hypothesis-testing統(tǒng)計學(xué)假設(shè)檢驗

1、Chapter 9,INTRODUCTION TOHYPOTHESIS TESTING,Statistics for Business(ENV),1,Hypothesis Testing,9.1Null and Alternative Hypotheses and Errors in Testing9.2z Tests about a Population with known s9.3t Tests about a Po

2、pulation with unknown s,2,Hypothesis testing-1,Researchers usually collect data from a sample and then use the sample data to help answer questions about the population. Hypothesis testing is an inferential statistical

3、process that uses limited information from the sample data as to reach a general conclusion about the population.,3,A hypothesis test is a formalized procedure that follows a standard series of operations.In this way, r

4、esearchers have a standardized method for evaluating the results of their research studies.,4,Hypothesis testing-2,5,The basic experimental situation for using hypothesis testing is presented here. It is assumed that the

5、 parameter ? is known for the population before treatment. The purpose of the experimentis to determine whether or not the treatment has an effect. Is the population mean after treatment the same as or different from th

6、e mean before treatment? A sample is selected from the treated population to help answer this question.,Procedures of hypothesis-testing,6,1. First, we state a hypothesis about a population. Usually the hypothesis concer

7、ns the value of a population parameter. For example, we might hypothesize that the mean IQ for UIC students is m = 110.2. Next, we obtain a random sample from the population. For example, we might select a random sample

8、 of n = 100 UIC students.3. Finally, we compare the sample data with the hypothesis. If the data are consistent with the hypothesis, we will conclude that the hypothesis is reasonable. But if there is a big discrepancy

9、between the data and the hypothesis, we will decide that the hypothesis is wrong.,Null and Alternative Hypotheses,The null hypothesis, denoted H0, is a statement of the basic proposition being tested. It generally repres

10、ents the status quo (a statement of “no effect” or “no difference”, or a statement of equality) and is not rejected unless there is convincing sample evidence that it is false.The (scientific or) alternative hypothesis,

11、 denoted Ha (or H1) , is an alternative (to the null hypothesis) statement that will be accepted only if there is convincing sample evidence that it is true.These two hypotheses are mutually exclusive and exhaustive.,7,

12、8,Determined by the level of significance or the alpha level,9,Alpha level of .05 -- the probability of rejecting the null hypothesis when it is true is no more than 5%.,Z,10,The locations of the critical region boundari

13、es for three different levels of significance,11,Example:Alcohol appears to be involved in a variety of birth defects, including low birth weight and retarded growth. A researcher would like to investigate the effect o

14、f prenatal alcohol on birth weight. A random sample of n = 16 pregnant rats is obtained. The mother rats are given daily doses of alcohol. At birth, one pup is selected from each litter to produce a sample of n = 16 newb

15、orn rats. The average weight for the sample is 15 grams. The researcher would like to compare the sample with the general population of rats. It is known that regular newborn rats (not exposed to alcohol) have an average

16、 weight of m = 18 grams. The distribution of weights is normal with sd = 4.,12,H0 : µ=18,13,1. State the hypothesesThe null hypothesis states that exposure to alcohol has no effect on birth weight.The alternative

17、hypothesis states that alcohol exposure does affect birth weight.2. Select the Level of Significance (alpha) levelWe will use an alpha level of .05. That is, we are taking a 5% risk of committing a Type I error, or,

18、the probability of rejecting the null hypothesis when it is true is no more than 5%.3. Set the decision criteria by locating the critical region,14,Alpha level of .05 -- the probability of rejecting the null hypothesis

19、 when it is true is no more than 5%.,Z,15,4. COLLECT DATA and COMPUTE SAMPLE STATISTICSThe sample mean is then converted to a z-score, which is our test statistic.,5. Arrive at a decisionReject the null hypothesis,Hyp

20、othesis Testing,Alternative Hypothesis H1: A statement that is accepted if H0 is falseWithout “=” signSay, “? ? 2” or “? < 2”,Null Hypothesis H0: A statement about the value of a population parameter (? and ?).

