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# METHODS OF ANALYSIS MATH 117

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This 21 page Class Notes was uploaded by Estevan Champlin Sr. on Thursday October 22, 2015. The Class Notes belongs to MATH 117 at University of California Santa Barbara taught by Staff in Fall. Since its upload, it has received 38 views. For similar materials see /class/226882/math-117-university-of-california-santa-barbara in Mathematics (M) at University of California Santa Barbara.

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Date Created: 10/22/15

393 23 t 8quot 9 DJ Review chp 6 8 9 10 Multiple Choice Circlethe bestanswer As a general rule the normal distribution is used to approximate the sampling distribution of the sample proportion only if a the sample size n is greater than 30 b the population proportion p is close to 050 c the underlying population is normal d hp and n1p are both greater than 5 Random samples of size 49 are taken from an infinite population whose mean is 300 and standard deviation is 21 The mean and standard error of the sample mean respectively are a 300 and 21 b 300 and 3 c 70 and 230 d 49 and 21 The central theorem is basic to the concept of statistical inference because it permits us to draw conclusion about the population based strictly on sample data and without having any knowledge about the distribution of the underlying population TRUE FALSE In the state of Alabama a random sample of 21 university students is taken The sample proportion of Alabama citizens is 081 If the process were repeated take another sample of size n 21 the sample proportion will always be 081 TRUE FALSE If a simple random sample of 300 observations is taken from a population whose proportion p06 then the expected value of the sample proportion is 04 True or False If all possible samples of size n are drawn from a population the probability distribution of the sample mean is referred to as a the normal distribution True or False Suppose a simple random sample of size n 40 is obtained from a population with mean 50 and standard deviation of 4 Is the sampling distribution of xbar approximately normally distributed Why What is the sampling distribution of xbar 8 The pvalue of a test of the null hypothesis is a the probability the null hypothesis is true b the probability the null hypothesis is false c the probability assuming the null hypothesis is false that the data will be as extreme or more extreme than your observation d the probability assuming the null hypothesis is true that the data will be as extreme or more extreme than your observation Part 11 Show all your work 1 Suppose that the daily high temperature in January in New York City s Central Park has a Normal distribution with the mean of 33 degree and standard deviation of 4 degrees a What is the probability that a day in January will have a high temperature between 34 and 40 degrees b What is the temperature such that 9 of days in January will have daily heights less than that temperature 2 According to the Energy Information Administration the mean age of a refrigerator in a home owned by the occupant is 85 years with standard deviation 66 years What is the probability that a random sample of 50 homes owned by the occupant results in a mean age of refrigerator of 57 years or less 3 According to the Energy Information Administration the mean household size in the United States in 1997 was 26 people with standard deviation 15 people What is the probability that a random sample of 100 households results in a mean household size of 24 people of less 4 Since 1900 the magnitude of earthquakes that measure 01 or higher on the Richter Scale in California is distributed approximately normally with a mean of 62 and standard deviation of 05 according to data obtained from the United States Geological Survey a Determine the 40 h percentile of the magnitude earthquakes in California b Determine the middle 85 of magnitudes 5 Mark Price holds the record for percentage of free throws made in the National Basketball Association at 904 a What is the probability that out of 200 free throws Mark Price makes 180 or more b What is the probability that out 200 free throws Mark Price makes between 180 and 392 6 United Airlines fight 1832 from Chicago to Orlando is on time 80 of the time according to United Airlines Suppose 70 flights are randomly selected what is the approximate probability that at least 60 fights are on time b What is the approximate probability that fewer than 50 flights are on time c What is the probability that between 50 and 55 flights inclusive are on time 7 In 2000 as reported by ACT research Service the mean ACT Math score was 207 If ACT Math scores are normally distributed with standard deviation of of 5 answer the following questions a What is the probability that a randomly selected student has an ACT math score less than 18 b What is the probability that a random sample of 10 ACT test takers had a mean math score of 20 or more c What is the probability that a random sample of 20 ACT test takers had a mean math score between 16 and 20 d If a student scores 17 on the ACT Math portion of the exam what is her percentile rank 8 The number of shark attacks per year in the United states is distributed approximately normally with mean of 318 and standard deviation of 10 according to data obtained from