Hypothesis (significance) Testing Using the Binomial Distribution.
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Transcript of Hypothesis (significance) Testing Using the Binomial Distribution.
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 distributionWe 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 are 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!!
QuestionA 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 rollX~B(20,p) we can assume that for a proper dice p = 1/6Now 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 HypothesisHo : P(6)=1/6 (probability of getting a six is unbiased~ B(20,1/6)
Step 2- state Alternative Hypothesis H1 : 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) ≤ 0.05i.e. The lower tail of the distribution - The Critical Value is Xc
Step 5 – Is our Observed value < The critical valueIf so Accept H1, otherwise Accept Ho
This is X-B(20,1/6)r P(X = r) P(X <= r) P(X >= r)
0 0.0241 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 0.0000These are the outcomes we observed 2
If Ho is true E(X) = np = 20x1/6 = 3.33We would expect to get around 3 sixes
From Tables Xc = 0 P(X≤0) = 0.0241 P(X ≤1) = 0.1227 Choose nearest result to 0.05 and ≤ 0.05
So............
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Xc the critical value
The Observed value
This is the expected value (μ) given X~B(20,1/6)
ConclusionThe 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
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 HypothesisHo: P>0.75 (New drug has improved cure rate)
This is another one tailed testBut 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) < 0.05Therefore 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 0.0000 1.0000
1 0.0000 0.0000 1.0000
2 0.0000 0.0000 1.0000
3 0.0000 0.0000 1.0000
4 0.0000 0.0000 1.0000
5 0.0000 0.0000 1.0000
6 0.0000 0.0000 1.0000
7 0.0002 0.0002 1.0000
8 0.0008 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
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Critical Value
≤ 1 %
Observed value
E(X) | X~B(20,.75) Expected value given Ho
Conclusion : So we Have No evidence to Reject HoThere is not enough evidence to say that the new drug is better
Binomial Hypothesis testingOne tailed tests
These are one tailed tests because H1 is P<p1 or P> p1
Write your answers formally....You MUST
1- Write down the Null Hypothesis and Alternative Hypothesis2- You should state the underlying Distribution on which Ho is based3- Ho always represents No change ; H1 always Change4- 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 TurnQuestion
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.