Basics of Statistical Hypothesis Tests 1

Transcription

1 Basics of Statistical Hypothesis Tests 1 Statistical hypothesis testin involves usin a sample test statistic to decide which of two competin claims to reject or fail to reject. One of the statements is called the null hypothesis and is denoted by. The other competin statement is called the alternative hypothesis and is denoted by. When we fail to reject the null hypothesis, we do not accept it in the sense that we necessarily believe it is true, but because there is not enouh evidence to reject. A hypothesis test is similar to a criminal jury trial where the null hypothesis states that the defendant is innocent and the alternative hypothesis states that the defendant is uilty. Juries have failed to convict a uilty criminal because the prosecutor did not provide sufficient evidence to convict even thouh they believed that the accused was probably uilty. Both statistical hypothesis statements, and, are about population parameters. The reason for doin a hypothesis test is that there is some reasonable belief that is false, the results of random sample indicate that the alternative hypothesis miht be true, or we suspect that somethin has chaned. The objective of an hypothesis test is to put the null hypothesis on trial with the intention of rejectin it by providin sufficient evidence and support for the alternative hypothesis. is always assumed to be true. In order to decide which of the two competin statements will be rejected or failed to be rejected, a simple random sample of sufficient size is constructed and the resultin sample data values are used to calculate the value of an appropriate sample test statistic. There are two methods of usin the value of the sample test statistic to determine whether or not to reject or fail to reject. Both methods require knowlede of the probability distribution of a sample statistic under the assumption that is true. The stratey behind both testin approaches is to ather enouh evidence to reject and at the same time keep the probability of makin a wron decision low. The four examples on pae 3 of this handout illustrate how to properly write null and alternative hypothesis test statements. Please carefully read and study these examples. In the traditional hypothesis test method, a rejection reion interval of low probability under a probability distribution curve is constructed so that the total area under the pd curve in the rejection reion equals the sinificance level of the test which is denoted by α. This is done under the assumption that is true. The sinificance level α is enerally set to 0.01, 0.05 or 0.10 and equals the probability of rejectin if it is true. Rejection reions for different types of hypothesis tests are described on pae 4. If the value of the test statistic falls in the rejection reion, we reject and accept with the understandin that we may have committed a type I decision error. If the value of the test statistic does not fall in the rejection reion, we fail to reject because the value of the test statistic does not provide sufficient evidence to reject. Failin to reject is done with the understandin that a type II error may have been committed. Rejection reions under a probability distribution curve represent very unusual events when is true. In the P-value hypothesis test method, the probability value or P-value of the test statistic is calculated under the assumption that is true. If the P-value of the test statistic is below a preset sinificance level α such as 0.01, 0.05, or 0.10, the P-value is considered to be sinificant and therefore we reject and accept with the understandin that we may have committed a type I decision error. If the P-value of the sample test statistic is above the sinificance level α, we fail to reject because the event correspondin to the P-value of the test statistic is not unusual enouh to cause rejection of. Keep in mind that P-values are probabilities and therefore P-values must rane from 0 to 1. Carefully read and study the examples on pae 6 of this handout.

2 Definitions for Hypothesis Tests Null Hypothesis A null hypothesis statement is a claim that a population parameter is equal to some specific value. A null hypothesis statement implies that there is no difference, nothin has chaned or everythin is as it was before some new procedure was implemented. Read and study the examples on paes 3 and 5. Alternative Hypothesis An alternative hypothesis statement is a claim that a population parameter is less than, reater than, or not equal to some specific value. An alternative hypothesis statement implies that there is a difference, somethin has chaned, or thins are different than they were before some new procedure was implemented. Study and read the examples on paes 3 and 5 of this handout. Sinificance Level of the Test α The symbol α is a number that represents the sinificance level of the test. Typical values assined to α are 0.01, 0.05, or When is rejected, it is rejected on the fact that the observed value of a sample test statistic is sinificant in the sense that its value represents an unusual or rare event when is true. Therefore α represents the probability of rejectin when in fact is true. When settin the value of α, one needs to consider the potential consequences of rejectin when it is true. Left-Tail Hypothesis Test: Left-tail tests come about when the alternative hypothesis claims that a population parameter is less than some specific value. The rejection reion interval of a left-tail test is a reion of low probability and is located on the far left side of the raph of the probability distribution. The riht end point of the left-tail rejection reion is a critical value because it separates reions of rejection and non rejection. The area below the pd curve and above the rejection interval equals α which is the sinificance level of the test. Left-tail rejection reions represent unusual or rare events when is true. Refer to the examples on pae 4 of this handout. Make sure that you know how to use your raphin calculator to find critical values. Riht-Tail Hypothesis Test: Riht-tail tests come about when the alternative hypothesis claims that a population parameter is reater than some specific value. The rejection reion interval of a riht-tail test is a reion of low probability and is located to the far riht side of the raph of the probability distribution. The left end point of the riht-tail rejection reion is a critical value because it separates reions of rejection and non rejection. The area below the pd curve and above the rejection interval equals α which is the sinificance level of the test. Riht-tail rejection reions represent unusual or rare events when is true. Refer to the examples on pae 4 of this handout. Make sure that you know how to use your raphin calculator to find critical values. Two-Tail Hypothesis Test: Two-tail tests come about when the alternative hypothesis claims that a population parameter is not equal to or is different than some specific value. The rejection reion of a two-tail test is the union of a left-tail reion and a riht-tail reion where the area of each reion equals α/. Two-tail rejection reions represent unusual or rare events when is true. The two critical values of a two-tail test are the upper bound of the left-tail reion and the lower bound of the riht-tail reion. Refer to the examples on the pae 4 of this handout. Make sure that you know how to use your raphin calculator to find critical values. P-Value of a Test Statistic Let the constant k equal the value of a sample test statistic derived from a simple random sample and let α = the sinificance level of the hypothesis test.. The P-value k is a probability that is less than α if and only if k is in the rejection reion of the hypothesis test. Let x equal the random variable of the probability distribution of the test statistic. Read and study the examples pae 6 of this handout. Left-tail test: P-value of k = P( x < k ) Riht-tail test: P-value of k = P( x > k ) Two-tail test and k is less than the median of the pd: P-value of k = P( x < k ) Two-tail test and k is reater than the median of the pd: P-value of k = P( x > k ) Because the area of each tail of a two-tail test = α/, we must double the probability in order to reject when the P-value < α.