Florida Museum of Natural History a determine the number of shark attacks per year that separates the top 2 from the bottom 98 b Determine the number of shark attacks per year that constitute the middle 80 of shark attacks per year 9 In a study of long distance runners the average weight was found to be 140 lbs with a standard deviation of 10 lbs assume the distribution of runners weights to be normal Find the probability that a random selected long distance runner weights a Describe the random variable X b More than 150 lbs c At most 135 lbs d Between 130 and 135 lbs 10 Psychomotor retardation scores for a large group of manicdepressive patients were found to be approximately normal with a mean of 930 and a standard deviation of 130 a What fraction of the patients scored between 800 and 1100 b Less than 800 c Greaterthan 1200 11 Suppose that the daily high temperature in January in New York City s Central Park has a Normal distribution with the mean of 33 degree and standard deviation of 4 degrees a What is the probability that a day in January will have a high temperature between 34 and 40 degrees b What is the temperature such that 9 of days in January will have daily heights less than that temperature 12 Ralph Carter is a high school history teacher who says that if student are not aware of the Holocaust then the curriculum should be reVised to correct that deficiency A Roper Organization surVey of 506 randomly chosen high school students showed that 268 of them did not know that the term Holocaust refers to the Nazi killing of about 6 million Jews during World War 11 Test the claim that the majority of high school students do not know what Holocaust refers to Find the P Value 0f the test What would you conclude at 5 significance leVel What would you conclude at 10 significance leVel 13 The Gallup Organization asked 1003 adults to answer the question If you could pick only one specific problem for science to solve in the next 25 years what would it bequot Thirty percent responded with a cure for cancer Use a test of significance to determine if it is plausible that if you could ask all adults in United States this question at most 25 would pick a cure for cancer 14 AIDA Personnel Services surveyed 942 human resource professionals and 32 said unauthorized longdistance calls were the leading way in which their company s workers cheated the company Assuming this represents a random sample construct and interpret a 90 confidence interval for the proportion of human resource professionals who have this opinion 15 In March 2001 a Gallup poll asked How would you rate the overall quality of the environment in this country today as excellent good only fair or poorquot Of 1060 adults nationwide 46 gave rating of excellent or good Use a test of significance to determine if this is convincing evidence that fewer than half of the nation s adults would give a rating of excellent or good Use 1 significance level 16 The editors of a magazine have noticed that people seem to believe that a successful life depends on having good friends They would like to have a story about this and use a headline such as Most adults believe friends are important for successquot So they commissioned a survey to ask a random sample of adults whether a successful life depends on having good friends In a random sample of 10 27 adults 53 said yes Should the editors go ahead and use their headline 17 In 1999 27 of Americans stated that reading was their favorite leisure time activity A researcher wants to know whether this percentage has declined since then She surveys 35 Americans and finds that 4 of them consider reading to be their favorite leisure time activity Is this evidence to support the claim that the proportion of Americans who consider reading to be their favorite leisure time activity has decreased at the 5 significance level 18 Suppose that the yield per tree of a certain fruit Variety has a population Variance of 800 pounds A random sample of 100 trees this season gaVe a mean yield of 250 pounds per tree a A harVest is considered poor if the true mean yield is less than 300 pounds of fruit per tree Test the claim that this season s harVest was poor at a 5 significance leVel b Find the 95 confidence interVal for the true mean yield of fruit per tree c Based on the interVal you found in part b would you belieVe the claim that this season s harVest was poor Explain 19 Last year a company s serVice technicians took an aVerage of 26 hours to respond to trouble calls from business customers who had purchased serVice contracts This year in an effort to imprOVe the company hired more serVice technicians The manager randomly sampled 15 calls and found the aVerage response time to be 22 hours with a sample standard deViation of 04 hours Construct a 90 confidence interVal for the true aVerage response time for this year 20 A tourist agency in Massachusetts claims the mean daily cost of meals and lodging for a family of four traveling in the state is 276 You work for a consumer protection advocate and want to test this claim In a random sample of 35 families of four traveling in Massachusetts the mean daily cost of meals and lodging is 285 and the standard deviation is 30 Do you have enough evidence to reject the agency39s claim Use a Pvalue and a 005 Adapted from American Automobile Association Review Solution for problems 12 to 17 Make sure that