S2.4b Hypothesis tests continued

1、 boardworks ltd 20061 of 65 these icons indicate that teachers notes or useful web addresses are available in the notes page. this icon indicates the slide contains activities created in flash. these activities are not editable. for more detailed instructions, see the getting started presentation. b

2、oardworks ltd 2006 1 of 65 a2-level maths: statistics 2 for ocr s2.4b hypothesis tests continued boardworks ltd 20062 of 65 hypothesis testing on population means one-sided hypothesis tests: normal mean two-sided tests: normal mean critical values and test statistics critical regions hypothesis test

3、s on a normal mean (large n) hypothesis tests on a poisson mean contents boardworks ltd 2006 2 of 65 hypothesis testing on population means boardworks ltd 20063 of 65 introduction to hypothesis testing has the installation of a new speed camera led to a reduction in the traffic speed? is a new drug

4、more effective than an existing treatment? its claimed that drinking a new drink makes you cleverer. is there evidence to support this? boardworks ltd 20064 of 65 hypothesis testing is concerned with trying to answer questions like these. hypothesis tests are crucial in many subject areas including

5、medicine, psychology, biology and geography. in s2, we usually deal with situations where we are testing the value of the population mean (specifically for a normal distribution). introduction to hypothesis testing boardworks ltd 20065 of 65 consider the following simple situation. iq (intelligence

6、quotient) scores are a measure of intelligence. the iq scores, x, of people in the uk are normally distributed with mean 100 and standard deviation 15. i.e. x n100, 15. a head teacher suspects that the students in her school could be more intelligent on average than the general population. to test h

7、er theory she randomly selects 20 pupils from her school and measures their iq. the mean iq of these 20 pupils is denoted by the random variable . if the students in her school had the same mean iq as in the general population then x a simple introductory example . 2 15 20 n 100, i.en 100, 11 25xx b

8、oardworks ltd 20066 of 65 so what values of the sample mean would be improbable if the children in the school had the same mean iq as the general population? a simple introductory example boardworks ltd 20067 of 65 so, if the pupils in the school had the same mean iq as the general population, the p

9、robability that the sample mean iq of 20 pupils would be at least 105.52 is about 0.05. this means that if the trial was repeated over and over again with different samples, and if the pupils had the same iq as the general population, the head teacher would expect to get a sample mean of 105.52 or o

10、ver just 1 time in 20 occasions. the figure of 1 in 20 (or 5%) is often taken as a cut-off point results with probabilities below this level are sometimes regarded as being unlikely to have occurred by chance. in situations where more evidence is required, cut-off values of 1% or 0.1% are typically

11、used. a simple introductory example boardworks ltd 20068 of 65 in hypothesis testing we are essentially presented with two rival hypotheses. examples might include: a formal introduction to hypothesis tests these rival hypotheses are referred to as the null and the alternative hypotheses. “the drug

12、has the same effectiveness as an existing treatment” or “the drug is more effective”. “the mean score on a s2 examination is 60%” or “the mean score is different from 60%”. “the vehicle speed along a stretch of road is the same as before” or “the vehicle speed is less than before”. boardworks ltd 20

13、069 of 65 the null hypothesis (h0) is often thought of as the cautious hypothesis it represents the usual state of affairs. the alternative hypothesis (h1) is usually the one that we suspect or hope to be true. hypothesis testing is then concerned with examining the data collected in the experiment

14、and deciding how likely or unlikely the data would have been, if the null hypothesis were true. the significance level of the test is the chosen cut-off value that divides results that might be plausibly obtained by chance, from results that are unlikely to have occurred if the null hypothesis were

15、true. a formal introduction to hypothesis tests boardworks ltd 200610 of 65 significance levels that are typically used are 10%, 5%, 1% and 0.1%. these significance levels correspond to different rigours of testing. a formal introduction to hypothesis tests note: it is important to appreciate that i

16、t is not possible to prove a hypothesis is true in statistics. hypothesis tests can only provide different degrees of evidence in support of a hypothesis. a 10% significance test can only provide weak evidence in support of a hypothesis. a 0.1% test is much more stringent and can provide very strong

