Hypothesis Testing - An Example Hypothesis Testing situation Sid says that he has psychic powers and can read people’s thoughts. To test this claim, a volunteer from the audience sits on the stage while Sid sits in a separate room off stage. The volunteer chooses a card from a well-shuffled pack and concentrates on the card for five seconds. At the same time, Sid writes down the suit of the card, either hearts, diamonds, spades or clubs. The card is replaced in the pack, the pack is shuffled and another card is drawn. The procedure is repeated until 20 cards have been drawn. 03/09/2015

Continued There are four suits, so Sid has a one in four chance of writing down the correct suit if he guesses the answer. If he isn’t guessing, you would expect him to get more than one in four correct. So if he gets five (or fewer) correct answers out of 20, you would definitely say that he is just guessing but if he gets as many as 19 or 20 correct you would have no hesitation in saying that he could read people’s thoughts. 03/09/2015

Null & Alternative Hypotheses Whenever we do a Statistical test we put forward a hypothesis. Null Hypothesis H 0 – the initial assumption that we want to test, shown mathematically. So for our example: We’re assuming that Sid is not psychic. Therefore the proportion of suits we expect him to get right is ¼ or So we say H 0 : p = /09/2015

Null & Alternative Hypotheses We can show that our assumption is right or wrong using an alternative Hypothesis H 1 So for our example: The alternative Hypothesis is that Sid is not guessing and therefore he should get more than 1 in 4 correct. Therefore: H 1 : p > /09/2015

One- & Two-Tailed Tests If in our Alternative Hypothesis (H 1 ) we require the value to be greater than or less than the Null Hypothesis (H 0 ) then we are doing a one-tailed test (i.e. the upper tail or the lower tail of the distribution respectively). If we require the H 1 value to be not equal to the H 0 value then we are doing a two-tailed test. 03/09/2015

Example If the null hypothesis stated that the proportion of boys in school is 0.5 what would be a suitable alternative hypothesis? Would this be a one or two tailed test? 03/09/2015

The p-value Once H 0 and H 1 have been set up we need to calculate the test statistic, T, upon which we decide whether to accept or reject the null hypothesis. The p-value tells us how likely or unlikely it is that the test statistic is true. 03/09/2015

Given the observed value t of the test statistic (T) the p-value is the probability that T will have a value at least as extreme as t if H 0 is true. 03/09/2015

Evidence to accept/reject p-value > 0.05Not sufficient evidence for rejecting H < p-value < 0.05Strong evidence for rejecting H < p-value < 0.01Very strong evidence for rejecting H 0 P-value < 0.001Overwhelming evidence 03/09/2015

Testing the mean of a distribution Example 1 The breaking strength of nylon cord (kg) is distributed: 25 modified cords are produced and X = 10.4kg Test the Hypothesis that modifications will not change the mean breaking strength. 03/09/2015

Example 2 In 1985 the heights of policemen were distributed: In 1995 the question asked was ‘Is the average height of a policeman now different to 1985?’ A sample of 60 were measured and the mean was found to be 182.5cm. Answer the question posed. 03/09/2015

Testing a probability / proportion (Binomial) 1. A manufacturer of seeding compost claims seeds sown in his compost have a higher germination rate than ordinary compost. Seeds in the ordinary compost have a germination rate of 70%. To test the manufacturer’s claim a sample of 20 seeds was planted in the new compost. 18 of these seeds germinated. Test the manufacturer’s claim.

Testing a probability / proportion (Binomial) 2. A supermarket claimed that 25% of customers pay by credit card. Test the claim in the following cases: a. In a 50 customer sample, 9 paid by credit card. b. In a 160 customer sample, 28 paid by credit card.

Testing the mean of a Poisson Distribution 1. The number of vehicles requiring assistance per day on a motorway has a Poisson distribution with mean 0.5. It was suggested that the mean will be greater on wet days. Test the suggestion in the following: a. On 10 wet days a total of 9 required assistance. b. On 55 wet days a total of 40 required assistance.