you also do problem 96 from the homework list If you find a error don t panic let me know and I will fix it Thank you l 12 We want to learn about the true proportion of high school students that do not know what the word Holocaust refers to So the method we would use is a One Sample Z Test for a Population Proportion a X The number of high school students that do not know what the word Holocaust refers to b We are told that we sampled 506 high school students so 11 506 We are told that 268 of the high school students that do not know what the word Holocaust refers to so E 05296 506 3 From our hypotheses we know that p0 050 c We are asked to test the claim that the majority of high school students does not know what the word Holocaust refers to A majority is over 50 So the claim we are testing is H0 p 050 H1 p gt 050 d Direction of the extreme is to the right e End PValue Check for normality Graph 13 p0 05296 050 z 33 po 1 p0 0501 050 n 506 p Value normcdfl33 E99 00918 f Since p value 0091 is greater than alpha 005 do not have enough evidence to reject the claim that the majority of high school students that do not know what the word Holocaust refers to If a 010 then we do Reject H0 since 00918 lt010 g Thus at 10 significance level we do have enough evidence to reject the claim that the majority of high school students that do not know what the word Holocaust refers to 13 The Gallup Organization asked 1003 adults to answer the question If you could pick only one specific problem for science to solve in the next 25 years what would it bequot Thirty percent responded with a cure for cancer Use a test of significance to determine if it is plausible that if you could ask all adults in United States this question at most 25 would pick a cure for cancer a Define the random variable X of people who This variable is a discrete variable and can be analyzed as a proportion The researcher is interested in at most so b H0 p 025 vs H1 p lt 025 or H0 p gt 025 vs H1 p S 025 c Decide what error are you willing to commit and decide the direction of extreme Direction of the extreme is to the left and 0 005 Note when the significance level is not given then use 005 d Collect and summarize the data n 1003 030 p0 0 25 p e Compute p value Check for normality Graph p value The chance of the observed 030 plus the chance of the more extreme In order to find this probability we need to check if it fulfills the normallity requirement np 2 5 and nl p 2 5 1003 025 25075 gt 5 true 1003 075 75225 gt5 true So now we need to standardize the observed value because that is the only way we can find probabilities for this data using the sampling distribution of the sample proportion p 030 025 005 00 130 n 3x 1251 4125 10137 1003 p value normalcdf oo 365 about 1 f Make a decision Since p Value is larger than alpha we don t have enough eVidence to reject H0 Therefore the data is not statistically significant g Conclusion We don t have enough eVidence to conclude that less than 25 will pick a cure for cancer as the only problem for science to solve 15 In March 2001 a Gallup poll asked How would you rate the overall quality of the environment in this country today as excellent good only fair or poorquot Of 1060 adults nationwide 46 gave rating of excellent or good Use a test of significance to determine if this is convincing evidence that fewer than half of the nation s adults would give a rating of excellent or good Use 1 significance level a Define the random variable X of people who believe that the overall quality of the environment in this country today 7 as excellent good This variable is a discrete variable and can be analyzed as a proportion The researcher is interested in fewer than half so b Hozp05 vs H12plt05 c Decide what error are you willing to commit and decide the direction of extreme Direction of the extreme is to the right and 0 001 d Collect and summarize the data x n 0461060 4876 since X is the number of people we need to round up x 448 n 1060 f2 046 p0 50 p e Compute p value Check for normality Graph p value The chance of the observed 046 plus the chance of the more extreme In order to find this probability we need to check if it fulfills the normallity requirement np 2 5 and nl p 2 5 1060 05530 gt 5 true So now we need to standardize the observed value because that is the only way we can find probabilities for this data using the sampling distribution of the sample proportion iquot 00 n 1060 p value normalcdf o 267 004 0 046 05 004 15 f Make a decision Since p value is less than alpha we have enough evidence to reject H0 Therefore the data is statistically significant g Conclusion We have enough evidence to conclude that less than half believe that the overall quality of the environment in this country today 7 as excellent good 16 The editors of a magazine have noticed that people seem to believe that a successful life depends on having good friends They would like to have a story about this and use a headline such as Most adults believe friends are important for successquot So they commissioned a survey to ask a random sample of adults whether a successful life depends on having good friends In a random sample of 1027 adults 53 said yes Should the editors go ahead and use their headline a Define the random variable X of people who believe that successful life depends on having good friends This problem deals with proportion and the