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# Elementary Statistical Methods STAT 30100

Purdue

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This 34 page Class Notes was uploaded by Bailey Macejkovic on Saturday September 19, 2015. The Class Notes belongs to STAT 30100 at Purdue University taught by Thomas Howell in Fall. Since its upload, it has received 19 views. For similar materials see /class/207927/stat-30100-purdue-university in Statistics at Purdue University.

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Date Created: 09/19/15

Lecture 8 Sections 71 amp 72 Inference for the Mean of a Population Previously we made the assumption that we know the population standard deviation 0 We then developed a confidence interval and used tests for significance to gather evidence foragainst an hypothesis all with a known a In normal practice a is unknown In this section we must estimate a from the data though we are primarily interested in the population mean u Con dence Interval for a Mean First Assumptions for Inference about a mean 0 Our data are a simple random sample SRS of size n from the population Observations from the population have a normal distribution with mean u and standard deviation 0 If population distribution is not normal it is enough that the distribution is unimodal and symmetric and that the sample size be large ngt15 Both u and a are unknown parameters Because we do not know a we make two changes in our procedure 1 The standard error 5 is used in place of 1 J Standard Error When the standard deviation of a statistic is estimated from the data the result is called the standard error of the statistic The standard error of the sample mean E is SELL x J17 Where s is the sample standard deviation n is the sample size 2 We calculate a different test statistic and use a different distribution to calculate our pvalue Lecture 8 Section 71 amp 72 Page 1 The tdistributions o The tdistribution is used when we do not know a The tdistributions have density curves similar in shape to the standard normal curve but with more spread The tdistributions have more probability in the tails and less in the center than does the standard normal This is because substituting the estimate s for the fixed parameter 5 introduces more variation into the statistic As the sample size increases the tdensity curve approaches the NOl curve Note This is because s estimates 5 more accurately as the sample size increases The t Distributions Suppose that an SRS of size n is drawn from a N ua population Then the onesample t statistic 37 S J13 has the t distribution with nl degrees of freedom The OneSample t Con dence Interval Suppose that an SRS of size n is drawn from a population having unknown mean u A level C confidence interval for u is gtXlt i x I J where t is the value for the tnl density curve with area C between t and t This interval is exact when the population distribution is normal and is approximately correct for large n in other cases Lecture 8 Section 71 amp 72 Page 2 Examples 1 J Suppose X Bob s golf scores are approximately normal distribution with unknown mean and standard deviation A SRS of n 16 scores is selected and a sample mean of E 77 and a sample standard deviation of s 3 is calculated Calculate a 90 confidence interval for u Example 71 in Textbook In 1996 the US Agency for International Development provided 238300 metric tons of corn soy blend CSB for development programs and emergency relief CSB is highly nutritious lowcost fortified food and can be incorporated into different food preparations worldwide As part of a study to evaluate appropriate vitamin C levels in this commodity measurements were taken on samples of CSB produced in a factory The following data are the amounts of vitamin C measured in milligrams per 100 grams of blend for a random sample of size 8 from a production run Compute a 95 confidence interval for u where u is the mean vitamin C content of the CSB 26 3123 22112214 31 By hand 225 s7191 n8 Lecture 8 Section 71 amp 72 Page 3 Using SPSS analyze gt descriptive statistics gt explore Move vitaminC to dependent list Click statistics and select clescriptives and changekeep a 95 con dence interval Click continue followed by OK Descriptives Statistic Std Error Vitamin C Mean 2250 2542 95 Confidence Lower Bound 1649 Interval for Mean Upper Bound 2851 5 Trimmed Mean 2267 Median 2250 Variance 51714 Std Deviation 1191 Minimum 11 Maximum 31 Range 20 Interquartile Range 14 Skewness 443 752 Kurtosis 631 1481 The OneSample t test 1 State the Null and Alternative hypothesis 2 Find the test statistic Suppose that an SRS of size n is drawn from a population having unknown mean ii To test the hypothesis H 0 u 0 based on a SRS of size n compute the onesample t statistic m s J17 3 Calculate the p value In terms of a random variable T having the tnl distribution the Pvalue for a test of H 0 against Ha ngtn0 is PT2t Ha nltn0 is PTSt Ha u u0 is 2PT2t These Pvalues are exact if the population distribution is