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1、,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.,Boardworks Ltd 2006,1 of 65,A2
2、-Level Maths:Statistics 2for OCR,S2.4b Hypothesis tests continued,Hypothesis testing on population meansOne-sided hypothesis tests:normal meanTwo-sided tests:normal meanCritical values and test statisticsCritical regionsHypothesis tests on a normal mean(large n)Hypothesis tests on a Poisson mean,Con
3、tents,Boardworks Ltd 2006,2 of 65,Hypothesis testing on population means,Introduction to hypothesis testing,Has the installation of a new speed camera ledto a reduction in thetraffic speed?,Is a new drug more effective than an existing treatment?,Its claimed that drinking a new drink makes you cleve
4、rer.Is there evidenceto support this?,Hypothesis testing is concerned with trying to answer questions like these.Hypothesis tests are crucial in many subject areas including medicine,psychology,biology and geography.In S2,we usually deal with situations where we are testing the value of the populati
5、on mean(specifically for a normal distribution).,Introduction to hypothesis testing,Consider the following simple situation.IQ(Intelligence 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
6、.A head teacher suspects that the students in her school could be more intelligent on average than the general population.To test her 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 sc
7、hool had the same mean IQ as in the general population then,A simple introductory example,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,So,if the pupils in the school had the same mean
8、 IQ as the general population,the probability 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
9、get a sample mean of 105.52 or over 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 o
10、f 1%or 0.1%are typically used.,A simple introductory example,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 has
11、 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”.,The null hypothesis(H0)is of
12、ten 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 and deciding how likely or unlikely the data
13、 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 true.,A formal introduction to hypothesis tests
14、,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 it is not possible to prove a hypothesis is true in statistics.Hypothesis tests c
15、an 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 evidence.,Hypothesis testing on population meansOne-sided hypothesis tests:normal me
16、anTwo-sided tests:normal meanCritical values and test statisticsCritical regionsHypothesis tests on a normal mean(large n)Hypothesis tests on a Poisson mean,Contents,Boardworks Ltd 2006,11 of 65,One-sided hypothesis tests:normal mean,Example:The times an athlete takes to run a 100 m race can be mode
17、lled 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 this 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
18、to decide whether or not the athlete has improved.,One-sided hypothesis tests:normal mean,11.54,11.31,11.47,11.69,11.46,11.71,11.49,11.61,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:=11.8H1:11.8,O
19、ne-sided hypothesis tests:normal mean,This hypothesis represents our cautious belief,i.e.that his 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 mathemati
20、cally,in terms of a parameter,.,One-sided hypothesis tests:normal 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.,Th
21、e mean time for his next 8 matches is,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 H0 is true and calculate,Standardize,One-si
22、ded hypothesis tests:normal mean,The significance level in this test was chosen to be 5%(0.05)the 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 pe
23、rformance.,This probability is called a p-value,Remember that statistics provide evidence,not proof.,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
24、extreme as those collected.,Step 3:State the distribution,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%.,On
25、e-sided hypothesis tests:normal mean,Step 1:Write out H0 and H1 in mathematical terms.,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
26、follow the diet and 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,Solution:Let represent the mean weight loss for the diet.H0:=12H
27、1:12,Significance level:10%Let X denote the random variable for the weight loss under this diet.If the null hypothesis is true,then X N12,16.If denotes the r.v.for the mean weight loss for 10 people,then under the null hypothesis,Examination-style question,The observed mean weight loss was lb.,Exami
28、nation-style question,We calculate,Since 0.3465 10%,we cannot reject the null hypothesis.,Standardize,The evidence does not suggest that the diets claims are exaggerated.,Hypothesis testing on population meansOne-sided hypothesis tests:normal meanTwo-sided tests:normal meanCritical values and test s
29、tatisticsCritical regionsHypothesis 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,The examples considered so far can all be classified as one-sided tests we have been testing for either an increase or a decrease in
30、 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-sided hypothesis test carried out at the%significance level is in
31、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,Example:A machine fills packets of rice.It dispenses W kg into each packet.The value of W follows a normal distribution,W N1.5,0.05.After the machine is
32、 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 line and weighed.It was found that,Conduct a test at the 5%significance level.,Two-sided tests:normal mean
33、,Solution:Let represent the mean amount of rice dispensed by the machine after servicing.The hypotheses can then be stated as follows:H0:=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.
