Chapter 23 Inferences About Means

Chapter 23 Inferences About Means Chapter 23 - Inferences About Means 391 Chapter 23 Solutions to Class Examples 1. See Class Example 1. 2. We want to know if the mean battery lifespan exceeds the 300-minute goal set by the manufacturer. We have 12 battery lifespans in our sample to test the claim. Hypotheses The null hypothesis is that the batteries have a mean lifespan of 300 minutes. The 300-minute goal has not been met. The alternative hypothesis is that the batteries have a mean lifespan greater than 300 minutes. The goal 300-minute goal has been met. H 0 : µ = 300 H A : µ > 300 Model Randomization Condition: This is not a random sample of batteries, but merely 12 batteries produced for preliminary testing. However, it is reasonable to assume that these batteries are representative of all batteries. Nearly Normal Condition: The distribution of battery lifespans is unimodal and symmetric, so it s reasonable to assume that the lifespans of all batteries could be described by a Normal model. Since the conditions have been met, we can do a one sample t-test for the mean, with 11 degrees of freedom. Mechanics - n = 12 df = 11 y = 306.25 s = 29.31 4 3 2 1 240 280 320 Battery Lifespans (min) t = y µ s n 306.25 300 t = 29.31 12 t 0.7387 P-value = P( y > 306.25) = P(t 11 > 0.7387) = 0.238 Conclusion Since the P-value is high, fail to reject the null hypothesis. There is no evidence to suggest that the mean battery lifespan exceeds 300 minutes. It does not appear that the company has met its goal. Confidence Interval The conditions have been met, so we can create a one-sample t-interval, with 90% confidence.

392 Chapter 23 - Inferences About Means y ± t * 11 SE( y ) = 306.25 ±1.796 29.31 12 = (291.05,321.45) I am 90% confident that the mean battery lifespan is between 291.05 and 321.45 minutes. Sample Size We want to know how many batteries to test s ME = t * to be 95% sure of estimating the mean lifespan to within n 15 minutes. First, do a preliminary estimate using z * = 1.96 as the 15 = 1.96 29.31 critical value. n Our first estimate is about 15 batteries. Now, do a better estimate, using t * 14 = 2.145 as the critical value. n = (1.96)(29.31) 15 n 14.67 We would need to sample about 18 batteries in order to estimate the mean battery lifespan to within 15 minutes, with 95% confidence. ME = t * s n 15 = 2.145 29.31 n n = (2.145)(29.31) 15 n 17.56 Finally, to estimate the mean battery lifespan to within 5 minutes, you could do the entire process again, perhaps using a critical value with much higher degrees of freedom. We know that it s going to take lots more batteries to cut the margin of error to a third of what it was. Alternatively, we know it will take a sample about 9 times as large, 18(9) = 162 batteries, since the margin of error was decreased to a third of its size. 3. See Class Example 3.

Investigative Task Chapter 23 - Inferences About Means 393 This is the first of a pair of tasks that use the same data set. The thrust of the tasks is more important than the particular data you use. We use the SAT scores reported for one college, but other data will work. You may want to find something more relevant to your students and revise the task accordingly. Ideally you need a reasonably large but manageable data set; 250 to 500 cases seems to work well. SAT s (or ACT s) are ideal because each individual has two scores (making paired comparisons possible) and lots of statistics are available. The first part of the task asks students to select a random sample, reviewing use of random numbers and sampling issues from Chapter 12. Based on the various samples, students construct confidence intervals, then use the interval to compare the local performance to state or national results. And this affords you the opportunity to look at all the confidence intervals together, noting that most hit the target while others miss. The second half of this task comes after Chapter 25, asking students to do two hypothesis tests one involving paired data and the other comparing two independent means.

394 Chapter 23 - Inferences About Means Intro Stats - Investigative Task Chapter 23 SAT Performance The data for this are found at the end of this document. This is the first of two related Tasks. In these Tasks you will investigate four questions about SAT scores. 1. What is the mean SAT-Math score at this college? 2. How do our SAT Math scores compare to those of another college? 3. Is there a significant difference between Verbal and Math scores for our students? 4. Nationally females tend to have higher Verbal scores than males. Is that true at our college? You can ask the registrar for a copy of the SAT scores of a class (no names of course). If possible, ask for Math scores, Verbal scores, and gender for a recent graduating class. You will not use all of these data, just a sample of 20 30 students. Your assignment: investigate the first two questions posed above. Draw a sample from these students, carefully explaining your procedure. Use your sample to create a 95% confidence interval for the mean SAT Math score. Based on your confidence interval, compare the performance of our students with national mean Math score of 503. Prepare a written report that includes complete demonstrations of the statistical procedures you use and your conclusions (in context, of course). Important: Save the Score Roster and your sample. You ll be using them to answer the other two questions in the next Investigative Task.