normal and are approximately correct for large n in other cases Lecture 8 Section 71 amp 72 Page 4 4 State the conclusions in terms of the problem Choose a significance level such as 1 005 then compare the Pvalue to the 1 level lfPvalue S a then reject H 0 lfPvalue gt 1 then fail to reject H 0 Examples 1 J Experiments on learning in animals sometimes measure how long it takes mice to find their way through a maze The mean time is 18 seconds for one particular maze A researcher thinks that a loud noise will improvedecrease the time it takes a mouse to complete the maze She measures how long each of 30 mice take to complete the maze with noise stimulus She finds their average time is 16 seconds and their standard deviation is s 3 seconds Do a hypothesis test to test the researchers assertions with 1 01 Example 72 in Textbook Suppose that we know that sufficient vitamin C was added to the CSB mixture to produce a mean vitamin C content in the final product of 40 mg 100 g It is suspected that some of the vitamin is lost or destroyed in the production process To test this hypothesis we can conduct a onesided test to determine if there is sufficient evidence to conclude that the CSB mixture lost vitamin C content at 1 005 level By hand Lecture 8 Section 71 amp 72 Page 5 Using SPSS analyze gt compare means gt One sample T test Move vitaminc into the test variable box and type in 40 for the test value To change the con dence interval Click options and change con dence interval from 95 to whatever I did not do this as I will keep the 95 default Click continue Lastly click OK OneSam ple Statistics Std Error N Mean Std Deviation Mean Vitamin C 8 2250 7191 2542 OneSample Test Test Val e 40 95 Confidence Interval of the D fference Mean t df Sig 2tailed Difference Lower Upper Vitamin C 6883 0 47500 2351 4149 Lecture 8 Section 71 amp 72 Page 6 Matched Pairs Design A common design to compare two treatments is the matched pairs design One type of matched pair design has 2 subjects who are similar in important aspects matched in pairs and each treatment is given to one of the subjects in each pair With only one subject 2 treatments are given in random order Another type of matched pairs is beforeand after observations on the same subject Paired t Procedures To compare the mean responses to the two treatments in a matched pairs design apply the onesample t procedures to the observed differences d Example Problem 731 is done by hand and using SPSS The researchers studying vitamin C in CSB in example 71 were also interested in a similar commodity called wheat soy blend WSB Both these commodities are mixed with other ingredients and cooked Loss of vitamin C as a result of this process was another concern of the researchers One preparation used in Haiti called gruel can be made from WSB salt sugar milk banana and other optional items to improve the taste Samples of gruel prepared in Haitian households were collected The vitamin C content in milligrams per 100 grams of blend dry basis was measured before and after cooking Here are the results Sample 1 2 3 4 5 Before 73 79 86 88 78 After 20 27 29 36 17 Set up appropriate hypotheses and carry out a significance test for these data It is not possible for cooking to increase the amount of vitamin C Lecture 8 Section 71 amp 72 Page 7 By hand Using SPSS gt Analyze gt Compare Means gt Paired 7 Sample T test Move before and after to paired variable box whichever variable is listed rst will come rst in the subtraction Click OK Paired Samples Statistics Std Error Mean N Std Deviation Mean Pair 1 Before 8080 5 6140 2746 After 2580 5 7530 3367 Paired Samples Test Pairs 1 Differences 95 Confidence Interval of the Diffe ence Std Std Error Mean Deviation Mean Lower Upper t df Sig 2tailed Pa 1 22 39 55000 3937 1761 50112 59888 31238 000 Lecture 8 Section 71 amp 72 Page 8 A confidence interval or statistical test is called robust if the confidence level or P value does not change very much when the assumptions of the procedure are violated The t procedures are robust against nonnormality of the population when there are no outliers especially when the distribution is roughly symmetric and unimodal Robustness and use of the OneSample t and Matched Pair t procedures 0 Unless a small sample is used the assumption that the data comes from a SRS is more important than the assumption that the population distribution is normal nlt15 Use t procedures only if the data are close to normal with no outliers ngt15 The t procedure can be used except in the presence of outliers or strong skewness o n is large in 40 The t procedure can be used even for clearly skewed distributions Lecture 9 Section 71 amp 72 Page 9 Comparing Two Means TwoSample Problems A situation in which two populations or two treatments based on separate samples are compared A twosample problem can arise o from a randomized comparative experiment which randomly divides the units into two large groups and imposes a different treatment on each group 0 From a comparison of