researcher is interested in the majority which is understood to be more than half b H0p05 vs H12pgt05 c Decide what error are you willing to commit and decide the direction of extreme Direction of the extreme is to the right and 0 005 as generally accepted value for the type I error d Collect and summarize the data x n 0531027 54431 since X is the number of people we need to round up x 545 n1027 13053 p0 50 p e Compute p value p value The chance of the observed 053 plus the chance of the more extreme In order to find this probability we need to check if it fulfills the normallity requirement np 2 5 and n1 p 2 5 1027 05513 gt 5 true So now we need to standardize the observed value because that is the only way we can find probabilities for this data using the sampling distribution of the sample proportion A p0 053 05 003 p 1875 M 0016 71 1027 p value normlcdf1875 00 003 f Make a decision Since p value is less than alpha we have enough evidence to reject H0 Therefore the data is statistically significant g Conclusion We have enough evidence to conclude that the majority of the population believe that successful life depends on having good friends therefore the editors should go ahead and use their headline 17 In 1999 27 of Americans stated that reading was their favorite leisure time activity A researcher wants to know whether this percentage has declined since then She surveys 35 Americans and finds that 4 of them consider reading to be their favorite leisure time activity Is this evidence to support the claim that the proportion of Americans who consider reading to be their favorite leisure time activity has decreased at the 5 significance level a Define the random variable X of Americans who stated that reading was their favorite leisure time activity This variable is a discrete variable and can be analyzed as a proportion The researcher is interested in at most so P H0 p 027 vs H1 p lt 027 c Decide what error you are willing to commit and decide the direction of extreme Direction of the extreme is to the left and 0 005 d Collect and summarize the data n35 x4 A 4 0ll p 35 p0027p e Compute p value p value The chance of the observed 011 plus the chance of the more extreme In order to find this probability we need to check if it fulfills the normallity requirement np 2 5 and nl p 2 5 35 027 945 gt 5 true 35 1 0272555 gt5 true So now we need to standardize the observed value because that is the only way we can find probabilities for this data using the sampling distribution of the sample proportion po 011 027 016 2133 0075 71 35 p value normalcdf o 2133 f Make a decision Since p Value is than alpha we 9 Conclusion 18 We want to learn about the true mean yield of fruit per tree and since the sample size is larger than 30 we would use is a One Sample Z Test for a Population Mean 1 X yield of fruit per tree 2 We need to find all the terms In this problem we are told that the random sample of 100 trees produced a mean yield of 250 pounds per tree So 11 100 x 250 From the hypotheses we know that Mo 300 In this problem we are told that the standard deViation of scores is V800 So 6 800 3 We are asked to test the claim that this season s harvest is poor That is we are testing H0 u300 H1 ult300 4 Direction of the extreme is to the left And a 005 5 End a PValue Zc yo 250 300 17 68 7 J yma p value normcdf E99 1768 Z 6 Since p value is less than a we can reject H0 at a 5 significance level 7 This means that we can conclude that this season s harvest was poor b We are asked to find a 95 confidence interval for the true mean yield of fruit per tree Since we decided that we were doing a One Sample Z Test for a Population Mean this means the formula for a confidence interval is given by 7 039 x i Z gtllt 7 J2 Now that we know what formula to use we just need to find all the terms In this problem we are told that the random sample of 100 trees produced a mean yield of 250 pounds per tree So 11 100 E 250 In this problem we are told we want a 95 confidence interval so we know 2 1960 In this problem we are told that the standard deviation is 4 800 So 6 V800 Once you find all four terms then you can plug them into the formula like this V800 250 i l 960 V100 250 i 55437 244 4563 255 5437 We are 95 confident that the true mean yield of fruit per tree is somewhere between 2444563 and 2555437 Review chp111214 Part Multiple Choice Circle the bestanswer 1 A 95 confidence interval for the mean uD is given by 05 25 Which ofthe following statements provides a valid interpretation of this interval a Ifthe procedure were repeated 95 ofthe time the population mean is in the interval 05 25 b Ifthe procedure were repeated 95 ofthe samples would fall in the interval 05 25 c Ifthe procedure were repeated 95 ofthe resulting confidence intervals would contain the population mean d Ifthe procedure were repeated 95 ofthe population mean would fall in the interval 05 25 2 Fill in the blank One assumption for ANOVA is that the random samples are a homogeneous b independent c mutually exclusive d normally distributed 3 Fill in the blank One assumption for ANOVA is that the distribution ofthe populations from which the samples come from is the distribution t b F c normal d uniform 4 Fill in the blank The basic concept underlying ANOVA is comparing the variation between the sample means MSB with the random variation of the observations within the samples MSW To