3 Type I Error A type I decision error is made when is rejected when in fact is true. The sinificance level α of a test equals the probability of committin a type I error. A jury makes a type I error when it convicts an innocent person. If a type I decision error can have serious consequences, set α = Type II Error A type II decision error is made when is not rejected when in fact is false. The probability of committin a type II error is denoted by β. A jury makes a type II error when it fails to find a uilty person uilty. Lowerin α causes β to increase. If a jury is overly concerned about renderin a uilty verdict, they increase the probability that a uilty criminal will be found not uilty. Power of the Test The power of the test equals 1 β which equals the probability of acceptin or supportin an alternative hypothesis that is true. The power of a test depends on a specific parameter. Increasin the sample size, increasin the level of sinificance α, or increasin the difference between the and parameters will cause the power of the test to increase. 3 Comment: The sinificance level α equals how much type I error one is willin to tolerate or risk. In real world situations, decision errors can have very serious consequences and determinin the consequences of decision errors can be extremely difficult. In introductory statistics courses, these issues are swept under the ru by simply tellin the student what level of sinificance to use and not worry about type II errors or the power of the test. A hypothesis test is a vastly superior method of makin a decision than usin a ut feelin or hunch. We must constantly remind ourselves that somewhat unusual events can happen by chance alone if is true and therefore the observed value of the sample test statistic must be unusual enouh (low P-value) to cause rejection of. Rejectin or failin to reject is not absolute proof of whether is true or false. Examples of Null Hypothesis and Alternative Hypothesis Statements : p = /9 The dice are not loaded in favor of rollin a 7 or 11. The dice are fair and therefore the probability of rollin a 7 or 11 equals /9. : p > /9 The dice are loaded in favor of rollin a 7 or 11 and therefore the probability of rollin a 7 or 11 is reater than /9. : μ = 9.84 The mean bounce heiht of newer baseballs equals 9.84 inches which is no different than the mean bounce heiht of standard reulation baseballs. : μ 9.84 The mean bounce heiht of the newer baseballs is different than the mean bounce heiht of standard reulation baseballs. : σ = 43.7 ft The standard deviation of altimeter readin errors of altimeters produced by the new production line is no different than the standard deviation of errors of the older production line. : σ < 43.7 ft The standard deviation of altimeter readin errors of altimeters produced by the new method is less than the standard deviation of error readins of altimeters produced by the old method. - p = 0 or p 1 = p Sendin reminders to potential voters in Macon County does not improve reistration rates in Macon County. - p < 0 or p 1 < p Sendin reminders to potential voters in Macon County improves voter reistration rates in Macon County.