Chapter 23 - Inferences About Means 395 Intro Stats - Investigative Task Chapter 23 Components Think Demonstrates clear understanding of construction and interpretation of confidence interval Show Sampling Procedure Randomization is correctly used Method of randomization is described Conditions Random sample Assesses normality Mechanics Identifies procedure Constructs correct interval Tell Interpretation Interprets interval in context Each component is scored as Essentially correct, Partially correct, or Incorrect. 1. The Sample E - Selects a random sample, explains the sampling process clearly. P Selects a random sample, but explanation of the process may be unclear or there may be mistakes in notation, or vocabulary. I Sample is not random or the process is not explained or there are several major mistakes in arithmetic, notation, or vocabulary. 2. The Conditions E Cites randomness, <10% of all possible students, and checks normality with a plot. P Discusses normality but omits the plot or makes minor mistakes in the other conditions. I Misunderstands or omits the conditions, or lists irrelevant issues (np 10). 3. The Mechanics E Identifies the procedure, shows the sample statistics and degrees of freedom, writes the formula using correct critical value and notation, and calculates the correct interval (perhaps with minor arithmetic or rounding errors). P Appears to be doing the proper procedure but omits important information, uses the wrong critical value, uses the wrong notation, or makes major errors in calculations. I Uses the wrong procedure or shows no work or makes several major mistakes. 4. The Interpretation E Correctly interprets the confidence interval in the proper context, and compares the local performance to statewide results. P Writes a conclusion that is correct but not in context or doesn t compare local performance to statewide results. I Does not interpret the confidence interval correctly. Scoring E s count 1 point, P s are 1/2 Score = sum of 4 components; rounding based on quality of P responses Comments Show plot

396 Chapter 23 - Inferences About Means Gender Verbal Math Gender Verbal Math Gender Verbal Math F 450 450 F 370 460 M 690 740 F 640 540 F 620 470 F 630 560 M 590 570 M 680 650 F 480 500 M 400 400 F 620 620 F 680 630 M 600 590 M 590 690 F 650 680 M 610 610 M 430 540 M 460 410 F 630 610 F 680 740 M 560 560 M 660 570 M 650 650 M 610 760 F 660 720 F 650 680 M 620 650 F 590 640 M 660 700 F 590 640 M 580 650 M 640 680 F 390 440 F 500 540 M 460 680 M 510 530 F 480 360 M 430 480 M 450 440 M 530 520 M 460 600 F 520 690 M 690 640 M 800 720 F 470 410 M 480 570 M 310 530 M 690 620 F 450 420 F 510 550 M 510 540 M 650 650 F 770 700 F 770 710 F 620 600 M 750 670 F 520 570 M 630 740 M 580 630 F 550 600 M 490 520 M 410 390 F 680 580 M 560 620 F 430 400 M 740 700 F 710 720 M 630 680 F 660 690 F 360 390 F 630 630 F 740 640 F 590 530 M 480 550 F 730 680 M 520 630 F 420 430 F 560 630 M 510 600 M 800 650 F 570 530 F 630 630 M 650 500 F 560 540 F 740 760 M 580 690 F 670 520 M 680 690 F 560 540 M 650 710 F 660 700 M 670 760 F 690 700 F 690 750 M 610 740 M 610 740 M 540 560 M 510 610 F 500 650 F 670 520 M 560 530 M 560 700 M 580 720 F 770 690 M 640 650 F 600 560 F 530 430 M 430 490 F 600 560 M 650 740 F 700 570 F 710 640 F 730 790 M 620 670 M 640 650 F 690 640 F 610 640 M 490 590 F 610 510 M 580 640 M 650 680 F 710 680 F 730 570 M 580 710 F 730 590 F 520 530 M 710 730 M 450 570 M 540 580 M 740 700 F 400 410 M 640 610 M 510 560 M 700 620 M 680 720 F 650 760 M 580 660 M 580 490 F 580 660 M 600 670 F 640 630 F 600 640 F 630 690 F 700 650 F 670 700 F 600 640 M 600 630 F 640 670 F 630 700 F 540 510

