Chapter 10 is an important chapter & sort of a capstone in the application of inferential statistics using the normal distribution.. Chapter 10 1 point each What follows is an example of how these problems should be done: In The Wizard of Oz, the wicked witch reported that, historically, an average of 15 of her evil flying monkeys crashed during each mission. She provided additional flight training for the monkeys and sampled the next 25 missions to determine if fewer crashes had occurred. Her sample revealed a mean of 13 crashes; the known population std dev is 5. At a LC of 99%, has their crash rate gone down? Ho: 15 H1: Please note: if you have not spent significant time studying the slides < 15 Excel calc: z = (13-15)/(5/25) = -2.0 z = - 2.33 and the solved problems for this chapter, your probability of successfully completing these problems is low. -2.0 Exercise 6: Ho is not rejected. Conclusion: 15, so the crash rate has NOT gone down. (Unlike the text, acceptance or rejection is not a correct conclusion in these problems.) 1. For a hypothesis test using the z distribution, find the critical value for both a one tail test & two tail test at LC =.99. One tail test, z= Two tail test, z= 2. Penndot wants to test their research theory that drivers are driving too close to the car in front of them, so they do a study of the distance between 100 cars on the PA Turnpike. They define \"too close\" as any distance under 50 feet; their sample has a mean of 45 with a of 22. At a 98.96% LC, is their theory correct Ho: z= Ho is H1: z = Conluusion: 3. The PA Assn of Brewers is attempting to confirm their research, at an LC of 93.7%, which theorizes that IUP has replaced PSU as the biggest party school in PA. PSU's average was 70 parties per week. A recent sample of 100 IUP students revealed a mean of 72 parties with known to be 12. Will IUP take over the top spot? Ho: z= Ho is H1: z = Conluusion: Exercise 12: [similar] Assignment Chapter 10 (b) (c) (d) (e) z = z= Ho is Conclusion: The management of Whinding Industries is considering a new method of assembling its golf carts. The present method requires 43 minutes, on the average, to assemble a cart. The mean assembly time for a random sample of 24 carts, using the new method, was 41.5 (a) Ho: H1: (b) (c) t = Exercise 16: [similar] t= Ho is Conclusion: A random sample of six resulted in a sample mean of 106 with a sample std dev of 6. Using the .05 significance level, can we conclude that the mean is different from 100? Ho: 10 H1: > 10 Test the following hypothesis: The sample mean is 12, and the sample size is 36. The population standard deviation is 5.5. Use = .01 a. z = b. z= Ho is c. d. Exercise 4: Ho: H1: (d) (e) In these problems, an incorrect conclusion represents 50% of the problem's value. Exercise 2: [similar] (a) minutes, and the standard deviation of the sample was 3.7 minutes. Using the .10 level of significance, can we conclude that the assembly time using the new method is faster? 4. How does the level of confidence relate to the level of significance? For Exercises 2 and 4, answer the questions: (b) What is the decision rule? (c) What is the value of the test statistic? (d) Accept or Reject H0? (e) Your conclusion. The MacInfarct restaurant chain claims that the mean waiting time of customers is about 3 minutes with a population standard deviatgion of 1 minute. The quality-assurance department found in a sample of 50 customers at the Plaque Road MacInfarct that the mean waiting time was 2.8 minutes. At the .04 significance level, can we conclude that the mean waiting time is less than 3 minutes? a. b. t = c. t= Ho is d. Conclusion: Ho: H1: (b) t = (c) (d) (e) Conclusion: A sample of 24 observations is selected from a normal population. The sample mean is 215, and the sample standard deviation is 11. Conduct the following test of hypothesis, using = .02 Ho: 220 H1: < 220 (a) Exercise 18: [similar] t= Ho is Conclusion: The powdered chlorine added to swimming pools to combat algae has a relatively short shelf life before it loses its effectiveness. Records indicate that the mean shelf life of a 5-gallon jug of chlorine is 2,000 hours. A new ingredient, Cobalt, was added to the chlorine to increase the shelf life. A sample of twenty jugs of