Download Chapter 8 Test Review The test will consist of both multiple choice

1 Chapter 8 Test Review The test will consist of both multiple choice and open-­‐response questions, and the questions will be AP style questions. The following is a list of sample questions from the test. Note: Not all questions are listed. Note: Prior material from other chapters is subject to showing up on the exam. 1) 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: a) Construct and interpret a 90% confidence interval to estimate the true mean weight of cereal in the boxes. 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. 10%: While the population of all boxes filled by the machine is not infinite, it is very large, satisfying the 10% condition. Normal/Large Sample: Since n = 30 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 0.200
17.92 ± 1.699
30
17.92 ± 0.062 (17.858, 17.982) Conclude: 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) 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. Since the claimed mean of 18 ounces is not in this confidence interval, we have evidence that the mean is not 18. 2 2) You want to calculate a 80% confidence interval for a population mean from a sample of n = 9. What is the appropriate critical t*? t* = 1.397 (use df = 8 and area of 0.10) 3) The government claims that students earn a mean of $4500 during their summer break from studies. A random sample of students gave a sample average of $3975, and a 95% confidence interval was found to be $3525 < μ < $4425. What does the phrase “95% confidence” mean? The method used to get the interval from $3525 to $4425, when used over and over, produces intervals which include the true population mean about 95% of the time 4) You want to estimate the mean SAT score for a population of students with a 90% confidence interval. Assume that the population standard deviation is σ = 100. How large should n be so that the margin of error is no more than ±10? 𝝈
𝒛∗
≤ 𝑴𝑬 𝒏
𝟏𝟎𝟎
𝟏. 𝟔𝟒𝟓
≤ 𝟏𝟎 𝒏
𝒏 ≈ 𝟐𝟕𝟏 5) An SRS of 50 students at a large middle school are asked, “Do you have a television in your bedroom? Twenty-­‐eight of the students responded “yes.” If 𝑝 = the proportion of students who answered “yes,” what is the standard error of 𝑝? 𝒑(𝟏 − 𝒑)
=
𝒏
𝟎. 𝟓𝟔(𝟏 − 𝟎. 𝟓𝟔)
= 𝟎. 𝟎𝟕𝟎𝟐 𝟓𝟎
6) The college newspaper of a large Midwestern university periodically conducts a survey of students on campus to determine the attitude on campus concerning issues of interest. Pictures of the students interviewed along with quotes of their responses are printed in the paper. Students are interviewed by a reporter “roaming” the campus selecting students to interview. On a particular day the reporter interviews five students and asks them if they feel there is adequate student parking on campus. Four of the students say, “no.” Which of the following conditions for inference about a proportion using a confidence interval are violated in this example? (multiple conditions may be violated) I. The data are an SRS from the population of interest. II. The population is at least ten times as large as the sample. III. 𝒏𝒑 ≥ 10 and 𝒏(𝟏 − 𝒑) ≥10 . 3 7) A 95% confidence interval for the mean play-­‐time for the population of all Tay-­‐Tay songs is (3.5, 4.5). Suppose you compute a 90% confidence interval using the same data. Which of the following statements is correct? a) The intervals have the same width b) The 90% interval is narrower c) The 90% interval is wider d) The 90% interval could be wider or narrower – it depends on the sample e) This question is a nightmare dressed as a daydream 8) A 90% confidence interval for p, the proportion of all South Pole Elves that eat from the 4 main food groups (candy, candy corns, candy canes, and syrup), was found to be (0.812, 0.886). The point estimate and margin of error for this interval are: 𝟎.𝟖𝟏𝟐!𝟎.𝟖𝟖𝟔
PE: = 𝟎. 𝟖𝟒𝟗 ME: 𝟎. 𝟖𝟖𝟔 − 𝟎. 𝟖𝟒𝟗 = 𝟎. 𝟎𝟑𝟕 𝟐
9) When the sample size is small, why is it necessary to inspect a graph of the sample data when calculating a confidence interval for a population mean? To verify that the population can be considered approximately Normal 10) A survey of a random sample of 180 credit card users found that the average credit card debt was $10,180 with a standard deviation of $1800. Set up the formula for a 95% confidence interval for the true mean debt amount. 𝟏𝟖𝟎𝟎
𝟏𝟎𝟏𝟖𝟎 ± 𝟏. 𝟗𝟕𝟑
𝟏𝟖𝟎
(use t*) 11) Note: Remember that the margin of error for a confidence interval covers only chance variation due to random sampling or random assignment. Margin of error does NOT account for bias due to a poor sampling technique. 4 12) A simple random sample of 1100 males aged 12 to 17 in the United States were asked whether they played massive multiplayer online role-­‐playing games (MMORPGs); 775 said that they did. Use this information to construct a 95% confidence interval to estimate the proportion of all U.S. males aged 12 to 17 who play MMORPGs. State: Construct a 95% confidence interval for the true population proportion p, the actual proportion of U.S. males aged 12 to 17 in the U.S. who play MMORPG’s. Plan: We will construct a one sample z-­‐interval. Conditions: Random: The problem states that an SRS was taken. 10%: The sample of 1100 U.S. Males is clearly less than 10% of the population. Large counts: 𝑛𝑝 ≥ 10 𝑛(1 − 𝑝) ≥ 10 !!"
!"#
1100 !!"" ≥ 10 1100 !!"" ≥ 10 Do: Use z* = 1.96 775
𝑝=
≈ 0.705 1100
0.705 ± 1.96
775 ≥ 10 !.!"#(!!!.!"#)
!!""
325 ≥10. 0.705 ± 0.02695 (0.678, 0.732) Conclude: We are 95% confident that the interval from 0.678 to 0.732 contains the true proportion of males aged 12 to 17 in the U.S. who play MMORPG’s.