Chapter 23 - Inferences About Means 397 Gender Verbal Math Gender Verbal Math Gender Verbal Math F 480 540 F 630 520 F 560 570 F 710 700 M 510 520 F 560 560 M 650 780 M 600 610 F 490 510 F 640 570 F 550 570 M 430 570 F 370 410 F 670 580 M 690 670 M 710 700 M 560 520 F 600 550 F 630 660 F 600 600 M 700 630 F 590 580 F 650 530 F 630 710 M 750 800 F 630 610 M 620 610 M 600 690 F 440 470 F 550 550 M 610 550 M 650 560 M 600 670 F 490 800 M 690 670 M 450 610 F 680 610 M 360 290 M 690 680 M 520 540 M 510 510 M 590 600 F 680 660 M 610 570 M 760 580 M 650 700 M 700 740 F 650 600 M 600 560 F 540 490 M 540 630 F 550 560 M 600 670 M 580 530 M 490 390 F 610 710 M 670 650 F 530 530 F 700 570 F 650 680 M 560 560 F 490 560 M 660 660 F 630 590 M 390 630 M 500 580 F 510 520 M 580 530 F 610 510 M 710 740 M 760 760 F 580 500 F 550 560 M 730 760 F 570 620 M 690 620 F 590 620 M 690 620 M 700 700 M 640 740 M 570 540 M 540 620 F 610 620 M 600 670 F 280 500 M 760 700 F 540 670 M 710 760 M 570 630 M 640 760 F 640 710 M 700 740 F 680 600 M 600 590 F 630 630 M 730 670 M 610 670 F 530 670 M 610 550 M 680 670 F 490 700 M 500 550 F 520 470 M 600 660 M 580 630 F 730 740 M 580 760 M 620 710 M 510 680 M 710 700 F 570 630 F 620 740 F 600 450 F 690 610 F 530 500 F 560 590 F 610 570 M 550 680 M 570 690 F 630 570 M 590 670 F 630 610 F 540 660 M 620 640 F 640 570 F 760 690 M 640 570 M 550 600 M 470 470 F 490 530 M 690 670 F 340 450 M 530 580 F 530 490 F 480 500 M 560 600 F 580 640 M 690 760 F 710 700 F 550 560 M 620 610 M 480 530 M 690 740 F 710 630 F 490 490 F 420 410 M 440 740 M 460 560 F 690 720 M 400 710

398 Chapter 23 - Inferences About Means Chapter 23 Investigative Task Sample Solution SAT Performance I want to determine the mean SAT-Math score at this college. I will take a simple random sample of 25 students by assigning a number 001-300 to each of the students, and then choosing 25 random numbers 001-300, ignoring repeats. Plan I want to find a 95% confidence interval for the mean SAT-Math score of all students at this college. I have data on the scores of 25 students, from a simple random sample of the 300 students at this college. Model Randomization Condition: The students were chosen by a simple random sample. 10% Condition: 25 students represent less than 10% of the population of 300 students. 10 8 6 4 2 Nearly Normal Condition: The distribution of SAT-Math scores in the sample is unimodal and symmetric. 375 600 Math The conditions are satisfied, so I will use a Student s t-model with (n 1) = 25 1 = 24 degrees of freedom to find a one-sample t-interval for the mean. Mechanics From my sample of 25 students: n = 25 scores y = 602.8 s = 74.3034 y ± t * 24 SE(y ) = 602.8 ± 2.064 74.3034 25 = 602.8 ± 30.672 = (572.13,633.47) Conclusion I am 95% confident that the interval from 572.13 to 633.47 contains the true mean SAT-Math score for students at this college. According to my confidence interval, students at this college had a higher mean SAT- Math score than students nationwide. The national mean of 503 was not included in my 95% confidence interval.

Chapter 23 - Inferences About Means 399 Intro Stats Quiz A Chapter 23 Name A professor at a large university believes that students take an average of 15 credit hours per term. A random sample of 24 students in her class of 250 students reported the following number of credit hours that they were taking: 12 13 14 14 15 15 15 16 16 16 16 16 17 17 17 18 18 18 18 19 19 19 20 21 1. Does this sample indicate that students are taking more credit hours than the professor believes? Test an appropriate hypothesis and state your conclusion.