random samples selected separately from different populations Note Do not confuse twosample designs with matched pair designs Assumptions for Comparing Two Means 0 Two independent simple random samples from two distinct populations are compared The same variable is measured on both samples The sample observations are independent neither sample has an in uence on the other Both populations are approximately normally distributed The means ul and u2 and standard deviations 01 and 02 of both populations are unknown Typically we want to compare two population means by giving a confidence interval for their difference 1 u2 or by testing the hypothesis of no difference Howl 220 The TwoSample t Con dence Interval Suppose that an SRS of size 111 is drawn from a normal population with unknown mean Ltl and that an independent SRS of size 112 is drawn from another normal population with unknown mean 2 The confidence interval for 1 u2 given by 2 2 s s xl x2il 1 2 quot1 quot2 has confidence level at least C no matter what the population standard deviations may be Here t is the value for the tk density curve with area C between t and t The value of the degrees of freedom k is approximated by software or we use the smaller of 111 l and 112 l Lecture 9 Section 71 amp 72 Page 10 TwoSample t Procedure 1 J U A Write the hypotheses in terms of the difference between means H 0 M1 2 0 H a 1 2 gt 0 or H a 1 2 lt 0 or H a 1 2 i 0 Calculate the test statistic A SRS of size 111 is drawn from a normal population with unknown mean ul and draw an independent SRS of size 112 from another normal population with unknown mean 2 To test the hypothesis H 0 ul 2 0 the twosample t statistic is x1 x 2 2 2 1 2 quot1 quot2 and use Pvalues or critical values for the tk distribution where the degrees of freedom k are either approximated by software or are the smaller of 111 l and I n l Note The twosample t statistic does not have a t distribution The software however uses a t distribution to do inference for twosample problems This is because it is approximately a t distribution with degrees of freedom calculated by a complex formula called the Welch Approximation Calculate the PValue Note Unless we use software we can only get a range for the Pvalue We use the following formulas Hazyl y gt0 is PT2I Haul u2lt0 is PTSI Ha ul u2 20 is 2PT2I Note Instead of using the degrees of freedom found by software you can use the smaller of n1 1 and n2 1 The resulting procedure is conservative State the conclusions in terms of the problem Choose a significance level such as 1 005 then compare the Pvalue to the 1 level If Pvalue S 1 then reject H 0 If Pvalue gt 1 then fail to reject H 0 Lecture 9 Section 71 amp 72 Page 11 Robustness and use of the TwoSample t Procedures The twosample t procedures are more robust than the onesample t methods particularly when the distributions are not symmetric They are robust in the following circumstances If two samples are equal and the two populations that the sample come from have similar distributions then the t distribution is accurate for a variety of distributions even when the sample sizes are as small as 111 2 n2 2 5 When the two population distributions are different larger samples are needed 171 n2 lt15 Use twosample t procedures if the data are close to normal If the data are clearly non normal or if outliers are present do not use I 11 11 gt15 The t procedures can be used except in the presence of outliers l 2 or strong skewness n1 n2 Z 40 The t procedures can be used even for clearly skewed distributions Lecture 9 Section 71 amp 72 Page 12 Examples 1 The US Department of Agriculture USDA uses many types of surveys to obtain important economic estimates In one pilot study they estimated wheat prices in July and in September using independent samples Here is a brief summary from the report Month n a S AM July 90 295 0023 September 45 361 0029 a Note that the report gave standard errors Find the standard deviation for each of the samples b Use a significance test to examine whether or not the price of wheat was the same in July and September Be sure to give details and carefully state your conclusion c Give a 95 confidence interval for the increase in price between July and September Lecture 9 Section 71 amp 72 Page 13 2 The survey for Study Habits and Attitudes SSHA is a psychological test designed to measure the motivation study habits and attitudes toward learning of college students These factors along with ability are important in explaining success in school Scores on the SSHA range from 0 to 200 A selective private college gives the SSHA to an SRS of both male and female firstyear students The data for the women are as follows 154 109 137 115 152 140 145 178 101 103 126 126 137 165 165 129 200 148 Here are the scores of the men 108 140 114 91 180 115 126 92 169 146 109 132 75 88 113 151 70 115 187 104 a Examine each sample graphically with special attention to outliers and skewness Is use of a t procedure acceptable for these data Lecture 