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HYPOTHESIS TESTING 1 Introduction The purpose of hypothesis testing is to determine whether there is enough statistical evidence in favor of a certain belief about a parameter And to permit generalizations from a sample to the population from which it came. The word hypothesis is just slightly technical or mathematical term for “sentence” or “claim” or “statement”. In statistics, a hypothesis is always a statement about the value of one or more population parameter(s). 2 Hypothesis: A statement that something is true concerning the population. Typical statistical hypotheses are: µ>5 cm P0.65 2>2.00 µ1-µ2>0 and so on. The following are not statistical hypothesis “ x 5“ because x is not a population parameter (it is a sample statistic) “µ is big enough” because though it is a statement about the population parameter µ, the statement is not quantitative. 3 There are two hypotheses (about one or more population parameter(s)) H0 - the null hypothesis H1 - the alternative hypothesis The hypothesis testing is the operation of deciding whether or not data obtained for a random sample supports or fails to support a particular hypothesis. The statistical procedure. hypothesis test is a five-step The first two steps of the hypothesis test procedure are to formulate two hypothesis. 4 STEP 1 Formulate the null hypothesis Null Hypothesis, H0: The hypothesis upon which we wish to focus our attention. Generally this is a statement that a population parameter has a specified value. STEP 2 Formulate the alternative hypothesis Alternative Hypothesis, Ha: A statement about the same population parameter that is used in the null hypothesis. Generally this is a statement which specifies that the population parameter has a value different from the value given in the null hypothesis. One-sided Two-sided H0: = 0 H0: = 0 H0: = 0 Ha: 0 Ha: > 0 Ha: < 0 5 At the conclusion of the hypothesis test, we will reach one of two possible decisions. We will decide in agreement with null hypothesis and say that we fail to reject H0. Or we will decide in opposition to null hypothesis and say that we reject H0. 6 There are four possible outcomes that could be reached as a result of the null hypothesis being either true or false and the decision being either “fail to reject” or “reject”. Null Hypothes is Decision True False Accept H0 Correct Decision (1- α) Type II Error β Reject H0 Type I Error α Correct Decision (1- β) 7 a = P(commit a type I error) = P(reject H0 given that H0 is true) = P(commit a type II error) = P(accept H0 given that H0 is false) Accept H0 Reject H0 8 STEP 3 Determine the test criteria. Test Criteria: Consist of 1. determining a test statistic, 2. specifying a level of significance a 3. and determinig the critical region. Test Statistic: A random variable whose value will be used to make the decision “fail to reject H0” or “ reject H0” Critical Region:The set of values for the test statistic that will cause us to reject the null hypothesis. Critical Value is the first value in the critical region. Level of significance: The probability of committing 9 the type I error, a. STEP 4 Obtain the value of the test statistic. The test statistic is some statistic that may be computed from the data of the sample. The test statistic serves as a decision maker, since the decision to reject or not to reject the null hypothesis depends on the magnitude of the test statistic. An example of a test statistic is the quantity xi z x z n 10 STEP 5 Make a decision and interpret it. Decision Rule: If the test statistic falls within the critical region, we will reject H0. If the test statistic does not fall in the critical region, we will fail to reject H0. The set of values that are not in the critical region is called the acceptance (noncritical) region. 11 Example Researchers are interested in the mean level of some enzyme in a certain population. They take a sample of 10 individuals, determine the level of enzyme in each and compute a sample mean 22. It is known that the variable of interest is approximately normally distributed with a variance of 45. Let us say that they are asking the following question: Can we conclude that the mean enzyme level in this population is different from 25? Step1-2 H0: = 25 Ha: 25 12 Step 3 Since we are testing a hpothesis about a population mean, we assume that the population is normally distributed, and the population variance is known, our test statistic is z. Let us say that we want the probability of rejecting a true null hypothesis to be a=0.05. Our rejection region is to consist of two parts. It seems reasonable that we should divide a equally and let a/2=0.025 be associated with small values and a/2=0.025 be associated with large values. 