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# Introduction to Statistics I STAT 1000

UCONN

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This 17 page Class Notes was uploaded by Blair Williamson on Thursday September 17, 2015. The Class Notes belongs to STAT 1000 at University of Connecticut taught by Vladimir Pozdnyakov in Fall. Since its upload, it has received 29 views. For similar materials see /class/205898/stat-1000-university-of-connecticut in Statistics at University of Connecticut.

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Date Created: 09/17/15

12 Introduction to Hypothesis Testing The purpose of hypothesis testing is to determine whether there is enough statistical evidence in favor of a certain belief about a parameter Examples 0 Is there statistical evidence in a random sample of potential customers that support the hypothesis that more than p of the potential customers will purchase new products Is a new drug effective in curing a certain disease A sample of patient is randomly selected Half of them are given the drug where half are given a placebo The improvement in the patients conditions is then measured and compared Concept of hypothesis testing There are two hypotheses about a population parameters H 0 the null hypothesis 0 t I H a the alternative research hypothesis Assume the null hypothesis is true Build a statistic related to the parameter hypothesized Pose the question How probable is it to obtain a statistic value at least as extreme as the one observed from the sample Two types of errors are possible when making the decision whether to reject H 0 O O N O E 0 Type I error a signi cance level reject H0 when it is true Type 11 error do not reject H0 when it is false or and are inversely related Make one of the following two decisions based on the test Reject the null hypothesis in favor of the alternative hypothesis Do not reject the null hypothesis in favor of the alternative hypothesis 04 Testing the Population Mean When the Population Standard Deviation is Known Normal Case The null hypothesis H0 I 0 Possible alternatives H a 1 y gt 0 H a 1 y lt 0 H a 1 0 The standardized test statistic Consider the test statistic Then the rejection region becomes 0 zaoo if alternative H a u gt yo 0 oo z1 in the case of Ha u lt 0 o oo z U 2oo in the case of Ha u 7i Lio The decision rule We reject the null at signi cance level a and accept the alternative if the test statistic falls into the rejection region If it does not then we say that there is not enough evidence to reject the null P Value method The p value provides information about the amount of statistical evidence that supports the alternative hypothesis By de nition the p value of a test is the probability of observing a test statistics at least as extreme as the one computed given the null hypothesis is true That is p value P gt Tc given that H0 is true Describing the pvalue o If the p value is less than 1 there is overwhelming evidence that support the alternative hypothesis If the p value lies between 1 and 5 there is strong evidence that support the alternative hypothesis If the p value lies between 5 and 10 there is weak evidence that support the alternative hypothesis If the p value exceeds 10 we say that there is no evidence that support the alternative hypothesis The pvalue can be used when making decisions based on rejection region methods as follows 0 De ne the hypotheses to test and the required signi cance level a 0 Perform the sampling procedure and calculate the test statistic and the p value associated with it 0 Compare the p value to a Reject the null hypothesis only if p lt11 otherwise do not reject the null hypothesis Calculating the Probability of a Type II Error To properly interpret the results of a test of hypothesis we have to 0 specify an appropriate signi cance level or judge the pvalue of a test 0 understand the relationship between Type I and Type II errors How do we compute a type II error Calculation of a type II error requires that o the rejection region be expressed directly in terms of the parameter hypothesized not standardized o the alternative value under H a be speci ed Effects on of changinga decreasing the significance level 1 increases the the value of and vice versa Judging the test A hypothesis test is effectively defined by the significance level a and by the sample size n Ifthe probability ofa type II error is judged to be too large we can reduce it by 0 increasing a 0 increasing the sample size In summary by increasing the sample size we reduce the probability of type II error Hence we will accept the null hypothesis when it is false less frequently Power of a test The power of a test is defined as l It represents the probability to reject the null hypothesis when it is false LargeSample Test of Hypothesis about u In real world the population standard deviation is usually unknown However if the sample size is large we can use the sample standard deviation 3 as an accurate estimate of 039 because of the law of large numbers Moreover in this situation we can omit the normality assumption because by CLT Y has approximately normal distribution When the sample size is large gt30 the tests statistics is X 0 s J2 Because of CLT it has approximately normal distribution No assumptions need to be made about the probability distribution of the population 2 Example 1 In a nationwide opinion poll based on a random sample of 2417 people one question is How do you rate the ethics of business executives of large companies A rating of 3 means no better or worse than most people a rating of l is much better than most people and 5 is much worse than most people the mean rating is 305 and the standard deviation is 062 Can we infer at signi cance level 5 that respondents rated the ethics of