Presentation on theme: "Poohsticks and hypothesis tests involving the binomial distribution"— Presentation transcript:
1Poohsticks and hypothesis tests involving the binomial distribution L.O.: To carry out one-sided and two-sided hypothesis tests involving a binomial distributionBased upon an idea published in Teaching Statistics
2The House at Pooh Corner (by AA Milne) Pooh believes that he is able to predict which of two fir-cones will emerge first from under a bridge in a game of poohsticks. To see whether he is right, he devises a simple experiment in which he plays the game of poohsticks 12 times.He is able to correctly predict the winning fir-cone on 10 of these occasions.Test whether Pooh seems able to predict the winning fir cone using a 5% significance test.
3The House at Pooh Corner (by AA Milne) H0: p = 0.5H1: p > 0.55% significance testLet X be the number of correct predictions.Under H0, X ~ B(12, 0.5).P(X ≥ 10) = 1 – P(X ≤ 9) = 1 – = or 1.93%As 1.93% < 5%, we can reject H0. There is evidence that Pooh has some ability to predict the winning fir-cone.This probability is called the p-value.
4One-sided and two-sided hypothesis tests Example:A drug is claimed to have a 40% success rate for curing patients with a certain disease. This is suspected of being an exaggeration. To test this claim, 16 people were treated with one drug and it was found that 3 were cured. Carry out a test at the 10% significance level.Solution:H0: p = H1: p < 0.410% significance testLet X be the number of people cured. Under Ho, X ~ B(16, 0.4)P(X ≤ 3) = or 6.51% (from tables).Since 6.51% < 10%, we can reject H0. There is weak evidence that the claimed success rate is exaggerated.
5One-sided and two-sided hypothesis tests Compare this example to the following slightly differently worded question:-
6One-sided and two-sided hypothesis tests Example:A drug is claimed to have a 40% success rate for curing patients with a certain disease. A doctor wishes to test whether this success rate is correct.The doctor gives the drug to 16 patients and finds that 3 are cured. Carry out a hypothesis test using a 10% level.Notice that we have the same data as before, i.e. 16 patients, 3 cured. However the hypotheses are now different:H0: p = 0.4H1: p ≠ 0.4This test is now called a two-sided (or two-tailed) test. We are essentially testing two different hypotheses, i.e. whether p < 0.4 and p > We use a 5% significance level for each test.Note: we are not indicatingwhether the success rateis believed to be toohigh or too low.
7One-sided and two-sided hypothesis tests Solution:H0: p = H1: p ≠ 0.410% significance testLet X be the number of people cured. Under Ho, X ~ B(16, 0.4)P(X ≤ 3) = or 6.51% (from tables).Since 6.51% > 5%, we cannot reject H0. There is no evidence that the claimed success rate is incorrect.
8One-sided and two-sided hypothesis tests Example:A magazine claims that 35% of all adults wear glasses.Anna wishes to test whether this figure is correct. She takes a sample of 18 adults and notices that exactly 13 of them wear glasses. Test at the 5% level whether the magazine claim is accurate.Solution:H0: p = H1: p ≠ (two-sided)Significance level: 5%Let X = number of people wearing glasses in sample.
9One-sided and two-sided hypothesis tests Under Ho, X ~ B(18, 0.35).P(X ≥ 13) = 1 – P(X ≤ 12) == 1 –= or 0.14%As this test is two-sided we compare this probability with 2.5%.As 0.14% < 2.5%, we can reject H0. We have some evidence that the proportion of adults wearing glasses is not 35%.
10Use of a normal approximation Example:There are many complaints from passengers about the late running of trains on a particular route. The railway company claims that the proportion of trains that are delayed is 10%. The Railway Passengers’ Association conducts a survey of a random sample of 200 trains and finds that 27 are delayed.Test at the 5% level whether the railway company is underestimating the proportion of trains that are delayed.
11Use of a normal approximation Solution:H0: p = 0.1H1: p > (one-sided test)Significance level: 5%Let X be the number of trains in the sample that are delayed.Under H0, X ~ B(200, 0.1).But X ≈ N[200 × 0.1, 200 × 0.1 × 0.9] = N[20, 18].P(X ≥ 27) = P(X ≥ 26.5) =Continuitycorrection
12Use of a normal approximation Conclusion:Since 6.28% > 5%, we cannot reject H0. There is no evidence that the train company is understating the proportion of delayed trains.