9 Section 71 amp 72 Page 14 Lecture 9 Section 71 amp 72 Page 15 b Most studies found that the mean SSHA score for men is lower than the mean score in a comparable group of women Test this supposition here That is state the hypotheses carry out the test and obtain a PValue and give your conclusions Using SPSS Note The data needs to be typed in using two columns In the rst column you need to put all the scores In the second column de ne the grouping variable as gender and enter women next to the women s scores and men next to the men s scores Analyze gt Compare means gt Independent Sample T test Move score to Test Variable box and gender to grouping variable box Click de ne groups and enter women for group I and men for group 2 Click continue followed by OK Group Statistics Std Error group N Mean Std Deviation Mean score women 18 14056 26262 6190 men 20 12125 32852 7346 Levene39s Test for Equality of Variances ttest for Equality of eans 95 Confidence Interval of the Mean Std Error D39 e ence Sig 2 Dif ferenc Dif ferenc F Sig t df tailed e e Lower Upper score Equal variances 1030 317 1986 36 055 19306 9721 410 39021 assumed Equal Var39ances 2010 35537 052 19306 9606 185 38797 assumed Independent Sam ples Test We use the second line Equal variances not assumed to get the ttest statistic p value etc Lecture 9 Section 71 amp 72 Page 16 c Give a 95 confidence interval for the mean difference between the SSHA scores of male and female firstyear students at this college 3 Suppose we wanted to compare how students performed on test 1 versus test 2 in stat 301 Below is data for a random sample of 10 students taking stat 301 along with the printout of the results from running the matched pairs test Paired Samples Statistics Std Error Mean N Std Deviation Mean Pair Test 1 82 80 10 14382 4 548 1 Test 2 80 70 10 14507 4 588 Paired Samples Test 95 Con dence Interval ofihe Sid Error Di erence Sid Deviation Lower U er i of Sig 2iailed Pair 1 Test 1 Test 2 456 4656 1859 9 096 Lecture 9 Section 71 amp 72 Page 17 Answer the questions below based on the test a Why was a matched pairs test used as opposed to a two sample t test b Is there a difference between test 1 and test 2 scores Write out the hypotheses and give the Pvalue Are our results significant at the 5 significance level c Suppose one of our friends thought test 2 was easier and the students generally did better on it We want to test whether the student is correct Write out the hypotheses to test this and give the Pvalue Are our results significant at the 5 significance level 4 An instructor thought the material for the second test was more difficult and hence the students may have done worse on the second test She decided to randomly sample 10 Exam 1 tests and 10 Exam 2 tests with each sample taken from all the test taken Below is data for the two random samples and the analysis Lecture 9 Section 71 amp 72 Page 18 G roup Statistics Std Error test N Mean Std Deviation Mean score 1 10 8280 14382 4548 2 10 8070 14507 4588 Independent Samplesrest 95 Con dence lrv Nal on e Std Error assumed 6 460 6 460 a What procedure was used and why b Write out the hypotheses to test this and give the PValue Are your results significant at the 5 significance level Lecture 9 Section 71 amp 72 Page 19 Lecture 7 Chapter 6 Statistical inference is the next topic we will cover in this course We have been preparing for this by describing and analyzing data graphs and plots descriptive statistics discussing the means to find andor generate data studies samples experiments and we have defined sampling distributions We are now ready for statistical inference The purpose of statistical inference is to draw conclusions from data It adds to the graphing and analyzing because we substantiate our conclusions by probability calculations Formal inference is based on the longrun regular behavior that probability describes When you use statistical inference the data should come from a random sample or a randomized experiment We will learn in this chapter the two most prominent types of formal statistical inference con dence intervals for estimating the value of a population parameter and tests of signi cance which assess the evidence for a claim Section 61 Estimating with con dence A confidence interval for a population parameter includes a point estimate and a margin of error The point estimate is single statistic calculated from a random sample of units For example E the sample mean is a point estimate of u the population mean Point estimates give us very little information So we add to our point estimate a margin of error making up a confidence interval Lecture 7 Chapter 6 Page 1 Example 1 You want to estimate the mean SAT Math score for high school seniors in California At considerable effort and expense you give the test to a simple random sample of 500 high school seniors The mean score for your sample is E 461 points The standard deviation of the SAT Math test is a known a 100 points Questions 1 Can we include a measure of