13 Step 4 z x n 22 25 45 / 10 1.41 Step 5 a/2=0.025 a/2=0.025 rejection region acceptance region 1.96 0 -1.96 -1.41 0.95 rejection region Step 6 Accept H0. We conclude that the mean enzyme level in this population is not different from 25 14 Example Suppose, instead of asking if they could conclude that 25, the researchers had asked: Can we conclude that the mean enzyme level in this population is less than 25? Step1-2 H0: = 25 Ha: < 25 15 Step 3 Since we are testing a hpothesis about a population mean, we assume that the population is normally distributed, and the population variance is known, our test statistic is z. Let us say that we want the probability of rejecting a true null hypothesis to be a=0.05. We will want our rejection region is to be where the small values are –at lower tail of the distribution. This time, since we have a one-sided test, all of a will go in the one tail of the distribution. 16 Step 4 z x n 22 25 45 / 10 1.41 Step 5 a=0.05 -1.645 -1.41 0.95 rejection region 0 acceptance region Step 6 Accept H0. We conclude that the mean enzyme level in this population is not less than (greater than 17 Hypothesis Testing Parametric Tests Nonparametric Tests • Sampling should be random. • Population should be distributed normally. • Variables sholud be continuous. • # of observations should be greater than 10. • No assumption on the distribution of the population. •No assumption on the type of the variable. • No assumption on the # of observations. 18 Non parametric tests are used as an alternative to parametric tests. Usually are used when the distribution of underlying population is nonnormal. Even the underlying population is normally distributed if the sample size is small (n<10), again nonparametric tests are used. While parametric tests are used to test the hypothesis based on population mean, proportion and standard deviation,nonparametric tests are used to test the hypothesis based on median or distribution of samples. 19 HYPOTHESIS TESTING: ABOUT A SINGLE POPULATION 20 A SINGLE POPULATION MEAN INTRODUCTION In this lecture we consider the testing of hypothesis about a population mean under three different conditions: 1. When sampling is from a normally distributed population of values with known variance, 2. When sampling is from a normally distributed population of values with unknown variance, 3. When sampling is from a population that is not normally distributed. 21 When is known and n is large When is unknown and n is small (n30) x z n x t s n If the population that we are sampling is approximately normal and n30, we will base our procedures on Student’s t distribution. Student’s t distribution is the distribution of t statistic. When n>30, t distribition approaches to the normal distribution. 22 Properties of t Distribution: 1. t is distributed with a mean of 0. 2. t is distributed symmetrically about its mean. 3. t is distributed with a variance greater than 1, but as the sample size n increases, the variance approaches to 1 4. t is distributed so as to be less peaked at the mean and thicker at the tails than the normal distribution. 5. t is distributed so as to form a family of distributions, a separate distribution for each sample size. The t distribution approaches to the normal distribution as the sample size increases. 23 Normal Distribution Student’s t, n=10 Student’s t, n=2 0 24 Example Researchers collected serum amylase values from a random sample of 15 apparently healty subjects. The mean and standard deviation computed from the sample are 96 and 35 units/100 ml, respectively. They want to know whether they can conclude that the mean of population from which the sample of serum amylase determinations came is different from 120. Hypothesis H0: = 120 Ha: 120 Test statistic Since the population variance is unknown, our test statistic is t Level of significance a=0.05 25 t x s n 96 120 35 15 2.65 t(a/2,n-1)= t(0.025,15-1)=2.14 a/2=0.025 a/2=0.025 0.95 2.65 2.14 Reject H0 0 2.1 4 Accept H0 Reject H0 26 t Distribution Table Amount of a in two-tail df. 0.20 0.01 0.05 0.025 0.01 Amount of a in one-tail 0.40 0.20 0.10 0.05 0.02 1 1.376 3.078 6.314 12.706 31.821 2 1.061 1.886 2.920 4.303 6.965 ... ... ... ... ... ... 30 0.854 1.310 1.697 2.042 2.457 40 0.851 1.303 1.684 2.021 2.423 ... ... ... ... ... ... 120 0.845 1.289 1.658 1.980 2.358 ... ... ... ... ... ... 0.842 1.282 1.645 1.960 2.326 Sonsuz 27 THE SIGN TEST When the normality assumptions can not be made or when the data at hand are ranks rather than measurments on an interval or ratio scale, an alternative test must be sought. A frequently used nonparametric test does not depend on the assumptions of the t test, or measurement beyond the ordinal scale is the sign test. This test focus on the median rather than the mean as a measure of central tendency or location. 