executives at different level in comparison to most people Solution 1 Setup Let u be the true nationwide mean rating H0 u 300 Ha u at 300 2 Test statistics X yo 305 300 s n 6 f V2417 3 Rejection Region RR 00 Z U 2oo 00 196 U l96oo Z 396 4 Conclusion Since the test statistic falls into the rejection region we reject the null at the signi cance level 5 and accept alternative That is we agree that there is a difference in rating of the executives ethics and general public However we have to note that from the practical point of View this di erence is negligible References 1 Chase and Bown General Statistics 2 Hildebrand and Ott Statistical Thinking for Managers 3 Keller and Warrack Statistics for Management anal Economics 4 McClave Benson and Sincich A First Course In Business Statistics 5 Conditional Probability and Independence Conditional Probability o The probability of an event when partial knowledge about the outcome of an experiment is known is called conditional probability 0 We use the notation PA l B for the conditional probability that event A occurs given that event B has occurred PA and B PAlB PB Example 1 The following data are characteristics of the votingage population regarding the 1992 presidential election in the United States Number of persons is measured in thousands Voted Did not vote Total Males 53312 35245 88557 Females 60554 36573 97127 Total 113866 71818 185684 For a randomly selected person from the population let A be the event that the person selected voted and Bbe the event that the person selected is a male Find each of the following 1 PA 2 PA 3 PAlB Solution 1 PA1138685684613 2 PA 1 PA 1 613 387 53312 3 PA BFW M wz 88557 PB A85684 Independent and Dependent Events 0 Two events A and B are said to be independent ifPAlBPA or PB l A PB Otherwise the events are dependent 0 Note that if the occurrence of one event does not change the likelihood of occurrence of the other event the two events are independent Note that independent events and mutually exclusive events are not the same Example 1 7 cont d Are eventsA and B independent Solution Since PA 613 at 602 PA B the events A and B are not independent Probability Rules Again Complement rule Each simple event must belong to either A or A Since the sum of the probabilities assigned to simple events is one we have for any event A PA 1 PA Addition rule For any two events A and B PA or B PA PB PA and B Multiplication rule For any two events A and B PA and B PAPB A PBPA B When A and B are independent PA and B PAPB Example 2 Let A and Bbe independent events with PA 3 and PB 4 What is PA or B Solution Let us use the addition law PA or B PA PB PA and B 3 4 PA and B Because of the independence of these events PA and B PAPB 3 X 4 12 Therefore PA or B 7 12 58 Probability Trees This is a useful device to build a sample space and to calculate probabilities of simple events and events Rules for constructing a probability tree 1 Events forming the rst set of branches must have known marginal probabilities must be mutually exclusive and should exhaust all possibilities so that the sum of branch probabilities is 1 2 Events forming the second set of branches must be entered at the tip of each of the sets of rst branches Conditional probabilities given the relevant rst branch must be entered unless assumed independence allows the use of unconditional probabilities Again the branches must be mutually exclusive and exhaustive 3 If there are additional sets of branches the probabilities must be conditional on all preceding events As always the branches must be mutually exclusive and exhaustive 4 The sum of path probabilities must be taken over all paths included in the relevant event Example 3 In a certain television game show a valuable prize is hidden behind one of the three doors You the contestant pick one of the three doors Before opening it the announcer opens one of the other two doors and you see that the prize isn t behind that door The announcer offers you the chance to switch the remaining door Should you switch or it does not matter Solution Call the door that you select A the others B and C Assuming that the prize is distributed randomly among the doors the probability that it s behind each of the doors is 13 If you picked a wrong door in A the announcer has no choice If B contains the prize the announcer must open C if C has the prize he must open B But if you picked correctly and A has the prize the announcer does have a choice Let us assume that the announcer picks B or C randomly each with probability 12 in this situation We can construct the following tree B 12 16 A 13 C 12 16 B 13 C 1 13 C 13 B 1 13 Door containing prize Announcer s choice Path probability Suppose that the announcer has chosen B and you chose A initially What is the probability that the prize behind C Pmehmd C chose B Pbehina C and chose B 13 Pchose B 1613 so Pbehina A l chose B 1 23 13 You have a better chance of winning if you switch to door C 23 Recommendedexercises 3867394 311073111 3113 311573116 311873121 References 1 Chase and Bown General Statistics Hildebrand and Ott Statistical Thinking for Managers Keller and Warrack Statistics for Management and Economics McClave Benson and Sincich A First Course In Business Statistics 59 Exercises 1 N E 4 Considered the tossing of two fair dice Consider the following events A sum is 7 or more B sum is even C sum is 7 andD sum is less than 11 a Verify that the only pair of mutually exclusive events isB C b Use the Addition Rule to nd PA or C c Let E sum is less than 4 and F 3 3 Find PA or E or F A card is to be randomly selected from an ordinary deck of 52 cards Consider the following events A ace B face card and C club a Verify that the only pair of mutually exclusive events is A B b Use the Addition Rule to find PA or B Assume that the