the precision associated with the point estimate 2 Can we include a measure of our confidence in our results Answer 0 Yes we can construct a confidence interval for u A con dence interval is calculated from the sample data and it represents an interval estimate of the population parameter A con dence interval includes an interval computed from the sample The interval is a measure of the variability of our point estimate a confidence level This confidence level measures the confidence that our inference is correct A J In this lesson we want to find a confidence interval for a population mean u Lecture 7 Chapter 6 Page 2 Given a SRS of n units from the population we can calculate the sample mean E This single calculated value E represents an outcome in the sampling distribution of E To obtain a 95 confidence interval for u based on this single observed value we treat our observed outcome E as though it is the true mean of the sampling distribution of E We then construct our interval about this observed value E For example we find the value for both the 25th percentile lower bound and 975th percentile upper bound of the normal distribution with a mean of E and a standard deviation of GJ Using the techniques for normal distributions we find that the 25th percentile of this distribution is 36 1961 and the 975th percentile is l96 n n J J These two percentiles form the lower and upper limits of our 95 confidence interval for u From Table A we find the following For a 95 Confidence Interval For a 90 Confidence Interval For a 99 Confidence Interval Lecture 7 Chapter 6 Page 3 Example 1 Suppose we wish to estimate u the driving time between Lafayette and Indianapolis We select a SRS of n 25 drivers The observed sample mean is 1 110 hours the first sample taken the sample mean calculated Let s assume that we know the standard deviation of X is a 05 hours Let E denote the distribution of the sample mean The standard deviation of E is CE 01 J3 Then x Nu01 Remember our first observed sample mean is 1110 hours An approximate 95 confidence interval for u can be constructed as follows 0 The lower endpoint is a x 1 96 W 110 19601 0904 The upper endpoint is c196 110 19601 1296 So our approximate 95 confidence interval for u is 09041296 Note We are not guaranteed that the true value of u is in the above interval Suppose we conducted another trial of our random phenomenon and obtained an E2 100 hours the second sample taken sample mean calculated If we calculate an approximate 95 confidence interval for u based on E2 we get 0804 1196 which is a different interval estimate If we repeatedly selected SRS of 25 drivers and for each SRS we calculated the approximate 95 confidence interval for u the population mean in the longrun 95 0f the intervals will contain the true value of u Lecture 7 Chapter 6 Page 4 Formal de nitions Con dence Interval A level C confidence interval for a parameter is an interval computed from sample data by a method that has probability C of producing an interval containing the true value of the parameter Con dence Interval for a Population Mean Choose a SRS of size n from a population having unknown mean u and known standard deviation 0 A level C confidence interval for u is 11 Here 2 is the value on the standard normal curve with area C between 2 and 2 This interval is exact when the population distribution is normal and approximately correct for large n in other cases o xiZ 0 Margln of error Z J17 So the confidence interval for a population mean can also be written as a i margin of error Example 1 Suppose X Bob s golf scores has a normal distribution with unknown mean and standard deviation 0 3 A SRS of nl6 units is selected and a sample mean of E 77 is observed a Calculate a 90 confidence interval for u b Calculate a 95 confidence interval for u c Calculate a 99 confidence interval for u Lecture 7 Chapter 6 Page 5 There is a tradeoff between the precision with which we estimate the unknown parameter and the confidence we have in the result Higher confidence larger C level requires a wider interval Another way in which the margin of error is changed is the sample size A larger sample size will result in a smaller margin of error Choosing the sample size Sample Size for Desired Margin of Error The confidence interval for a population mean will have a specified margin of error In when the sample size is 2 2039 n m 1 You are planning a survey of starting salaries for recent liberal arts major graduates from your college From a pilot study you estimate that the standard deviation is about 8000 What sample size do you need to have a margin of error equal to 500 with 95 confidence Example Lecture 7 Chapter 6 Page 6 Some Cautions The above interval estimation method applies only to SRS Slightly different methods are required when the data are obtained from more complicated surveys and experiments The above formulas do not correct the data for any unknown bias Consequently if the data are biased then ANY inferences based on those data are