21、With “=” signSay, “? = 2” or “? ? 2”,17,Step 1: State the null and alternate hypotheses,Three possibilities regarding means,H0: m = m0H1: m = m0,,H0: m m0,H0: m > m0H1: m < m0,The null hypothesis alwa

22、ys contains equality.,3 hypotheses about means,18,a constant,Step 1: State the null and alternate hypotheses,Step Two: Select a Level of Significance, ?,Measures the max probability of rejecting a true null hypothe

23、sis,H0 is actually true but you reject it (false positive).,H0 is false but you accept it (false negative).,Level of Significance, ?,Type I Error,Type II Error,19,? ? too high,Level of Significance: the maximum allowabl

24、e probability of making a type I error,,,Risk table,20,Step Two: Select a Level of Significance, ?,Step 3: Select the test statistic,A test statistic is used to determine whether the result of the research study (the di

25、fference between the sample mean and the population mean) is more than would be expected by chance alone.,We will only consider statistics Z or t, for the time being.Since our hypothesis is about the population mean.,2

26、1,Test Statistic,The term test statistic simply indicates that the sample mean is converted into a single, specific statistic that is used to test the hypotheses.The z-score statistic that is used in the hypothesis test

27、 is the first specific example of what is called a test statistic.We will introduce several other test statistics that are used in a variety of different research situations later.,22,Reject the H0 if,Computed z >

28、 Critical z,Computed z < - Critical z,Decision Rule,,H0: ? ? ?0,Computed z > Critical zOr Computed z < - Critical z,H0: ? ? ?0,H0: ? = ?0,23,Determined by level of significance,Step 4: Formulate the dec

29、ision rule.,Critical value: The dividing point between the region where H0 is rejected and the region where H0 is accepted, determined by level of significance.,,From the table, with statistic z, one tailed test and sign

30、ificance level 0.05, we found the critical value 1.65.,24,H0: ? ? ?0,Reject if z > Critical z,One-Tailed Test of Significance,,.,If H0: ? ? ?0 is true, it is very unlikely that the computed z value is so large.,,25,

31、26,H0: ? ? ?0,Computed z < - Critical z,Reject the H0 if,If H0: ? ? ?0 is true, it is very unlikely that the computed z value (from the sample mean) is so small.,,If H0: ? = ?0 is true, it is very unlikely that the

32、 computed z value is extremely large or small.,,,,Two-Tailed Tests of Significance,27,Step 5: Make a decision.,28,,Reject !,,Accept !,An insurance company is reviewing its current policy rates. When originally setting th

33、e rates they believed that the average claim amount was $1,800. They are concerned that the true mean is actually higher than this, because they could potentially lose a lot of money. They randomly select 40 claims, and

34、calculate a sample mean of $1,950. Assuming that the population standard deviation of claims is $500, and set level of significance ? = 0.05, test to see if the insurance company should be concerned.,29,Step 1: Set the

35、null and alternative hypotheses,Example One Tailed (Upper Tailed),30,Step 2: Calculate the test statistic,Example One Tailed (Upper Tailed),Step 3: Set Rejection RegionLooking at the picture below, we need to put all

36、of alpha in the right tail. Thus,R : Z > 1.96,31,Step 4: ConcludeWe can see that z=1.897 < 1.96, thus our test statistic is not in the rejection region. Therefore we fail to reject the null hypothesis. We canno

37、t conclude anything statistically significant from this test, and cannot tell the insurance company whether or not they should be concerned about their current policies.,Example One Tailed (Upper Tailed),32,Trying to en

38、courage people to stop driving to campus, the university claims that on average it takes people 30 minutes to find a parking space on campus. John does not think it takes so long to find a spot. He calculated the mean

39、time to find a parking space on campus for the last five times and found it to be 20 minutes. Assuming that the time it takes to find a parking spot is normally distributed, and that the population standard deviation = 6

40、 minutes, perform a hypothesis test with level of significance alpha ? = 0.10 to see if his claim is correct.,Example: One Tailed (Lower Tailed),33,Step 1: Set the null and alternative hypotheses,Example: One Tailed (Low

41、er Tailed),Step 2: Calculate the test statistic,Step 3: Set Rejection RegionLooking at the picture below, we need to put all of alpha in the left tail. Thus,R : Z < -1.28,34,Example: One Tailed (Lower Tailed),Step 4

42、: ConcludeWe can see that z=-3.727 < -1.28, thus our test statistic is in the rejection region. Therefore we reject the null hypothesis in favor of the alternative.We conclude that the mean is significantly less th

43、an 30, thus John has proven that the mean time to find a parking space is less than 30.,35,Example: Two Tailed,A sample of 40 sales receipts from a grocery store has mean = $137 and population standard deviation = $3

44、0.2. Use these values to test whether or not the mean in sales at the grocery store are different from $150 with level of significance alpha ? = 0.01.,Step 1: Set the null and alternative hypotheses,Step 2: Calculate the

45、 test statistic,36,Example: Two Tailed,Step 3: Set Rejection RegionLooking at the picture below, we need to put half of alpha in the left tail, and the other half of alpha in the right tail. Thus,R : Z 2.58,Step 4: C

46、oncludeWe see that Z= -2.722 < -2.58, thus our test statistic is in the rejection region. Therefore we reject the null hypothesis in favor of the alternative. We can conclude that the mean is significantly different