17、 evidence. boardworks ltd 200611 of 65 hypothesis testing on population means one-sided hypothesis tests: normal mean two-sided tests: normal mean critical values and test statistics critical regions hypothesis tests on a normal mean (large n) hypothesis tests on a poisson mean contents boardworks l

18、td 2006 11 of 65 one-sided hypothesis tests: normal mean boardworks ltd 200612 of 65 example: the times an athlete takes to run a 100 m race can be modelled by a normal distribution with mean 11.8 seconds and standard deviation 0.3 seconds. the athlete changes his coach and wants to know whether thi

19、s has improved his performance. he records the times (in seconds) that he runs in his next 8 races: carry out a hypothesis test using a 5% significance level to decide whether or not the athlete has improved. one-sided hypothesis tests: normal mean 11.54, 11.31, 11.47, 11.69, 11.46, 11.71, 11.49, 11

20、.61 boardworks ltd 200613 of 65 solution: we begin by writing down the 2 rival hypotheses. let represent the mean time that the athlete runs the 100 m in after changing his coach. h0: = 11.8 h1: 11.8 one-sided hypothesis tests: normal mean this hypothesis represents our cautious belief, i.e. that hi

21、s performance has not improved after changing his coach. this hypothesis represents what is suspected to be true, i.e. that his performance has improved. note that the hypotheses have been written mathematically, in terms of a parameter, . boardworks ltd 200614 of 65 one-sided hypothesis tests: norm

22、al mean significance level: 5% let x denote the time that the athlete can now run the 100 m in. if the null hypothesis is true, then x n(11.8, 0.3). if denotes the random variable for the mean time in 8 races, then under the null hypothesis. x . . 2 0 3 8 n 11 8,x the mean time for his next 8 matche

23、s is . .x 11 5411 3111 6192 28 11 535 seconds 88 boardworks ltd 200615 of 65 one-sided hypothesis tests: normal mean this mean time is less than we would have expected if the null hypothesis were true. however could a mean as low as this have occurred just through chance? to decide, we assume that h

24、0 is true and calculate (.)p11 535x . . 2 0 3 n 11 8, 8 x ,zn 0 1 . . . 2 0 3 8 11 535 11 8 2 498 standardize boardworks ltd 200616 of 65 one-sided hypothesis tests: normal mean (.)(.)xz p11 535p2 498 (.)z 1 p2 498 . 1 0 9938 . 0 0062 the significance level in this test was chosen to be 5% (0.05) th

25、e probability calculated is much lower than this. we conclude: there is evidence to reject h0 at the 5% significance level. the data provides some evidence to suggest that the new coach has led to improved performance. this probability is called a p-value remember that statistics provide evidence, n

26、ot proof. boardworks ltd 200617 of 65 step 5: compare the p-value with the significance level and make conclusions can h0 be rejected? interpret in context. step 4: calculate the p-value, i.e. the probability (under h0) of obtaining results as extreme as those collected. step 3: state the distributi

27、on, assuming the null hypothesis to be true. the steps involved in carrying out a test on a normal mean using the p-value method are as follows: step 2: state the significance level if none is mentioned in the question, it is usual to choose 5%. one-sided hypothesis tests: normal mean step 1: write

28、out h0 and h1 in mathematical terms. boardworks ltd 200618 of 65 examination-style question: it is claimed that people who follow a new diet over a 6 week period will have a mean weight loss of 12 pounds. a dietician suspects that this is an exaggeration. she selects 10 people to follow the diet and

29、 finds that their mean weight loss was 11.5 pounds. assuming that the weight losses follow a normal distribution with variance 16 lb, carry out a hypothesis test at the 10% significance level. examination-style question boardworks ltd 200619 of 65 solution: let represent the mean weight loss for the

30、 diet. h0: = 12 h1: 10%, we cannot reject the null hypothesis. standardize . 1 0 65350 3465 the evidence does not suggest that the diets claims are exaggerated. boardworks ltd 200621 of 65 hypothesis testing on population means one-sided hypothesis tests: normal mean two-sided tests: normal mean cri

31、tical values and test statistics critical regions hypothesis tests on a normal mean (large n) hypothesis tests on a poisson mean contents boardworks ltd 2006 21 of 65 two-sided tests: normal mean boardworks ltd 200622 of 65 the examples considered so far can all be classified as one- sided tests we