400 Chapter 23 - Inferences About Means 2. Find a 95% confidence interval for the number of credit hours taken by the students in the professor s class. Interpret your interval.

Chapter 23 - Inferences About Means 401 Intro Stats Quiz A Chapter 23 Key A professor at a large university believes that students take an average of 15 credit hours per term. A random sample of 24 students in her class of 250 students reported the following number of credit hours that they were taking: 12 13 14 14 15 15 15 16 16 16 16 16 17 17 17 18 18 18 18 19 19 19 20 21 1. Does this sample indicate that students are taking more credit hours than the professor believes? Test an appropriate hypothesis and state your conclusion. H : 15 0 µ = credit hours; Students in the professor s class are taking an average of 15 credit hours. H A : µ > 15 credit hours; Students in the professor s class are taking more than 15 credit hours, on average. Conditions: * Randomization condition: Students from the class were randomly sampled. * 10% condition: The sample is less than 10% of the class population. * Nearly Normal condition: The histogram of credit hours is unimodal and reasonably symmetric. This is close enough to Normal for our purposes. 5 Histogram of Credit Hours 4 Frequency 3 2 1 0 12 14 16 Credit Hours 18 20 Under these conditions, the sampling distribution of the mean can be modeled by Student s t with degrees of freedom: df = n 1 = 24 1 = 23. We will use a one-sample t-test for the mean. 2.22 We know: n = 24, y = 16.6, and s = 2.22. So, SE( y ) = = 0.453. 24 y µ 0 16.6 15 t = = = 3.532 The P-value is Pt ( 23 > 3.532) = 0.0009. SE( y) 0.453 The P-value of 0.0009 says that if the true mean credit hours for students in the professor s class is 15, samples of 24 students can be expected to have an observed mean of 16.6 credit hours or more less than 0.1% of the time. This is rare enough for us to reject the null hypothesis. This sample indicates that the professor s students are taking more than 15 credit hours, on average.

402 Chapter 23 - Inferences About Means 2. Find a 95% confidence interval for the number of credit hours taken by the students in the professor s class. Interpret your interval. With the conditions satisfied (from Problem 1), we can find a t-interval for mean credit hours. We know: n = 24, y = 16.6, s = 2.22, and 2.22 SE( y ) = = 0.453. 24 Our confidence interval has the form n 1 interval is then s y± t. We have t 23 = 2.069. Our 95% confidence n 2.22 16.6 ± 2.069 = 16.6 ± 0.94, or 15.66 to 17.54. 24 We are 95% confident that the interval 15.66 to 17.54 contains the true mean number of credit hours that students in the professor s class are taking.

Chapter 23 - Inferences About Means 403 Intro Stats Quiz B Chapter 23 Name Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. The insurance company knows that, last year, the life expectancy of its policyholders was 77 years. They want to know if their clients this year have a longer life expectancy, on average, so the company randomly samples some of the recently paid policies to see if the mean life expectancy of policyholders has increased. The insurance company will only change their premium structure if there is evidence that people who buy their policies are living longer than before. 86 75 83 84 81 77 78 79 79 81 76 85 70 76 79 81 73 74 72 83 1. Does this sample indicate that the insurance company should change its premiums because life expectancy has increased? Test an appropriate hypothesis and state your conclusion.

404 Chapter 23 - Inferences About Means 2. For more accurate cost determination, the insurance companies want to estimate the life expectancy to within one year with 95% confidence. How many randomly selected records would they need to have?