4 Rejection reions when the sample test statistic has a standard normal probability distribution with µ = 0 and σ = 1. Left-Tail Test (α = sinificance level) If α = 0.01, critical z-score value = -.36 If α = 0.05, critical z-score value = If α = 0.10, critical z-score value = -1.8 Riht-Tail Test (α = sinificance level) If α = 0.01, critical z-score value =.36 If α = 0.05, critical z-score value = If α = 0.10, critical z-score value = 1.8 Two-Tail Test (Area of each tail = α/) If α = 0.01, critical z-score values = If α = 0.05, critical z-score values = If α = 0.10, critical z-score values = α= 0.05 = the area of the rejection reion for a lefttail test. 4 α= 0.01 = the area of the rejection reion for a rihttail test. is rejected if the value of the sample test statistic falls in either rejection reion of a two-tail test. Critical value = for a left-tail test when α = if z < Critical value =.36 for a riht-tail test when α = if z >.36. Rejection reions when the sample test statistic has a Student s t-distribution with n-1 derees of freedom. The critical value of a rejection reion depends on the derees of freedom parameter of the t- distribution and the sinificance level α. The raph to the riht shows a t-distribution havin rejection reions with α equal to 0.01 and The critical values correspondin to α = 0.01 and 0.05 are and respectively when df = 30. These values can be found in a table of critical values for the t-distribution or by usin the invt function command built into the TI raphin calculator. For a two-tail test, the area of each tail must equal α/. The critical value = when the area of the rejection reion for a left-tail =.01 and df = 30. if t < Critical value = The critical value = when the area of the rejection reion for a riht-tail =.05 and df = 30. if t > Critical value = Rejection reions when the sample test statistic has a chi-square distribution with n-1 derees of freedom. The critical value of a rejection reion depends on the derees of freedom parameter of the chisquare distribution and the sinificance level α. The χ distribution shown to the riht equals has 1 derees of freedom and critical values equal to 5.6 and 1.06 when α = Critical values can be found in a table of critical values for the chi-square distribution. Be careful when readin values from a χ table. For a two-tail test, the area of each tail must equal α/. Mean μ = df = 1 The area of each rejection reion = a =.05. When df = 1, the critical values are c = 5.6 and c = L R

5 Example 1: Let z = where the z-statistic is derived from a simple random sample. For a left-tail test, P-value of = P(x ) = normalcdf(-9999, ) = For a riht-tail test, P-value of = P(x ³ ) = normalcdf(-1.985, 9999) = For a two-tail test, P-value of = P(x ) = normalcdf(-9999, ) = Example : Let z =.135 where the z-statistic is derived from a simple random sample. For a left-tail test, P-value of.135 = P(x.135) = normalcdf(-9999,.135) = For a riht-tail test, P-value of.135 = P(x ³.135) = normalcdf(.135, 9999) = For a two-tail test, P-value of.135 = P(x ³.135) = normalcdf(.135, 9999) = Example 4: Let t = where the t-statistic is derived from a simple random sample and derees of freedom = 64. For a left-tail test, P-value of = P(x 1.577) = tcdf(-9999, 1.577, 64) = For a riht-tail test, P-value of = P(x ³ 1.577) = tcdf(1.577, 9999, 64) = For a two-tail test, P-value of = P(x ³ 1.577) = tcdf(1.577, 9999, 64) = (because > median = 0 ) Example 5: Let c = where the c -statistic is derived from a simple random sample and derees of freedom = 30 = m. For a left-tail test, P-value of = P(x ) = c cdf(0, , 30) = For a riht-tail test, P-value of = P(x ³ ) = c cdf(17.046, 9999, 30) = For a two-tail test, P-value of = P(x ) = c cdf(0, , 30) = Example 6: Let c = where the c -statistic is derived from a simple random sample and derees of freedom = 36 = m. Calculatin the P-Value of a Test Statistic Examples Backround: Most of the statistics that we use in this course can be rouped into three eneral cateories. The z-statistics have a standard normal probability distribution with μ = 0 and σ = 1. The t- statistics have a t probability distribution with df derees of freedom and the chi-square-statistics have a χ probability distribution with df derees of freedom and mean μ = df. Refer to the handout that describes the probability distribution of various samplin statistics. ( because < median = 0 ) ( because.135 > median = 0 ) Example 3: Let t = where the t-statistic is derived from a simple random sample and derees of freedom = 36. For a left-tail test, P-value of = P(x ) = tcdf(-9999, , 36) = For a riht-tail test, P-value of = P(x ³ ) = tcdf(-1.675, 9999, 36) = For a two-tail test, P-value of = P(x ) = tcdf(-9999, , 36) = (because < median = 0 ) (because is closer to the left-tail of the pd curve ) For a left-tail test, P-value of = P(x 41.88) = c cdf(0, 41.88, 36) = For a riht-tail test, P-value of = P(x ³ 41.88) = c cdf(41.88, 9999, 36) = For a two-tail test, P-value of = P(x ³ 41.88) = c cdf(41.88, 9999, 36) = ( besause is closer to the riht tail of the pd curve ) 6