mother is a carrier for colorblindness and the father is normal Assume also that when a parent contributes a gene from a gene pair either gene is equally likely to be contributed Let event A carries gene for colorblindness and event B is colorblind a Assign probabilities to each outcome b Are the events A B mutually exclusive c Find the probability that an offspring will either carry the colorblind gene or be colorblind The following data are characteristics of the votingage population regarding the 1992 presidential election in the United States Number of persons is measured in thousands A A Voted Did not vote B Males 53312 35245 E Females 60554 36573 For a randomly selected person from the population let A be the event that the person selected voted and B be the event that the person selected is a male Find each of the following 20133 bPA cPA and B There are 2000 voters in a town Consider the experiment of randomly selecting a voter to be interviewed The voters in the town are the possible outcomes of the experiment The event A consists of being in favor of more stringent building 0 gt1 9 0 codes the event B consists of having lived in the town less than 10 years The following table gives the numbers of voters in various categories A A B 100 700 E 1000 200 Find each of the following a PA b PB c PA and B Two cards are to be selected from an ordinary deck of 52 cards Assume the first card is replaced before the second card is selected Consider the following events A first card is an ace B second card is an ace C second card is a king a Find PA and B b Find PB or C c Now suppose a third card is selected after replacement of the first and second cards Let D third card is not an ace Find PA and B and D A sporting goods store has a large batch of cans of tennis balls on hand Ten percent of the cans are unacceptable that is contain at least one defective ball a A customer decides to purchase one can What is the probability that the customer will be satisfied b A customer is to purchase two cans Find the probability that 1 Both cans will be satisfactory ii Exactly one can will be satisfactory iii At least one can will be satisfactory Suppose that the probability that a child produced by a couple will have a particular disease is 1 If they plan to have four children what is the probability that one or more children will have the disease A large shipment contains 2 defective items Five items are to be selected What is the probability of getting one or more defective items Assume that we tossed two fair dice Consider the following events A sum is 7 or more B sum is even C sum is 7 and D sum is less than 11 Find a PA or B b PA or C c PA or D Two cards are to be selected from an ordinary deck of 52 cards Assume that the first card is not replaced before the second one is drawn Consider the following events A first card is an ace B second card is an ace a Find PA and B b Find PB Are the events A B independent c Find PA or B 12 If the mother is a carrier for colorblindness and the father is normal a Find the probability that a child will be colorblind if it is male b Find the probability that a child will be colorblind if it is female 13 A committee of seven consists of two males and five females Two members are to be chosen randomly to look into a specific problem What is the probability that both males will be chosen Hint Imagine the selection as a twostage processselect one member then another without replacement 14 A business employs 600 men and 400 women Five percent of the men and 10 of the women have been working there for more than 20 years If an employee is selected by chance what is the probability that the employee is male given that the length of employment is more tan 20 the years 15 A change was proposed in the mathematics curriculum at a college The mathematics majors were asked whether they approved of the proposed change The results of the survey follow Approved No opinion Did not approve Female 21 6 12 Male 14 10 7 Suppose that a mathematics major is selected by chance Find the probability that a The student is female given no opinion b The student approves of the proposed change given the student is male c The student is male given the student does not approve of the proposed change d The student is male and approves of the proposed change 7 Continuous Probability Distributions Cumulative Distribution Function We de ne a cumulative distribution function F x of a random variable X as FxPX ltx Properties of the cdf 1 F oo0 Foo1 2 Fx T Probability Density Function If the cdf F x is differentiable then we can de ne a probability density function f x by f x F 39 x The density function satisfies the following conditions 0 f x is nonnegative o The total area under the curve representing f x equals 1 The probability that X falls between a and b is found by calculating the area under the graph of f x between a and b Pa lt X lt b jfxdx The expected value of the random variable X is EX j xfxdx Properties of the expected value Ecc EraXaEX EXYEXEY EXY EXEY if random variables X and Y are independent ie PXlta and YltbPXltaPXltb for all possible a and b The expected value of the random variable gX is EgX j gxfxdx all x In particular EX jxfxdx VarX EX EX2 x EX2 f 90656 01 2 mm EX 2 EX2 Ix2fxdx jx xmx all r all 6 Properties of the variance Varconst 0 VaraX a2 VarX VarXY VarX VarY2COVX Y where COVX Y EX EXY EY If X and Y are independent then COVX Y 0 and VarXY VarX VarY If VarX 0 then X const Uniform Distribution A random variable X is said to be uniformly distributed if its density function is l fx7 anSb b a The expected value and the variance of the uniform distribution is EXLb Varm The Normal Distribution This is the most important continuous distribution considered here 0 Many random variables can be properly modeled as