32、have been testing for either an increase or a decrease in the value of the parameter, . sometimes we are not looking specifically for an increase or decrease in , but instead we may want to examine whether the value of has changed. in these situations we use a two-sided (or a 2-tailed) test. a two-s

33、ided hypothesis test carried out at the % significance level is in a sense two separate one-sided tests. the significance level is therefore shared between these two tests, % for each tail. two-sided tests: normal mean boardworks ltd 200623 of 65 example: a machine fills packets of rice. it dispense

34、s w kg into each packet. the value of w follows a normal distribution, w n1.5, 0.05. after the machine is serviced, the quality control manager wishes to test whether this has resulted in a change in the mean amount of rice dispensed. a random sample of 25 packets of rice is taken off the production

35、 line and weighed. it was found that . i i w 25 1 37 95 conduct a test at the 5% significance level. two-sided tests: normal mean boardworks ltd 200624 of 65 solution: let represent the mean amount of rice dispensed by the machine after servicing. the hypotheses can then be stated as follows: h0: =

36、1.5 (i.e. no change) h1: 1.5 (i.e. a change in the mean) 5% significance level, so 2.5% for each tail. let represent the r.v. for the mean mass of rice in 25 packets of rice. then under the null hypothesis, w two-sided tests: normal mean . . 2 0 05 n 1 5, n 1 5, 0 0001 25 w boardworks ltd 200625 of

37、65 two-sided tests: normal mean . . 37 95 1 518 25 w this is more than usual, so we calculate(.)p1 518w .n 1 5, 0 0001w n 0, 1z . . . 1 518 1 5 1 8 0 0001 the observed mean weight dispensed standardize boardworks ltd 200626 of 65 two-sided tests: normal mean (.)(. )p1 518p1 8wz the test is two-sided

38、, so we compare the p-value with 2.5%. as 0.0359 2.5% we cannot reject the null hypothesis. the evidence does not suggest that the mean amount of rice dispensed has changed. .0 0359 .1 0 9641 boardworks ltd 200627 of 65 hypothesis testing on population means one-sided hypothesis tests: normal mean t

39、wo-sided tests: normal mean critical values and test statistics critical regions hypothesis tests on a normal mean (large n) hypothesis tests on a poisson mean contents boardworks ltd 2006 27 of 65 critical values and test statistics boardworks ltd 200628 of 65 hypothesis tests on a normal mean can

40、be conducted without calculating the p-value. case 1: one-sided tests h0: = 0 h1: 0 suppose the test is to be conducted based on a sample of size n and with sample mean . x if the original data follows a normal distribution, x n, , then the distribution of the sample mean is critical values and test

41、 statistics , 2 nx n suppose the hypotheses to be tested are boardworks ltd 200629 of 65 the test proceeds by standardizing the sample mean: 2 n x z this quantity z is often referred to as the test statistic. critical values and test statistics rejection region acceptance region for a test at the 5%

42、 level, the critical value of z is 1.645: if z 1.645, we can reject h0. if z 1.645, we cannot reject h0. for a test at the 1% level, the critical value of z = 2.326. n x z or boardworks ltd 200630 of 65 case 2: one-sided tests: h0: = 0 h1: 1.645, we cannot reject h0. for a test at the 1% level, the

43、critical value of z = 2.326. suppose the hypotheses to be tested are n x z rejection region acceptance region boardworks ltd 200631 of 65 case 3: two-sided tests: h0: = 0 h1: 0 critical values and test statistics the test statistic is again given by for a test at the 5% level, the critical value of

44、z is 1.96: if |z| 1.96, we can reject h0. if |z| 0 one-sided h1: 37 using a 10% significance level. let t be the r.v. for the journey time to work. under h0, t n37, 122. so under h0, . 2 12 n 37, 10 t the observed value of is t . . 4039428 42 8 1010 t critical values and test statistics boardworks l

45、td 200636 of 65 the test statistic is: critical values and test statistics n x z the critical value for a one-sided significance test at the 10% level is 1.282. x the test statistic lies in the rejection region. therefore we can reject the null hypothesis. there is some evidence that nus journey tim