6 How to Properly State the Conclusion of a Hypothesis Test Examples 5 Every hypothesis test involves usin a simple random sample(s) taken from the parent population(s) and the calculated value of a test statistic derived from a random sample(s). The P-value of the test statistic depends on the probability distribution of the test statistic. In this course, the probability distributions of test statistics are the normal distribution, t-distribution, χ -distribution and F-distribution. Each conclusion below is based on some iven problem situation and the results of a hypothesis test. Example 1: Left-tail test where : μ = 100, : μ < 100 and α = The value of the test statistic was and the P-value of = Conclusion: With a P-value = and the level of sinificance = 0.05, we fail to reject the null hypothesis. There is not enouh evidence to support the claim that the mean number of cars sold per month at Honda dealerships is less than 100 cars. Example : Riht-tail test where : μ = 100, : μ > 100 and α = The value of the test statistic was.175 and the P-value of.175 = Conclusion: With a P-value = and the level of sinificance = 0.05, we reject the null hypothesis. There is enouh evidence to support the claim that the mean number of cars sold per month at Honda dealerships is reater than 100 cars. Example 3: Left-tail test where - p = 0, p < 0 and α = The value of the test statistic was -.44 and the P-value of -.44 = Conclusion: With a P-value = and the level of sinificance = 0.01, we reject the null hypothesis. There is enouh evidence to support the claim that infants who sleep under bednets experience a lower incidence of malaria than infants who do not sleep under bednets. Example 4: Two-tail test where : σ 1 = σ, : σ 1 σ, and α = The value of the test statistic was and the P-value of = Conclusion: With a P-value = and the level of sinificance = 0.05, we fail to reject the null hypothesis. There is not enouh evidence to support the claim that there is a difference in the standard deviations of test scores on memory recall tests for students who do not use marijuana and students who are heavy marijuana users. Example 5: Riht-tail test where : χ = 0, : χ > 0 and α = 0.05 for a Goodness-of-Fit test. ( In this example, the χ statistic is a measure of the difference between expected and experimental outcomes of a sample.) The value of the χ test statistic was 38.3 and the P-value of 38.3 = Conclusion: With a P-value = and the level of sinificance = 0.05, we fail to reject the null hypothesis. There is no evidence to support the claim that the roulette wheel is not fair. With a fair roulette wheel, we should expect relatively lare P-values such as because the pd of the χ statistic has 37 derees of freedom, μ of the distribution = 37 and the value of the χ test statistic was 38.3.

7 Alorithm for Conductin a Hypothesis Test Every hypothesis test involves calculatin the value of an appropriate test statistic. The value of the test statistic depends on a sinle random sample which is one of the infinitely many possible random samples that can be taken from a parent population. The value of the test statistic is used to decide if there is enouh evidence to support the alternative hypothesis. The steps listed below describe the eneral procedure to properly conduct a hypothesis test. Carefully read and study the problem to determine what type of hypothesis test to conduct and the sinificance level α of the test. Some of the eneral types of hypothesis tests are described below. Is the hypothesis test about the mean μ or the population proportion p of a population? Is the hypothesis test about the difference of two population proportions p 1 and p? Is the hypothesis test about matched data pairs taken from two dependent populations? Write proper null and alternative hypothesis statements. Your and statements should ive the reader a clear understandin of the purpose of the test. The examples on pae 3 of this handout illustrate how to write proper hypothesis statements. This step is the most difficult part of the hypothesis test process, If this part is done well, the remainin parts of the process are relatively easy. Take an appropriate random sample from the parent population. If necessary, check the sample for normality and sample size requirements. Determine what test statistic is needed in order to carry out the test. Some examples of the many different samplin statistics are shown below. Then find the values of the appropriate basic sample statistics such as n, x, p or s. 7 x - m p ˆ - p x - m ( n -1) s æ ( Ei - O ) ö i z =, z =, t =, c = or c = i s p q s åç s è E ø n n n Many different samplin statistics have the same probability distribution. Describe the probability distribution of the sample test statistic. See the examples below. An aproximately standard normal distribution with μ = 0 and σ = 1. A t-distribution with 39 derees of freedom, μ = 0 and σ = A chi-square distribution with 4 derees of freedom, μ = 4 and σ = Remark: For a t-distribution, μ = 0 and σ = (df/(df )). For a χ distribution, μ = df and σ = (*df) If you doin a traditional hypothesis test, use the sinificance level α of the test to calculate the critical value(s) of the rejection reion under the raph of the probability distribution of the test statistic. Rejection reions will be either left-tail, riht-tail, or two-tail. See pae 4 of this handout.. Use the values of the basic statistics derived from the random sample to calculate the value of the test statistic. Calculate the P-value of the statistic found in the previous step. See pae 6 of this handout. If you are doin a traditional hypothesis test, state the test criteria for rejectin. See the examples below. if z < if χ > if t < or t >.0639 Write a complete and succinct conclusion of the results of the hypothesis test. In most cases, it should take no more than two sentences to write a ood conclusion. Read and study the examples on pae 5 of this handout. The P-value method of statin a conclusion simplifies writin a ood conclusion.