47、from $150, thus I have proven that the mean sales at the grocery store is not $150.,Example: credit manager,Lisa, the credit manager, wants to check if the mean monthly unpaid balance is more than $400. The level of sig

48、nificance she set is .05. A random check of 172 unpaid balances revealed the sample mean to be $407. The population standard deviation is known to be $38.,Should Lisa conclude that the population mean is greater than $

49、400, or is it reasonable to assume that the difference of $7 ($407-$400) is due to chance? (at confidence level 0.05),37,Step 1H0: µ $400,Step 2The significance level is .05.,Step 3 Since ? is known, we can find

50、 the test statistic z.,Step 4H0 is rejected if z > 1.65 (since ? = 0.05),Step 5Make a decision and interpret the results. (Next page),,,,,Example: Lisa, the credit manager,38,The p-value is .0078 for a one-tailed

51、test. (ref to informal ans.),Computed z of 2.42 > Critical z of 1.65, p of .0078 < a of .05. Reject H0.,Step 5Make a decision and interpret the results.,We can conclude that the mean unpaid balance is greater t

52、han $400.,39,Limitation of z-scores in hypothesis testing,The limitation of z-scores in hypothesis testing is that the population standard deviation ? (or variance) must be known.What if you don’t know the µ and ?

53、of the population?Answer: use the sample variability instead,40,41,Sample variance s2 = sum of squares of deviation/ (n-1) = sum of square of deviations/df= SS/df,Since you must know the sample mean before you can

54、compute sample variance, this places a restriction on sample variability such that only n-1 scores in a sample are free to vary. The value n-1 is called the degrees of freedom (or df ) for the sample variance.,42,If you

55、select all the possible samples of a particular size (n), the set of all possible t statistics will form a t distribution.,Z statistic,t statistic,,Unknown ?,Good for: (i) large sample n>30, with the underlying distr

56、ibution may or may not be Normal(ii) small sample n<30 with the underlying distribution is Normal,43,Distributions of the t statistic for different values of degrees of freedom are compared to a normal distribution

57、.,44,45,46,47,The t distribution with df = 3. Note that 5% of the distribution is located in the tails t>2.353 and t<2.353.,The label on Fries’ Catsup indicates that the bottle contains 16 ounces of catsup. A sam

58、ple of 36 bottles from last hour’s production revealed a mean weight of 16.12 ounces per bottle and a sample standard deviation of 0.5 ounces. At the 0.05 significance level, test if the process out of control? That i

59、s, can we conclude that the mean amount per bottle is different from 16 ounces?,48,Step 1 State the null and the alternative hypotheses H0: m = 16H1: m = 16,Step 3Since the sample size is large enough and the popula

60、tion s.d. is unknown, we can use the test statistic is t.,Step 2 Select the significance level. The significance level is .05.,,Step 4 State the decision rule. Reject H0 if z > 1.96 or z < -1.96 (since ? = 0.0

61、5),Step 5Make a decision and interpret the results.(Next page),,,,,49,Computed z of 1.44 a of .05, Do not reject the null hypothesis.,The p-value is .1499 for a two-tailed test.,Step 5: Make a decision and interp

62、ret the results.,We cannot conclude the mean is different from 16 ounces.,50,The test statistic is the t distribution.,Testing for a Population Mean: Unknown (Population) standard deviation , Small sample. But the under

63、lying distribution is Normal,The critical value of t is determined by its degrees of freedom which is equal to n-1.,51,The current rate for producing 5 amp fuses at a Electric Co. is 250 per hour. A new machine has been

64、 purchased and installed. According to the supplier, the production rate are normally distributed. A sample of 10 randomly selected hours from last month revealed that the mean hourly production was 256 units, with a sa

65、mple s.d. of 6 per hour.,At the 0.05 significance level, test if the new machine is faster than the old one?,52,Step 1 State the null and alternate hypotheses.H0: µ 250,Step 2 Select the level of significance

66、. It is .05.,Step 3 Since the underlying distribution is normal, s is unknown, use the t distribution.,Step 4 State the decision rule.degrees of freedom = 10 – 1 = 9 . Reject H0 if t > 1.833,,,,53,Computed t of

67、3.162 >Critical t of 1.833 p of .0058 < alpha of .05Reject Ho,The p-value is 0.0058. (obtained from t, need a software to find it.),Step 5 Make a decision and interpret the results.,The mean number of fuse

68、s produced is more than 250 per hour.,54,If the p-value is less than alpha , then reject the null hypothesis.,Amount of time UIC students spend in library from surveyMean 41.72 minutesStandard deviation 40.179 minutes