28 The sign test gets its name from the fact that pluses and minuses, rather than numerical value, provide the raw data used in the calculation. The null hypothesis to be tested concerns the value of the population median M. H0: Population median is M0. (M=M0) HA: Population median is not M0. (MM0) The data converted to (+) and (-) signs A plus sign will be assigned to each piece of data larger than M0, a minus sign to each piece of data smaller than M0 and the zero to those data equal M0 29 The sign test uses only the plus and minus signs. When n < 25, the sign test table are used where k is the number of the less frequent sign and n is number of observations. Reject the null hypothesis whenever the number of the less frequent sign is extremely small. When n 25 Accept H0 Reject H0 p<a z za or p < a/2 or z n k 2 z n /2 za / 2 p> a or p > a/2 Otherwise 30 Example: Researchers wished to know if instruction in personal care and grooming would improve the appearance of mentally retarded girls. In a school for the mentally retarded, 10 girls selected at random received special instruction in personal care and grooming. Two weeks after completion of the course of instruction the girls were interwieved by a nurse and a social worker who assigned each girl a score based on her general appearance. We wish to know if we can conclude that the median score of the population from which we assume this sample to have been drawn is different from 5. Girl 1 2 3 4 5 6 7 8 9 10 Score 4 5 8 8 9 6 10 7 6 6 31 H0: The population median is 5. HA: The population median is not 5. Girl 1 2 3 4 5 6 7 8 9 10 Score 4 5 8 8 9 6 10 7 6 6 - 0 + + + + + + + + <5 =5 >5 >5 >5 >5 >5 >5 >5 >5 Since # of (-)=1 and # of (+)=8 and # of (0)=1; k=1, n=10-1=9 32 From the sign test table p=0.0195 Since p<0.025, we reject H0. We conclude that the median score is not 5. 33 A SINGLE POPULATION PROPORTIO The proportion or the percentage of a population and the probability associated with the occurence of a particular event all involve the binomial parameter p. Recall that p was defined to be the theoretical or population probability of success on a single trial in a binomial experiment. Testing hypotheses about population proportions is carried out in much the same way as for means when the conditions necessary for using the normal curve are met. 34 The mean of x is np, thus the mean of p, p should be np/n=p Standard error of p p npq n pq n An observed value of p belongs to a sampling distribution that • is approximately normal • has a mean p p , equal to p • has a standard error z pq n , equal to pP PQ / n 35 Example Suppose we are interested in knowing what proportion of automobile drivers regularly wear sealt belts. In a survey of 300 adult drivers, 123 said they regularly wear seat belts. Can we conclude that from these data that in the sample population the proportion who regularly wear belts is not 0.50? H0: p=0.50 p=123/300=0.41 HA: p0.50 z pP PQ / n 0.41 0.50 (0.50)(0.50) / 300 3.11 Critical z values are -3.11<-1.96 Reject H0. 1.96 We conclude that in the population the proportion who regularly wear belts is not 0.50. 36 ONE SAMPLE CHI SQUARE TEST There are many problems for which the information is categorized and the results are shown by the way of counts. Suppose that we have a number of celss into which n observations have been sorted. The observed frequencies in each cell are denoted by O1, O2, O3, ..., Ok. Note that the sum of all observed frequencies is equal to O1+ O2+ O3+ ...+ Ok=n What would like to do is to compare the observed frequencies with some expected or theoretical frequencies, denoted by E1, E2, E3, ..., Ek, for each of these cells. Again, the sum of these expected frequencies must be exactly E1+ E2+ E3+ ...+ Ek=n37 We will decide whether the observed frequencies seem to agree or seem to disagree with the expected frequencies. This will be accomplished by a hypothesis test using the chi-square distribution, 2. The calculated value of the test statistic will be 2 ( O E ) i 2 i Ei i 1 k 2 ( df k 1,a ) reject H 0 2 38 Chi-square table a df 0.05 0.01 0.001 1 3,841 6,635 10,827 2 5,991 9,210 13,815 3 7,815 11,340 16,268 ... ... ... 43,770 50,890 59,703 30 39 wearing sealt belts Observed Expected (O-E) (O-E)2 (O-E)2/E Yes 123 150 -27 729 4.86 No 177 150 27 729 4.86 Total 300 300 0 9.72 H0: The proportion of drivers wearing sealt belts is equal to 0.50. HA: The proportion of drivers wearing sealt belts is not equal to 0.50. 2 ( O E ) i 2 i 9.72 Ei i 1 2 2 (1, 0.05) 3.84 reject H 0 2 40