also biased This includes biases arising from nonresponse undercoverage and response error or hidden bias in experiments Because the sample mean is not resistant confidence intervals are not resistant to outliers If the sample size is small eg nlt15 and the sampling distribution of X is not well approximated by the normal density curve then the above formulas should not be used Typically we do not know the population standard deviation 0 Lecture 7 Chapter 6 Page 7 Section 62 A Confidence interval is one of the two common types of statistical inference it is used to estimate a population parameter from a sample statistic with certain confidence The second common type is a signi cance test which assesses evidence provided by gathered data in favor of some claim about the population an hypothesis about the population Tests of Signi cance Often we have specific questions regarding a particular value of a population parameter Example 1 Bob golf scores are historically normally distributed with u 77 strokes and c5 3 strokes Bob has recently made two improvements to his game he has taken a lesson from a local Professional and he has read a book on the mental approach to putting With these improvements Bob thinks his game is better Let s examine the truth of Bob s thinking using a test of significance Bob has played several rounds after these improvements His scores after the improvements were 77 73 74 78 78 75 75 74 71 We now ask if Bob s average score u 77 is still a reasonable value or has he improved so his average would be less We ask Question 1 Is u 77 Question 2 Is u lt 77 Lecture 7 Chapter 6 Page 8 Example 2 Bob has a driver s license that gives his weight as 190 pounds Bob did not change this information the last two times he renewed his license but has been eating too well in the interim Recently Bob decided to go on a diet and feels he has changed his eating habits Let s test the truth of the weight on Bob s license using a test of significance We know over time Bob s weight is approximately normally distributed with a standard deviation of 3 pounds Bob has weighed himself during the last month and his weights are 193 194 192 191 We wonder if the weight on Bob s license is still correct Question 1 Is u 190 Question 2 Is u 3395 190 If we look at the two examples we see that the question 1 s and 2 s are similar in form In the first question a particular value of the parameter is specified We call this statement about the parameter the null hypothesis or H0 In the second question an alternative set of possible parameter values is specified This second statement about the parameter is called the alternative hypothesis or H a Note The parameter value specified in the null hypothesis usually represents an established standard or generally accepted norm The suspected change in the parameter value is summarized by the alternative hypothesis Lecture 7 Chapter 6 Page 9 Now let s express our above examples in terms of the null and alternative hypothesis Example 1 H 0 u 77 H a u lt 77 Note This is a onesided significance test Example 2 H 0 u 190 H a u 190 Note This is a twosided significance test The goal of a significance test is the following Based on a random sample from the population we want to determine if a change has resulted in a shift in the mean Because the null hypotheses represents the established or accepted mean value and we believe a change has occurred as specified in the alternative hypothesis we want to use the data to determine statistically if we can Reject the null hypothesis in favor of the alternative Let s look at the first example based upon the statistical techniques we already know X Bob s golf score We use the scores Bob has recorded after his lesson and book read as a sample of X We know that Bob s golf scores are normally distributed hence any sampling distribution of his scores is also normally distributed If his scores were not normally distributed we would need to assume that our sample size is large enough so that the Central Limit theorem applies and the sampling distribution of E can be approximated by the normal density curve The standard deviation of E is 7E 3 5 1 stroke Lecture 7 Chapter 6 Page 10 Now let us consider our mean of the sampling distribution in context of our two hypotheses If the null hypothesis is true then the mean of the sampling distribution is Mx77 If H0 is true then J NN771 If the null hypothesis is false and H a is true then the mean is some value lt 77 strokes If H0 is false and Ha is true then YN Nul for some value u lt77 Values of E close to 77 would tend to support H 0 and values that are less than 77 would provide evidence against H 0 With the sample of Bob s last 9 scores we get a sample mean of f 75 Can we conclude that we should reject H 0 in favor of H a To figure this out we calculate a pvalue For example we calculate the PYlt75 PZ lt 751772PZ lt 200 00228 Note that to calculate this probability we assumed H 0 was true and 7 N N 77 1 Because the probability of obtaining a sample