Hypothesis significance Testing Using the Binomial Distribution Hypothesis

Hypothesis (significance) Testing Using the Binomial Distribution

Hypothesis the idea: We use random samples to estimate population parameters – If we assume that our random sample comes from a particular background distribution we can test to see if the result is likely or unlikely, given that distribution If the test result is very unlikely we may decide that the data does not come from our background Distribution If the test result is likely we will decide that the data comes from our background distribution We determine how likely through our significance level

We can answer questions like this - Is a dice fair ? (comparing our sample result against the expected result) -Does a new drug give better results than an old drug ? (by comparing our sample result against expected result) -Can you taste the difference between Coke and Pepsi? (by comparing our sample result against expected result) IN all the questions this week we will assume a population from the Binomial Distribution and we will compare our OBSERVED data VS EXPECTED Data to draw a conclusion

The binomial Distribution — 2 possible outcomes —Probability of Success Failure does not change during Trials —There a fixed number of trials —Trials are Independent Plan of action We will learn through a question You should have understood how to use Binomial tables yesterday if nothing else!!

Question A six sided dice is thrown 20 times and the number of sixes recorded. Only 2 sixes come up. Is the dice loaded? –The number of sixes is our Discrete Random Variable (X) –It follows a Binomial because You either get a six or you do not –The probability of getting a six does not change for each dice roll X~B(20, p) we can assume that for a proper dice p = 1/6 Now we set up a test to see if our sample fits a X~B(30, 1/6) model

We always write the process formally!! Step 1 – state Null Hypothesis Ho : P(6)=1/6 (probability of getting a six is unbiased~ B(20, 1/6) Step 2 - state Alternative Hypothesis H 1 : P(6)<1/6 (probability of getting six is less than 1/6 – biased) Step 3 - Decide on a significance level to test at Given at 5% usually Step 4 – From Tables ->Find the CRITICAL VALUE of P(X≤Xc) that CUTS OFF ≤ 5% This is find P(X≤xc) ≤ 0. 05 i. e. The lower tail of the distribution - The Critical Value is Xc Step 5 – Is our Observed value < The critical value If so Accept H 1, otherwise Accept Ho

This is X-B(20, 1/6) If Ho is true E(X) = np = 20 x 1/6 = 3. 33 We would expect to get around 3 sixes r P(X = r) r From Tables Xc = 0 P(X≤ 0) = 0. 0241 P(X ≤ 1) = 0. 1227 These are the outcomes we observed 2 P(X = r) P(X <= r) P(X >= r) 0 0. 0241 1. 0000 1 0. 0986 0. 1227 0. 9759 2 0. 1919 0. 3146 0. 8773 3 0. 2358 0. 5504 0. 6854 4 0. 2053 0. 7557 0. 4496 5 0. 1345 0. 8902 0. 2443 6 0. 0689 0. 9591 0. 1098 7 0. 0282 0. 9873 0. 0409 8 0. 0094 0. 9967 0. 0127 9 0. 0026 0. 9993 0. 0033 10 0. 0006 0. 9999 0. 0007 11 0. 0001 1. 0000 0. 0001 12 0. 0000 1. 0000 0. 0000 13 0. 0000 1. 0000 0. 0000 14 0. 0000 1. 0000 0. 0000 15 0. 0000 1. 0000 0. 0000 16 0. 0000 1. 0000 0. 0000 17 0. 0000 1. 0000 0. 0000 18 0. 0000 1. 0000 0. 0000 19 0. 0000 1. 0000 0. 0000 20 0. 0000 1. 0000 0. 0000 Choose nearest result to 0. 05 and ≤ 0. 05

So. . . Conclusion The observed value is greater than the critical value so do not reject Ho (The observed value is not significantly different from the expected value) P(X≤ 1) is >5% so getting a result of 1 six is not unusual enough to reject Ho WE SAY THERE IS NOT SUFFICIENT EVIDENCE TO SAY THE DICE IS BIASED This is the expected value (μ) given X~B(20, 1/6) 0 1 2 3 4 5 6 The Observed value Xc the critical value 7 8 9 10 11 12 13 14 15 16 17 18 19 20 One tailed test looking at the lower end of the tail

Example II A drug for curing headaches has been found to be 75% effective from research A new drug has been tested on 20 people and 18 have reported that their headaches have been cured. The drug company state in their TV advert that their new drug is better than the old one. Are they justified in saying this? The drugs could be said to fit a Binomial X~B(20, p) For the old drug P=0. 75, For the new drug the claim is P>0. 75 Step 1 – State the NULL Hypothesis Ho: P=0. 75 (Nothing has changed new drug same as old) Step 2 – State the Alternative Hypothesis Ho: P>0. 75 (New drug has improved cure rate)

This is another one tailed test But it looks at the other end of the distribution Ho: P=0. 75 (Nothing has changed new drug same as old) Ho: P>0. 75 (New drug has improved cure rate) Step 3 – State the significance level (given as 5% here = 0. 05) Step 4 - From tables find the critical value (cuts off 5% of the tail of the distribution) That is P(X≥Xc) < 0. 05 We rearrange this because our tables only give P(X≤r) P(X≥Xc)= 1. 0 - P(X≤Xc-1) so 1. 0 - P(X≤Xc-1) < 0. 05 Therefore P(X ≤Xc-1) > 0. 95 From Tables P(X ≤ 17) = 0. 9087 P(X ≤ 18) = 0. 9757 ( This one) So Xc-1 = 18 Xc = 19 (This is our Critical Value) r P(X = r) P(X <= r) P(X >= r) 0 0. 0000 1. 0000 2 0. 0000 1. 0000 3 0. 0000 1. 0000 4 0. 0000 1. 0000 5 0. 0000 1. 0000 6 0. 0000 1. 0000 7 0. 0002 1. 0000 8 0. 0009 0. 9998 9 0. 0030 0. 0039 0. 9991 10 0. 0099 0. 0139 0. 9961 11 0. 0271 0. 0409 0. 9861 12 0. 0609 0. 1018 0. 9591 13 0. 1124 0. 2142 0. 8982 14 0. 1686 0. 3828 0. 7858 15 0. 2023 0. 5852 0. 6172 16 0. 1897 0. 7748 0. 4148 17 0. 1339 0. 9087 0. 2252 18 0. 0669 0. 9757 0. 0913 19 0. 0211 0. 9968 0. 0243 20 0. 0032 1. 0000 0. 0032

Xc = 19 P(X = r) Critical Value Observed value E(X) | X~B(20, . 75) Expected value given Ho r Conclusion : So we Have No evidence to Reject Ho There is not enough evidence to say that the new drug is better 0 1 2 3 4 5 6 7 8 ≤ 1% 9 10 11 12 13 14 15 16 17 18 19 20

Binomial Hypothesis testing One tailed tests These are one tailed tests because H 1 is P<p 1 or P> p 1 Write your answers formally. . You MUST 1 - Write down the Null Hypothesis and Alternative Hypothesis 2 - You should state the underlying Distribution on which Ho is based 3 - Ho always represents No change ; H 1 always Change 4 - The question will give you the Significance level otherwise assume 5% 5 – Calculate the Critical Value either P(X ≤ Xc) < 0. 05 for left tail OR P(X ≥ Xc) < 0. 05 for Right tail 6 - If the Observed value is Not in the Critical region Accept Ho 7 – Write your conclusions clearly in English

Your Turn Question Mrs De Silva is running for President. She claims to have 60% of the population voting for her. She is suspected of overestimating her level of support. You want to test this, so you take a random sample of 20 people and find that only 9 will vote for Mrs De Silva. Test at the 5% level that she has overestimated her support.