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Hypothesis Testing. In 2007, 1.3 million Canadians (4.8% of Canadians – 4.2% of girls and women and 5.3% of boys and men 12 years of age and older) reported having heart disease. From HeartAndStroke.com. A new (fictional) drug has been developed called 'Healthy Heart'
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Hypothesis Testing • In 2007, 1.3 million Canadians (4.8% of Canadians – 4.2% of girls and women and 5.3% of boys and men 12 years of age and older) reported having heart disease. • From HeartAndStroke.com
A new (fictional) drug has been developed called 'Healthy Heart' • A study has shown that a random sample of 140 Canadians were given the drug for many years and at the end of that period only 5 of the recipients had heart disease. • 5/140=3.57% • This is just over a 25% reduction.
Legitimate Conclusion? • It appears that this drug reduces the risk of developing heart disease by over 25%. • How confident can the company be in presenting their data as proof that they can help prevent heart disease?
The Null Hypothesis • The null hypothesis Ho is that the drug has no affect. • The alternate hypothesis h1 is that the drug reduces heart disease. • How can we check to see if Ho can be refuted?
The Assumption • Lets assume that Ho is true. If it is then we got our test results by chance. • If we can show that getting such a low rate of heart disease by chance is unlikely, we have shown that the alternate hypothesis (H1), that the drug works, is likely.
How to Pose the Question • If the probability of a drug recipient having heart disease is 4.8%, what is the probability that out of 140 randomly selected people less than 6 have heart disease? • This is a binomial experiment. • Each member of the sample either has heart disease or not. • Under our assumption p=4.8%, n=140 and we want P(X<6)
1% • Because a claim about a drug is very important, we don’t want to make a claim if there is a 1 percent chance that we are wrong. • In a study of less importance we may accept a 5 or 10 percent chance of being wrong.
P(X<6) • µ =140(.048) =6.72 • σ=sqrt(np(1-p)) =sqrt(140(.048)(.952)) = 2.53 Z=(5.5-6.72)/2.53= - 0.48 P (x<6) = 31.56%
Stop the Press • Since P (x<6) = 31.56% the probability that the results of our study is just coincidental is too high to make a medical claim. • This is far above the 1% criteria that we have established.
Re-CapTo test a Hypothesis: • Assume the null hypothesis is true • Determine how strict your criteria is • Calculate the probability of getting your observed results if Ho is true. • Compare the calculated probability to your criteria. If it is too high, accept the null hypothesis. • If the probability of coincidence is lower than your criteria reject the null hypothesis.
Terms • The criteria that you establish is called the significance level α. • The probability that Ho is untrue will be 1- α if your test lets you reject the null hypothesis. • We call 1- α the confidence level.
Testing a Sample Mean • In the case of a mean calculated from a sample, the means of samples are normal distributed around the true mean µ with a standard deviation σ/√n. • If a sample mean x is to be tested against an assumed mean µ, assume Ho (the true mean is µ) and find the probability of getting the sample mean x from your sample. Use σ/√n as the standard deviation.
Repeated Sample Example: • Bearings should have a mean diameter of 39 mm with a standard deviation of 3 mm when a machine works right. • A sample of 50 bearings has a mean of 44m. • With significance level α=10%, should the machine be fixed?
Test Ho • If the true mean u = 39 mm, what is the probability that the sample mean is greater than or equal to 44 mm? • P(x>44) =1-P(x<44) • P(x<44): Z=(44-39)/(3/√50)=11.8 • P(Z<11.8) = near 100% P(x>44)=1-near 100%=near 0% 0<10%, Accept H1, the machine must be fixed.
Practice • Page 457 3 to 7