34、Then under the null hypothesis,Two-sided tests:normal mean,Two-sided tests:normal mean,This is more than usual,so we calculate,The observed mean weight dispensed,Standardize,Two-sided tests:normal mean,The test is two-sided,so we compare the p-value with 2.5%.As 0.0359 2.5%we cannot reject the null
35、hypothesis.The evidence does not suggest that the mean amount of rice dispensed has changed.,Hypothesis testing on population meansOne-sided hypothesis tests:normal meanTwo-sided tests:normal meanCritical values and test statisticsCritical regionsHypothesis tests on a normal mean(large n)Hypothesis
36、tests on a Poisson mean,Contents,Boardworks Ltd 2006,27 of 65,Critical values and test statistics,Hypothesis tests on a normal mean can 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
37、.,If the original data follows a normal distribution,X N,then the distribution of the sample mean is,Critical values and test statistics,Suppose the hypotheses to be tested are,The test proceeds by standardizing the sample mean:,This quantity z is often referred to as the test statistic.,Critical va
38、lues and test statistics,Rejection region,Acceptanceregion,For a test at the 5%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.,OR,Case 2:One-sided tests:,H0:=0 H1:0,Critical values and test st
39、atistics,The test statistic is again given by,For a test at the 5%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.,Suppose the hypotheses to be tested are,Rejection region,Acceptanceregion,Case
40、 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 z is 1.96:If|z|1.96,we can reject H0.If|z|1.96,we cannot reject H0.,For a test at the 1%level,the critical value of z=2.576.,Suppose the hypotheses
41、to be tested are,Rejection region,Acceptanceregion,Rejection region,Step 5:Compare the test statistic with the critical value and makeconclusions can H0 be rejected or not?Interpret in context.,Step 4:Calculate the value of the test statisticusing the formula:,Step 3:State the distribution assuming
42、the null hypothesis to be true.,Step 2:State the significance level if none is mentioned in the question,it is usual to choose 5%.,The steps involved in carrying out a test on a normal mean using a test statistic are as follows:,Step 1:Write out H0 and H1 in mathematical terms.,Critical values and t
43、est statistics,The critical values for various hypothesis tests are given in the following table:,Critical values and test statistics,10%,5%,1%,1.282,1.645,2.326,1.282,1.645,2.326,1.645,1.96,2.576,Example:Nu knows from past experience that her journey times to work follow a normal distribution with
44、mean 37 minutes and standard deviation 12 minutes.The council changes the phasing on several sets of traffic lights.Nu suspects that this will lengthen her journey times.She records the times,T minutes,for 10 journeys to work:40424735454941444639,Conduct a test using a 10%significance level.,Critica
45、l values and test statistics,Solution:The hypotheses to be tested are as follows:,H0:=37H1:37,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,.,The observed value of is,Critical values and test statistics,The test statistic is:,Critical valu
46、es and test statistics,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 time to work has increased.,Rejection region,Acceptanceregion,E
47、xamination-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 thriving colony of birds on a remote island and he wishes to know whether they are of the same species.He collects a
48、 sample of 15 eggs from these birds and measures their length,X mm.A summary of his results is given by,Form suitable hypotheses and carry out a test using a 5%significance level.,Examination-style question,Solution:The hypotheses to be tested are as follows:,H0:=26H1:26,5%two-sided test(2.5%for eac
49、h tail).Let X be the r.v.for the length of an egg.Under H0,X N26,2.42.So under H0,The observed value of is,Examination-style question,The test statistic is:,Examination-style question,The critical values for a two-sided significance test at the 5%level are 1.96.,x,The test statistic does not lie in
50、the rejection region.Therefore we cant reject the null hypothesis.,The evidence does not suggest that the colony of birds is of a different species.,Rejection region,Acceptanceregion,Rejection region,Hypothesis testing on population meansOne-sided hypothesis tests:normal meanTwo-sided tests:normal m