have more support for the conclusion that the population means are different the M88 should be as compared to the MSW a large b small c the same d can t tell 5 i in the blank An Ftest statistic value of 18 would imply that the variation between the sample means is very unlikely and we should reject 1 the variation within the samples is very unlikely and we should accept 116 the null hypothesis is false the alternative hypothesis is false an error occurred in computing F 82 e V88 6 The symbol d refers to the difference in the mean of two dependent populations the difference in the means of two independent populations the matched pairs differences the mean difference in the pairs of observations taken from two dependent samples 7 A political analyst in Texas surveys a random sample of registered Democrats and compares the results with those obtained from a random sample of registered Republicans This would be an example of e independent samples f dependent samples g independent samples only if sample sizes are equal h dependent sample only if the sample size are equal 3833 Part II SHOW ALL NECESSARY WORK ON THESE PAGES FOR CREDIT 1 Popular wisdom is that eating presweetened cereal tends to increase the number of dental caries cavities in children A sample of children was with parental consent entered into a study and followed for several years Each child was classi ed as a sweetenedcereal lover or a nonsweetened cereal lover At the end of the study the number of cavities was measured Here is the summary data Group n mean Sweetenedcereal lover 10 641 50 Nonsweetened cereal lover 15 520 150 Test whether children who eat sweetened cereal have more cavities than children who don t 2 A random sample of 15 student s cars found that on average student s cars were 789 years old and had a standard deviation of 367 years A random sample of 20 faculty member s cars yielded an average age of 599 years and a standard deviation of 365 years Test the claim that student s cars are older on average than faculty cars at a 1 signi cance level 3Paramedics in a large city say they need more staff at night because they get more emergency calls then A random sample of seven days showed that the paramedics received the following number of emergency calls during the day 65 54 81 67 75 83 79 A random sample of seven nights showed that the paramedics received the following number of emergency calls during the day 72 85 80 88 82 56 Use 1 signi cance level to test the claim that there is a difference between the mean number of calls during the day and the mean number of calls during the night Also nd the pvalue of the test and make conclusion based on this pvalue You may assume that the number of calls during the day and during the night follow normal distribution 4 A new program has been developed to emich the kindergarten experience of children in preparation for rst grade Pupils in each classroom are tested at the beginning of the school year pretest and again at the end of the school year posttest The following table gives the scores of 9 randomly selected students eXposed to the new curriculum higher score better performance Pupil 1 7 8 9 Pretest 9 6 14 12 9 8 12 8 11 Postest 16 11 14 10 14 12 15 11 14 Test at a 5 level of signi cance if the new curriculum increased performance on the average 5 A mathematics pro ciency test as given to 1810 randomly selected 13yearold American students The following table gives that mean scores of 905 male and 905 female students along with the standard deviations Male Students Tc 4746 sx 1925 Female Students 7 4732 sy 1534 Assume that true population standard deviations are not equal Estimate with 90 con dence the true difference in average pro ciency score in mathematics for male and female students Based on your estimate can you argue that there is a difference between average pro ciency score in mathematics for male and female students What level of signi cance would you use 6 a A random sample of 40 women is partitioned into three categories with ages of below 20 20 through 40 and over 40 The systolic blood pressure levels are obtained The partial ANOVA results for the study are given below Fill in the blanks to complete the table 8 pts Groups b What would the hypotheses be for this test c Would the results ofthis test be significant at the a 05 level Provide a oneor twosentence conclusion for this test 7 A researcher selected a sample of 150 seniors from each of three area high schools and asked each senior Do you drive to school in a car owned by either you or your parentsquot The data are shown in the table At a 005 test the claim that the proportion of students who drives their own or their parents cars is the same at all three schools School 1 School 2 School 3 Tom Yes 18 22 16 56 No 33 38 34 94 50 50 50 150 Suppose a eld botanist nds two different morphs of a rare specie2 pink 3 blue owers 7 8 Test the claim that the proportion of blue and pink flowers is the same for both kind of soil The obsen ecl frequency point data are E P Total Normal soil 34 12 46 Calcareous soill 16 38 5 Total 50 50 100 t Solutions Part 1 c 2 b 3 c 4 a 5 e 6 d 7 e PARTquot We want to learn about the difference in true mean cavities and our two samples are independent children that eat sweet cereal are independent of children that don t So the method we would use is a 2Sample Independent T Test for the difference between