Chapter 23 - Inferences About Means 405 Intro Stats Quiz B Chapter 23 Key Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. The insurance company knows that, last year, the life expectancy of its policyholders was 77 years. They want to know if their clients this year have a longer life expectancy, on average, so the company randomly samples some of the recently paid policies to see if the mean life expectancy of policyholders has increased. The insurance company will only change their premium structure if there is evidence that people who buy their policies are living longer than before. 86 75 83 84 81 77 78 79 79 81 76 85 70 76 79 81 73 74 72 83 1. Does this sample indicate that the insurance company should change its premiums because life expectancy has increased? Test an appropriate hypothesis and state your conclusion. H0 : µ = 77years The average life expectancy of the insurance company s policy holders is 77 years. HA : µ > 77 years The average life expectancy of the insurance company s policy holders is more than 77 years. Conditions: * Randomization condition: The records from the insurance company were randomly sampled. * Nearly Normal condition: The histogram of the ages at death is unimodal and reasonably symmetric. This is close enough to Normal for our purposes. y µ 0 78.6 77 t = = = 1.597 SE( y) 1.002 P= P( t > 1.597) = 0.063 19 Life Expectancy 4 Under these conditions, the sampling distribution of the mean can be modeled by Student s t with degrees of 1 freedom: df = n 1 = 20 1 = 19. We will use a one-sample t-test for the mean. 70 75 80 DeathAge 85 90 We know: n = 20, y = 78.6 years, and s = 4.48 years. SE( y) = s 4.48 = = 1.002 years. n 20 The P value of 0.063 says that if the true mean life expectancy for a person had increased to 77 years, samples of 20 records can be expected to have an observed mean life expectancy of 78.6 years or more 6.3% of the time. This is not rare enough for us to reject the null hypothesis. This sample does not indicate that the insurance company needs to increase their premiums because there was not enough evidence to indicate that people who buy their policies are living longer than before. Count 3 2 Histogram

406 Chapter 23 - Inferences About Means 2. For more accurate cost determination, the insurance companies want to estimate the life expectancy to within one year with 95% confidence. How many randomly selected records would they need to have? We wish to find the sample size, n, that would allow a 95% confidence level for the mean life expectancy, µ, of a policy holder from the insurance company to have a margin of error of only one year. The 95% critical value of the Student s t-statistic with 19 degrees of freedom (df = n 1) is t * 19 = 2.093. * 19 ( ) ME = t SE y 2 ( )( ) 2 * t 2.093 4.48 19 s n = = = 87.92 88 ME 1 records. We would need to sample at least 88 records from the insurance company to estimate the life expectancy of a policy holder to within one year with 95% confidence.

Chapter 23 - Inferences About Means 407 Intro Stats Quiz C Chapter 23 Name Textbook authors must be careful that the reading level of their book is appropriate for the target audience. Some methods of assessing reading level require estimating the average word length. We ve randomly chosen 20 words from a randomly selected page in Stats: Modeling the World and counted the number of letters in each word: 5, 5, 2, 11, 1, 5, 3, 8, 5, 4, 7, 2, 9, 4, 8, 10, 4, 5, 6, 6 1. Suppose that our editor was hoping that the book would have a mean word length of 6.5 letters. Does this sample indicate that the authors failed to meet this goal? Test an appropriate hypothesis and state your conclusion. 2. For a more definitive evaluation of reading level the editor wants to estimate the text s mean word length to within 0.5 letters with 98% confidence. How many randomly selected words does she need to use?

408 Chapter 23 - Inferences About Means Intro Stats Quiz C Chapter 23 Key Textbook authors must be careful that the reading level of their book is appropriate for the target audience. Some methods of assessing reading level require estimating the average word length. We ve randomly chosen 20 words from a randomly selected page in Stats: Modeling the World and counted the number of letters in each word: 5, 5, 2, 11, 1, 5, 3, 8, 5, 4, 7, 2, 9, 4, 8, 10, 4, 5, 6, 6 1. Suppose that our editor was hoping that the book would have a mean word length of 6.5 letters. Does this sample indicate that the authors failed to meet this goal? Test an appropriate hypothesis and state your conclusion. Hypotheses: H 0 :µ = 6.5 H a :µ 6.5 where µ represents the population mean of word lengths Plan: We have a random sample of less than 10% of the words in the book. A histogram of theobserved word lengths looks roughly unimodal and symmetric, so the population of all word lengths may be approximately normal. It is appropriate to use a one sample t-test. Mechanics: n = 20 x = 5.5 s = 2.685 df = 19 t = P = 2 P(t 19 < 1.67) = 0.11 5.5 6.5 2.865 20 = 1.67 Collection 1 6 4 2 Histogram 0 4 8 12 letters Conclusion: Because the P-value is so high we do not reject H 0. This sample does not provide evidence that the average word length differs from the goal of 6.5 letters. 2. For a more definitive evaluation of reading level the editor wants to estimate the text s mean word length to within 0.5 letters with 98% confidence. How many randomly selected words does she need to use? 2 2 zs 2.326 2.685 2 = = = 12.49062 = 156.02 n= 157 ME 0.5 First estimate: n ( ) Based on first estimate, 2 2 t157s 2.353 2.685 2 n= = = ( 12.63561) = 159.66 n= 160 ME 0.5 Table T., using df = 140 from