f lt 75 is small we would rej ect H 0 Let s say Bob wanted quicker feedback on his golf improvements and wanted it after his first 5 scores In this case the sample size is smaller and as a result we get a sample mean of f 76 and a 3 1342 then PYlt76 PZ ltJ 02266 which is not very small This suggests that it is reasonable that Y 76 as an observed outcome from N77l 342 So we would fail to reject H 0 As you can see the sample size often affects whether we reject H 0 That is why we never accept H 0 but rather fail to reject H 0 This failure to reject H 0 may be because we did not select a large enough sample Lecture 7 Chapter 6 Page 11 Test of Signi canceHypothesis Tests formal steps Step 1 State the Null and Alternative Hypothesis Null Hypothesis H 0 The statement being tested in a statistical test is called the null hypothesis The test is designed to assess the strength of the evidence against the null hypothesis Usually the null hypothesis is a statement of no effect or no difference H 0 y 0 Alternative Hypothesis H a The claim about the population that we are trying to find evidence for Choose one of the three potential hypotheses Onesided tests H a y gt 0 H a y lt 0 Twosided test H a y i 0 Step 2 Find the test statistic lf uo is the value of the population mean Lt specified by the null hypothesis the onesample 2 statistic is Ux Step 3 Calculate the p value For one sided tests pvalue PZ E z or PZ 2 2 For two sided tests pvalue 2PZ Z z Step 4 State conclusions in terms of the problem One way to do this is to choose a significance level that defines how much evidence we desire39 we choose an d Then compare the pvalue to the 1 level lfpvalue lt 1 then reject H 0 Z lfpvalue gt 1 then fail to reject H 0 Your conclusion should be in this form At a rejectfail to reject H 0 evidenceno evidence that restate the null hypothesis in language layperson can understand Even though H a is what we hope or believe to be true our test gives evidence for or against H 0 only We never prove H 0 true we can only state whether we have enough evidence to reject H 0 which is evidence in favor of H a but not proof that H a is true or that we don t have enough evidence to reject H 0 Lecture 7 Chapter 6 Page 12 Example 1 A shipment of machined parts has a critical dimension that is normally distributed with mean 12 centimeters and standard deviation 01 centimeters The acceptance sampling team believes that the measurements are less than 12 centimeters Consequently they take a random sample of 25 of these parts and obtain a mean of 1199 Is the acceptance sampling team correct in their assertions Use an 1 level of 001 Lecture 7 Chapter 6 Page 13 Con dence Intervals and TwoSided Tests A level 1 twosided significance test rejects a hypothesis H 0 u uo exactly when the value uo falls outside a level 11 confidence interval for u Examples 1 Bob has a driver s license that gives his weight as 190 pounds Bob did not change this information the last two times he renewed his license but has been eating too well in the interim Recently Bob decided to go on a diet and feels he has changed his eating habits Let s test the truth of the weight on Bob s license using a test of significance We know over time Bob s weight is approximately normally distributed with a standard deviation of 3 pounds Bob has weighed himself during the last month and his weights are 193 194 192 191 Lecture 7 Chapter 6 Page 14 Section 63 Use and Abuse of Tests Choosing a Level of Significance If we want to make a decision based on our test we choose a level of significance in advance We do not have to do this however if we are only interested in describing the strength of our evidence If we do choose a level of significance in advance we must choose 1 by asking how much evidence is required to reject H 0 The choice of 1 depends on the type of study we are doing Practical Significance Statistically significant does not mean practically significant If we use a large sample we may get a statistically significant result but it may not be practically significant Don t Ignore Lack of Significance Often only research that shows statistically significant results is published This is a problem because other researchers may repeat the experiment thinking that there is a difference when there may not be Sometimes a lack of Significance may be due to a small sample size In this case the power of the test is low Some Cautions about Statistical tests 0 As with Cl s badly designed surveys or experiments often produce invalid results Formal statistical inference cannot correct basic flaws in data collection As with Cl s tests of significance are based on laws of probability Random sampling or random assignment ensures that these laws apply Statistical significance is not the same thing as practical significance There is no sharp border between significant and non significant only increasingly strong evidence as the P value gets smaller Lecture 7 Chapter 6 Page 15

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