chlorine was tested & the mean was 2050 hours with a population std dev of 100 At the .02 level of significance, has the Cobalt increased the shelf life of the chlorine? Ho: H1: For the following Exercises : (a) State the null hypothesis and the alternate hypothesis. (b) State the critical value (c) Compute the value of the test statistic. (d) Accept or Reject H0? z = z= Ho is Conclusion: Chapter 8 I use the z-table linked to my homepage as z Table Lind 1 point each 1. Explain what a sampling distribution is. 2. The symbol sigma sub x bar [hopefully appearing at right] has 2 names, what are they? x 3. If x bar =10, = 14.6, = 9 & n = 17, find z, using Excel to calculate it. z = z= 4. Explain the value the Central Limit Theorem brings to statistics. x n 5. A machine used to fill beer cans fills them to a mean amount of 16 oz. If a sample of 25 cans has a mean of 13.3 oz, & the population std dev is 5, how likely is it, if the machine is working properly, that it would fill to 13.5 oz or less? z = P(z) = Assignment Chapter 8 Exercise 6: x The following exercises are similar to those in the text. A population consists of the following values: 2, 2, 4, 4, 7 a. List all samples of size 2, and compute the mean of each sample. Exercise 6 [part 2]: b. Compute the mean of the distribution of sample means and the population mean. Compare the two values. = x bar = c. Compare the standard deviation the population with that of the sample means. = Exercise 16: s= A normal population has a mean of 75 and a standard deviation of 5. You select a sample of 40. Compute the probability the sample mean is: a. Less than 74.8 b. Between 73 and 76. Exercise 16 [part 2]: c. Between 76 and 77.9 d. Greater than 76. Exercise 18: According to an IRS study, it takes a mean of 330 minutes for taxpayers to prepare, copy, and electronically file a 1040 tax form. This distribution of times follows the normal distribution and the standard deviation is 70 minutes. A consumer watchdog agency selects a random sample of 40 taxpayers. b. What is the likelihood the sample mean is greater than 320 minutes? c. What is the likelihood the sample mean is between 320 and 350 minutes? n Chapter 9 1 point each The calculations are done using Excel commands. 1. If xbar = 50, n = 49 & = 7 find the 95% Margin of Error for . 50 2. In a distribution of fingernail length where n= 196, = 7 & x bar = .75, compute a 92% Interval Estimate. & determine if Zelda, whose nails are 1.87 long, will be within the interval? Interval Estimate: .75 Is Zelda within the interval? 3. If the sample mean is 100, z = 1.85 & sigma sub x bar = 20, what is the interval estimate for ? 100 4. What is the level of confidence for question 3. above? 5. When is the t distribution used in lieu of the z distribution? Assignment Chapter 9 Exercise 2: Similar to those in the text. A sample of 71 observations is taken from a normal population with a standard deviation of 5. The sample mean is 40. Determine the 95 percent confidence interval for the population mean. 40 Exercise 4: Suppose you want an 80 percent confidence level. What z value would you use? z= z = 1.44 z = -1.44 Exercise 8: Prof. Participle is a professor of English. Recently she counted the number of misspelled words in a group of student essays. She noted the distribution of misspelled words per essay followed the normal distribution with a standard deviation of 2.44 words per essay. For her 10 A.M. section of 40 students, the mean numer of misspelled words was 6.05. Construct a 90 percent confidence interval for the mean number of misspelled worlds in the population of student essays. 6.05 Exercise 12: The American Sugar Producers Associatin wants to estimate the mean yearly sugar consumption. A sample of 16 people reveals the mean yearly consumption to be 55 pounds with a standard deviation of 20 pounds. a. What is the best estimate of the population mean? c. For a 98 percent confidence interval, what is the value of t? Exercise 12 [part 2]: d. Develop the 95 percent confidence interval for the populatin mean. e. Would it be reasonable to conclude that the population mean is 67 pounds? 55 Please see that it does not contain question it actually says about the steps for Question 6,12,16 and 18.Please let me know asap