normally distributed 0 Many distributions can be approximated by a normal distribution 0 The normal distribution is the cornerstone distribution of statistical inference Normal distribution A random variable X with mean u and varianceEIO392 is normally distributed if its probability density function is given by 1 706707 2 fx e 2quot 00 S x S 00 O392 39 Where It 314159 and e 271828 Normal random variable has a bell shaped distribution symmetrical around u How does the standard deviation affect the shape of f x 7 005 1 000 1 005 1 000 1 Functions of Normal Random Variables o If X is normally distributed with mean u and the variance 0392 then X a is also normally distributed with the mean u a and the variance 0392 o If X is normally distributed with mean u and the variance 0392 then aX is also normally distributed with the mean a and the variance 61202 o If X and Y are normally distributed independent random variables with the means uX and y and the variances 039 and 039 then aX bY also has the normal distribution with mean ayX byy and the variance 612039 bio Finding Normal Probabilities Two facts help calculate normal probabilities o The normal distribution is symmetrical 0 Any normal distribution can be transformed into a speci c normal distribution called STANDARD NORMAL DISTRIBUTION or Z Every normal variable with some Li andI039 can be transformed into the standard normal random variable Z X X 0X By the properties mentioned above we have that E Z 0 and VarZ l Therefore once probabilities for Z are calculated probabilities of any normal variable can found The symmetry of the normal distribution makes it possible to calculate probabilities for negative values of the random variable Z De nition Let us consider a numbera such that0 lt a lt 1 then critical value of standard normal distribution 21 represents that value for which the area under the standard normal curve to the right of 21 is equal to a ie PZ gtzaa Exponential Distribution The exponential distribution can be used to model 1 the length of time between telephone calls 2 the length of time between arrivals at a service station 3 the lifetime of electronic components When the number of occurrences of an event follows the Poisson distribution the time between occurrences follows the exponential distribution A random variable is exponentially distributed if its probability density function is given W fx te39lx x20 where A is a parameter of the distribution the average number of occurrences of the events EX 1 VarX 1 PX gt x e Exercises p 240 4847496 p 241 410274107 References 1 Chase and Bown General Statistics 2 Hildebrand and Ott Statistical Thinking for Managers 3 Keller and Warrack Statistics for Management and Economics 4 McClave Benson and Sincich A First Course In Business Statistics Exercises 1 The lengths X of the nails in a large shipment received by a carpenter are approximately normally distributed with mean of 2 inches and standard deviation 1 inch a If a nail is randomly selected nd P18 lt X lt 207 b What proportion of nails has lengths that lie within one standard deviation of the mean c The carpenter cannot use a nail shorter than 175 inches or longer than 225 inches What percentage of the shipment of nails will the carpenter be able to use N Scores of males on the 1974 Mathematical Scholastic Aptitude Test MSAT were normally distributed with mean 500 and standard deviation 100 a What score indicates a percentile rank of 95 b The middle 40 of the distribution is bounded by what two scores c If 1000 of these students are randomly selected how many are expected to score higher than 650 E 4 V39 0 gt1 The length of time it takes for a ferry to reach a summer resort from the mainland is approximately normally distributed with mean 2 hours and standard deviation 12 minutes Over many past trips what proportion of times has the ferry reached the island in a Less than 1 hour 45 minutes b More than 2 hours 5 minutes c Between 1 hour 50 minutes and 2 hours 20 minutes Some auto companies design emission sensors so that they must be replaced after 100000 miles One such company found that the service life X in months of these sensors is approximately randomly normal with mean 48 months and standard deviation 9 months a The company decided to guarantee the sensors for 3 years What percentage of the sensors will not satisfy the guarantee b The company decided to replace only 1 of all sensors What should be the length in months of the guarantee LetX be the number of minutes after 11 o clock that a bus leaves the bus station Assume that the distribution of times is approximately normal with mean 15 and standard deviation 4 minutes a If a person gets to the bus station at 11 10 what is the probability that the person has missed the bus b If a person is willing to risk a 20 chance of not making the bus what is the maximum number of minutes after 11 o39clock that the person can reach the station c What time should the person reach the station to have a 5050 chance of catching the bus A diastolic blood pressure reading of less than 90 mm is considered normal a reading of 90 or more indicates hypertension Assume that the distribution of diastolic blood pressure readings of people 30 to 39 years old in the Framingham Heart Study is approximately normal with mean 79 and standard deviation 11 What proportion of this population has a Normal diastolic blood pressure b Hypertensive diastolic readings Assume that the number of hours a product will function before needing service is approximately normally distributed a If the standard deviation is 70 and 10 of the products will break down before 700 hours what is the mean b If the mean time is 800 hours and 20 will function for more than 850 hours what is the standard deviation

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