1 Outline 1. Hypothesis Tests – Introduction 2. Technical vocabulary Null Hypothesis Alternative Hypothesis α (alpha) β (beta) 3. Hypothesis Tests – Format 4. Hypothesis Tests - Examples Lecture 9

2 Hypothesis Tests - Introduction Hypothesis tests ask questions about population means (or variances). These are always questions about probability. E.g., if population mean on some dimension is  0, what is the probability that a random sample drawn from that population would have a sample mean in some range relative to  0 ? Lecture 9

3 Hypothesis Tests - Introduction If that probability turns out to be very small, we conclude that the sample was not drawn from the population with mean  0. This is useful when  0 is the mean for an untreated population and is the mean for a treated sample. In that case, we conclude that the treatment had an effect (i.e., that  T ≠  0 ) Lecture 9 X

4 00 If the population mean is really μ, we would be surprised to find in this range here L We wouldn’t be surprised to find in this range here if μ is the true mean.

5 Hypothesis Tests - Introduction If the sample mean is in the tail of the sampling distribution, we could conclude either that something unlikely happened, or that our sample was not drawn from the population with mean  0 (the treatment worked) If that “something unlikely” is sufficiently unlikely, we prefer the conclusion that the treatment worked. So, how unlikely is “sufficiently unlikely?” Lecture 9

6 Hypothesis Tests – technical vocabulary Null Hypothesis (H 0 ) Hypothesis of no effect – that the treatment didn’t work Hypothesis that the historical population mean  0 is still true. Alternative Hypothesis (H A ) Hypothesis that treatment worked – so treated population mean is  A ≠  0. Lecture 9

7 Hypothesis Tests – technical vocabulary α (Alpha) Probability of rejecting the null hypothesis when you should not α ≤.05 means that, if you repeatedly sampled from the population with mean  0, you would get a sample mean that far from  0 0. So, it’s possible for you to conclude that there was a treatment effect when in fact you just got one of those “outlier” samples. That is called a Type I Error. Lecture 9

8 00 The null hypothesis is true and we just were unlucky? Or… The null hypothesis is not true, which is why is so far from μ 0 ?

9 Hypothesis Tests – technical vocabulary Question – why settle for 5 chances in 100 of making a Type I error? Why not go for 1 chance in 100? Or 0 chances in 100? Answer – because as we move the critical value of out into the tail of the distribution around  0, we increase the probability of Type II error – not rejecting H 0 when we should. Lecture 9

10 Lecture 9 might be below the critical value L even if the treatment worked X 00 AA L α β

11 Hypothesis Tests – technical vocabulary β (Beta) is the probability of a Type II error – not rejecting H 0 when you should. In order to compute β, you must have a specific alternative hypothesis. On previous slide, this was shown as  A. In order to compute β you have to do two steps: (1) compute L; (2) compute probability that is smaller (or larger) than L given  A. Lecture 9 X

12 Lecture 9 1. Given α, compute L. 2. Knowing L, compute β. 00 AA L α β

13 Hypothesis Tests - Format H 0 :  =  0 (this is the historical population mean) H A :  <  0 H A :  ≠  0 or H A :  >  0 (One-tailed test)(Two-tailed test) Test Statistic:Z = -  -   s Lecture 9 XX

14 Hypothesis Tests - Format Rejection Region: One-tailed test:Two-tailed test: Zobt > Zα│Zobt│ > Zα/2 orZobt < -Zα Always report your decision explicitly!! (Did you reject H 0 or not reject H 0 ?) You NEVER “accept” H 0 – only fail to reject it. Lecture 9

15 Hypothesis Tests – Example 1 The average length of classical music CDs is known to be 70.6 minutes. You suspect pop music CDs are shorter. To test this hypothesis, you test a random sample of 100 pop music CDs. Shown below are some calculations on the data you collect. Σx = 6250Σx 2 = a. What should you conclude about the relative lengths of classical and pop music CDs? (α =.01) Lecture 9

16 Hypothesis Tests – Example 1 H 0 :  = 70.6 H A :  < 70.6 Test Statistic:Z = -  0 S X Rejection region:Z obt ≤ Lecture 9 X Why negative? Why S X ? Why not S? Why one-tailed?

17 Hypothesis Tests – Example 1 Z obt = 62.5 – /√100 = -2.7 Decision: Reject H 0. There is evidence that pop music CDs are shorter than classical music CDs. Lecture 9 Why 30?

18 Hypothesis Tests – Example 2 At this time of year, half of the tomatoes at Fonzie’s Fine Foods have a shelf life of 5 days or more once they arrive at the store. Fonzie is considering a new type of genetically-modified tomato, hoping these will have a longer shelf life. He plans to order a sample of 25 of these new tomatoes and will conclude that they last longer if 19 or more of them stay fresh for at least 5 days on the shelf. Lecture 9

19 Hypothesis Tests – Example 2 a. Suppose that these genetically-modified tomatoes, in fact, do not last any longer than the other tomatoes Fonzie was buying. What is the probability that he will incorrectly conclude that the new tomatoes do have a longer shelf life? Lecture 9

20 Hypothesis Tests – Example 2 P(tomato stays fresh ≥ 5 days) =.5 We want P(X ≥ 19 │ p =.50) when n = 25. That value = 1 – P(X ≤ 18│p =.50) when n = 25. From table, desired p =.993. So the probability of an incorrect conclusion is =.007 Lecture 9

21 Hypothesis Tests – Example 2 b. Suppose that Fonzie buys a case of 50 of the new genetically-modified tomatoes and tests the hypothesis that these tomatoes will remain fresh longer than 5 days using α =.05. What is the probability that he will conclude that the new tomatoes do last longer if the new tomatoes actually last for 8 days, on average, with a standard deviation of 8.75 days? Lecture 9

22 Hypothesis Tests – Example 2 This is a question about Power. The power of a test is the complement of β. β is the probability that you do not reject H 0 when you should. Power is the probability that you do reject H 0 when you should (power = 1 – β) Lecture 9

23 Lecture 9 1. Given α, compute L. 2. Knowing L, compute C. 3. Add.5 + C to get power. 5 8 L α β C

24 Hypothesis Tests – Example 2 α =.05. Therefore Z = L =  0 + Z α (S X /√n) L = (8.75/√50) = = Lecture 9

25 Lecture α β C

26 Hypothesis Tests – Example 2 Now, find area C: Z = – 8 (8.75/√50) Z = Associated p =.2823 (from Table). Lecture 9

27 Lecture 9 1. Given α, compute L. 2. Knowing L, compute C. 3. Add.5 + C to get power. 5 8 α β

28 Hypothesis Tests – Example 2 The power of this test is = Note that this is the power of the test calculated for the specific H A that  = 8. If H A specified a different value of , this procedure would produce a different value for the power of the test. Lecture 9