1. HYPOTHESIS TESTING

2. A hypothesis test is a procedure for deciding if a null hypothesis should be accepted or rejected in favour of an alternative hypothesis. A statistic is calculated from a test and is analysed to determine if it falls within a pre-set acceptance region. If it does the null hypothesis is accepted if not the alternative hypothesis is accepted. The null hypothesis is the statistical hypothesis that states that there are no differences between observed and expected data. In statistics a result is called statistically significant if it is unlikely to have occurred by chance. A hypothesis is an initial explanation for an observation, that can be tested by further investigation Hypothesis: Hypothesis Test:

3. Example 1: A teacher thinks that only 20% of the pupils in the school watch the news regularly. He chooses 25 pupils at random and finds that 10 of them watch the news regularly. Test at the 5% level of significance, whether or not there is evidence that the percentage of pupils who watch the news is more than 20%. State your hypotheses clearly. H 0 : p = 0.2 H 1 : p > 0.2 This is a one tailed test. We are only asked to test if the proportion is more than 20%. Under H 0, X ~ B( 25, 0.2) P(X ≥ 10) = We need to know how likely it is to have at least 10 pupils watching the news when we only expect 20% of the to watch it. (i.e. 5 of them). 1 – P(X ≤ 9) =1 – 0.9827 =0.0173 So, there is only a 1.73% chance of this happening if the mean really is 5. This is too unlikely to believe H 0. ( It is less than 5%). i.e. accept that more than 20% watch the news regularly. We reject H 0, in favour of H 1

4. Example 2: Past records from a supermarket show that 15% of people who buy a bag of frozen chips buy the low fat bag. On one particular day a random sample of 40 people was taken from those that had bought frozen chips and 2 of them were found to have bought a low fat bag. Test, at the 5% significance level, whether or not the proportion p of people who bought a low fat bag of chips that day was different from 15%. State your hypotheses clearly. This is a two tailed test. We are asked to test if the proportion is different from 15%. H 0 : p = 0.15 H 1 : p ≠ 0.15 Under H 0, X ~ B( 40, 0.15) We need to know how likely it is to have at most 2 buying low fat chips when we expect 15% of them to buy low fat chips. (i.e. 6 of them). P(X ≤ 2) = So, there is a 4.86% chance of this happening. This is not too unlikely to believe H 0. ( It is more than 2.5%). Note that for a two tailed test the 5% significance is split into 2.5% in each tail. We accept that 15% do buy the low fat chips. 0.0486 There is insufficient evidence to reject H 0. We reject H 1 in favour of H 0.

5. Example 3: A single observation x is taken from a Binomial distribution, B(25, p). This observation is used to test H 0 : p = 0.45 against H 1 : p ≠ 0.45. Using a 10% level of significance, find the critical region for this test. The probability of rejecting either tail should be as close as possible to 5%. State the actual significance of this test. Under H 0, X ~ B( 25, 0.45) From the tables: P(X ≤ 7) = P(X ≤ 15) = 0.9560 P(X ≥ 16) = = 0.0440 ( 6.39%) ( 4.40%) The actual significance of this test is ( 10.79% ). Hence the critical region is: X ≤ 7  X ≥ 16 0.0639 1 – 0.9560 0.0639 + 0.0440 = 0.1079

6. Example 4: The mean number of a certain type of chocolate bar sold by a shop each day follows a Poisson distribution with mean 8. After a promotion, the following day 15 are sold. Test at the 5% level of significance whether the promotion has improved the sales of the chocolate bar. This is a one tailed test. We are asked to test if the number of bars sold each day is more than 8. H 0 : λ = 8 H 1 : λ > 8 Under H 0, X ~ P o (8) P(X ≥ 15) =1 – P(X ≤ 14) =1 – 0.9827 =0.0173 So, there is a 1.73% chance of this happening. This is too unlikely to believe H 0. ( It is less than 5%). We accept that the promotion has increased the sales. i.e. since 0.0173 < 0.05 there is evidence to reject H 0 in favour of H 1.

7. Example 5: In an office incoming calls are thought to occur at a rate of 0.35 per minute. To test this, the number of calls during a random 20 minute interval was recorded and found to be 2. Test at the 5% level of significance, whether or not there is evidence that the rate of incoming calls is different from 0.35 per minute. In a 20 minute interval the expected number of calls is0.35 × 20 = 7 This is a two tailed test. We are asked to test if the number of calls is different from 0.35 per minute. H 0 : λ = 7 H 1 : λ ≠ 7 Under H 0, X ~ P o (7) P(X ≤ 2) = 0.0296 So, there is a 2.96% chance of this happening. This is too likely to reject H 0. ( It is more than 2.5%). We accept that the rate is 0.35 calls per minute. i.e. since 0.0296 > 0.025 there is insufficient evidence to reject H 0.

8. Summary of key points: This PowerPoint produced by R.Collins ; Updated Feb. 2014 The null hypothesis is the statistical hypothesis that states that there are no differences between observed and expected data. A one tailed test is a test on an observed value to see if it is more than (or less than), an expected value. A two tailed test is a test on an observed value to see if it is different from an expected value. We always believe that the null hypothesis is true unless there is significant evidence to suggest otherwise. The critical region is the set of observed values for which the null hypothesis is rejected.