46、e to work has increased. . . 12 10 42 837 1 528 rejection region acceptance region boardworks ltd 200637 of 65 examination-style question: a particular species of bird lays eggs which have lengths that are normally distributed with mean 26 mm and standard deviation 2.4 mm. a biologist discovers a th

47、riving colony of birds on a remote island and he wishes to know whether they are of the same species. he collects a sample of 15 eggs from these birds and measures their length, x mm. a summary of his results is given by . 15 1 385 4 i x form suitable hypotheses and carry out a test using a 5% signi

48、ficance level. examination-style question boardworks ltd 200638 of 65 solution: the hypotheses to be tested are as follows: h0: = 26 h1: 26 5% two-sided test (2.5% for each tail). let x be the r.v. for the length of an egg. under h0, x n26, 2.42. so under h0, . 2 2 4 n 26, 15 x the observed value of

49、 is x . . 385 4 25 693 15 x examination-style question boardworks ltd 200639 of 65 the test statistic is: examination-style question n x z the critical values for a two-sided significance test at the 5% level are 1.96. x the test statistic does not lie in the rejection region. therefore we cant reje

50、ct the null hypothesis. the evidence does not suggest that the colony of birds is of a different species. . . . 2 4 15 25 69326 0 495 rejection region acceptance region rejection region boardworks ltd 200640 of 65 hypothesis testing on population means one-sided hypothesis tests: normal mean two-sid

51、ed tests: normal mean critical values and test statistics critical regions hypothesis tests on a normal mean (large n) hypothesis tests on a poisson mean contents boardworks ltd 2006 40 of 65 critical regions boardworks ltd 200641 of 65 example: a random variable x follows a normal distribution with

52、 mean and variance 36. a random sample of 9 observations is drawn from the distribution and the sample mean is found. the value of is used to carry out a test of the hypotheses = 42 against the alternative 42. find the critical region for a test using a 1% significance level. x the critical region f

53、or a hypothesis test is the range of values for which the null hypothesis could be rejected. x critical regions boardworks ltd 200642 of 65 solution: the hypotheses to be tested are as follows: h0: = 42 h1: 42 under h0, x n42, 36. 36 n 42, 9 x critical regions n x z the test statistic is: we reject

54、h0 if z 2.576 .i.e. if422 576 2 or422 576 2xx so the critical region is (to 3 s.f.) .or36 847 2xx 1% two-sided test (0.5% for each tail) 42 2 x 6 9 42x rejection region acceptance region rejection region so under h0, boardworks ltd 200643 of 65 example 2: a gardener buys a bag of fertilizer that cla

55、ims to produce larger vegetables than the fertiliser that she currently uses. she decides to try the fertilizer out on her carrots. she knows that carrots grown using her old fertilizer had lengths (in cm) that could be modelled using a n19, 3.52 distribution. she records the lengths, x cm, of a ran

56、dom sample of 20 carrots grown using the new fertiliser. critical regions x a) explain why a one-sided hypothesis test is appropriate in this situation and formulate appropriate hypotheses. b) find a critical region for if a 5% significance level is adopted. boardworks ltd 200644 of 65 solution: a)

57、a one-sided hypothesis test is appropriate because we are looking for an increase in the length of carrots (the new fertilizer claims to produce larger vegetables than her current fertilizer). h0: = 19 h1: 19 critical regions let represent the mean length of a carrot grown with the new fertiliser. t

58、he hypotheses to be tested then, are as follows: boardworks ltd 200645 of 65 b) under h0, x n19, 3.52. so under h0, 2 3.5 n 19, 20 x critical regions n x z the test statistic is: we reject h0 if z 1.645 . . 3 5 20 19 i.e. if1 645191 29 x x so the critical region is.20 29x .3 5 20 19x rejection regio

59、n acceptance region boardworks ltd 200646 of 65 hypothesis testing on population means one-sided hypothesis tests: normal mean two-sided tests: normal mean critical values and test statistics critical regions hypothesis tests on a normal mean (large n) hypothesis tests on a poisson mean contents boa

60、rdworks ltd 2006 46 of 65 hypothesis tests on a normal mean (large n) boardworks ltd 200647 of 65 recap the central limit theorem: if a random sample of n observations is taken from any distribution with mean and variance then, provided n is large, the sample mean approximately follows a normal dist