 level, we expect to make a Type I error and reject a true H0 about once in every 20 tests we perform. If we perform 10 such tests, the chances are quite good that we will make at least one Type I error. (In fact, the probability is [pic]).
PTS: 1 REF: Test 10C
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