population means Let ux the true mean number of cavities for sweetened cereal lover Let W the true mean number of cavities for non sweetened cereal lover We are asked to test the claim that on average children that eat sweet cereal have more cavities than those that don t So we are testing H0 ux uy vs H1 uxgtuy OR H0ux uy0 vs H1ux uygt0 This is a 2sample problem Two samples are independent True standard deviations ax and OJ are NOT known and NOT equal Therefore we will use t distribution with df 9 The test statistic for this test is We need to find all the terms We are told that 10 sweet children are sampled and they had an average of 641 cavities and a standard deviation of 5 cavities Thus nx 10 641 sx 50 We are told that 15 nonswee children are sampled and they had an average of 520 cavities and a standard deviation of 15 cavities Thus F3 ny 15 5m sy 15 So the test statistic is 641 520 20289 52 152 77 10 15 Since the signi cance level a is not provided we will nd a pvalue The H1 is rightsided thus pvalue will be on the right tail p value tcdf0289E999 038 Thus p valuegt010 We would Reject H0 if p value lt a and would NOT Reject H0 if a lt p value Since Pvalue is very big bigger than the smallest possible value of the signi cance level of 10 we CANNOT reject H0 at a 10 signi cance level This means that we CANNOT conclude that the children that eat sweet cereal get more cavities than those that don t We want to learn about the difference in true mean age of cars and our two samples are independent student cars are independent of faculty cars So the method we would use is a 2 Sample Independent TTest for the difference between population means Let M the true mean age of student s cars Let W the true mean age of faculty cars We are asked to test the claim that on average student s cars are older than faculty cars So we are testing Ho M Hy H1 uxgtuy OR H0ux uy0 H1ux uygt0 The test statistic for this test is We need to nd all the terms We are told that 15 student cars are sampled and they had an average of 789 years and a standard deviation of 367 years Thus nx 15 EIw sx 367 We are told that 20 faculty cars are sampled and they had an average of 599 years and a standard deviation of 365 years Thus ny 20 E 599 sy 365 This is a 2sample problem Two samples are independent True standard deviations ax and OJ assumed to be equal Therefore we will use t distribution with df2015 233 So the test statistic is I xl y 1 789 15991 152 Sp 77 367 77 nx my 15 20 pvalue cdfl52E99 33 0068 Since pvalue 0068 gt than alpha 001 we CANNOT reject H0 at a 1 signi cance level This means that we CANNOT conclude that the student cars are older than faculty cars X 7 of calls received during the day Y 7 of calls received during the night th The true mean of calls received during the day uy The true mean of calls received during the night From given samples we can compute the following summary statistics nx7 ny7 x72 7777 Sx lO47 Sy lO78 H0izuxzuy HAU x z y OR H0I x1 y0 HAU x y 0 This is a twosided test thus the rejection region and pvalue will be on both tails 4 This is a 2sample problem Two samples are independent True standard deviations ax and O39y are NOT known but ARE equal Therefore we will use t distribution with dfnx ny 277 212 Before computing the test statistic we need to estimate the true standard deviation ie to compute Sp nx ls my 1 V nxny 2 7 110472 7 110782 77 2 10626 t qoo4 Sp ii 106261ll nx my 7 7 pvalue 2 tcdfl004E9912 168 Test statistics is Since pvalue 0168 gt than alpha 005 we fail to reject H0 Thus there is not evidence to suggest that there is a difference between the average number of calls during the day and night There is not need to hire more night personal We want to learn about the difference in true mean scores and our two samples are paired each pupil has two scores a pretest and a posttest So the method we would use is a Matched Pairs TTest for the difference between population means Let M the true mean pretest score Let W the true mean postest score We are asked to test the claim that the new curriculum has increased performance This means that posttest scores should be higher than pretest scores we are testing H0 Px lquot39y I3911 x ltP y OR H0ux py0 H1ux pylt0 OR Hotud0 H12Pdlt0 The test statistic for this test is We need to find all the terms The rst thing to do in Matched Pairs problems is to take all the differences between the pairs Pupll 2 3 4 5 6 7 8 9 XPretest 9 6 14 12 9 8 12 8 11 YPostest 16 11 14 10 14 12 15 11 14 d X 7 y 7 5 0 2 5 4 3 3 3 We know we sampled 9 pupils so 11 9 7 d Then d73llll 32 nZdZ Zd2 9146 7282 d 73611 nn 71 9 8 sd 4st 7361 27131 So the test statistic is i d 3111 I 344 sd 2713 J 15 pvalue tcdf E99 344 8 0004 Since pvalue 0004 is less than alpha 005 we can reject H0 at a 5 signi cance level This means that we can conclude that the new curriculum increases performance on average 5 X 7 Scores on a mathematics pro ciency test for females Y 7 Scores on a mathematics pro ciency test for males th The true average score on a mathematics pro ciency test for females uy The The true average score on a mathematics pro ciency test for females Since population standard deviations are not known and assumed equal 0x 039 039 we will y estimate 039 with Sp 1 2 1 2 2 2 Vnx Sxny y 905 11925 905 11534 174051 nxny72 905905 2 90 Con dence Interval for x uy

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