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Unformatted text preview: Hypothesis Testing Utku Suleymanoglu UMich Utku Suleymanoglu (UMich) Hypothesis Testing 1 / 39 Introduction The Idea Statistical inference is about arriving at conclusions about unknown population parameters using data. Example: we want to know the yearly income of public university graduates in the USA at the age 30. An unknown mean µ, unknown variance, σ 2 , unknown distribution. Can’t interview all young people, so we get a sample of 100: x = 55. Let’s say we know σ 2 = 9 ¯ x is our best guess for µ. We can also build a confidence interval around it: something like ¯ x ± zα/2 σx . ¯ ¯ These are our best answers to the question “what is µ?”. Our estimate, x and σ 2 , and what we know of the sampling distribution of x , can also help us ¯ ¯ evaluate claims about µ. Suppose somebody claims that µ ≥ 60. Utku Suleymanoglu (UMich) Hypothesis Testing 2 / 39 Introduction We are going to learn how to evaluate the validity of these types of hypotheses regarding the unknown population parameters (µ, or p .) ¯ We just measured that with 100 observations that x = 55. The spirit of the testing would be ¯ evaluating the chances of this sample being drawn from a population with µ ≥ 60. Is it possible to have an x = 55 from a distribution with µ = 60? Yes, techically, but highly ¯ unlikely. We use the sampling the distribution of the sample mean under the assumption that the claim is right: Then X ∼ N (60, 9/100) = N (60, 0.09). We know x will change from sample to sample, but how likely it is that it is less than 55, or 54? ¯ Very very very . . . very unlikely. Then, this claim is not really credible, the data does not support it. This is the spirit of hypothesis testing. Let’s formalize the idea of “claims.” Utku Suleymanoglu (UMich) Hypothesis Testing 3 / 39 Null and Alternative Hypotheses The Null and Its Alternative We will devise two complementary hypotheses to formalize the idea of a claim: null hypothesis and alternative hypothesis. Null Hypothesis: Holds the claim to be challenged, to be refuted if possible. H0 : µ ≥ 60. Alternative Hypothesis: Alternative theory to be maintained if the null hypothesis is rejected: H1 (HA ) : µ < 60. As a researcher if I think that the average yearly income at 30 for university graduates (µ) is less than 60 thousand dollars a year, to back up my claim, I challange the opposite claim, that it is higher than 60. If I reject the H0 , then I get evidence for my theory. Alternatively, you might also be asked a hypothesis directly. “Test the claim that µ is less 50”. Then this claim should be put in the null hypothesis. Utku Suleymanoglu (UMich) Hypothesis Testing 4 / 39 Null and Alternative Hypotheses Examples 1. You are unhappy with your car: you think your new car travels less than 30 mpg. To get it serviced for free, you want to find evidence that MPG < 30. You record your consumption for a month and build a sample. You challange the claim that E (MPG ) = µ ≥ 30 H0 :µ ≥ 30 H1 :µ < 30 2. You are an engineer and have a machine in your factory that produces pencils. You think the machine is broken, and it produces pencils of incorrect lenghts. It supposed to produce pencils of an average length of 5 inches. You collect a sample and test: H0 :µ = 5 H 1 :µ = 5 KEY: In all hypothesis testing, you look at the evidence and decide whether you have enough evidence to reject the null hypothesis. If you don’t have it, you fail to reject the null hypothesis: you don’t accept the null hypothesis. Jury analogy. . . Utku Suleymanoglu (UMich) Hypothesis Testing 5 / 39 Null and Alternative Hypotheses Type I and II Errors The hypothesis testing we will do is not perfect: there could be mistakes. Reality Testing Result H0 is True H0 is False Reject H0 Type I Error Correct Fail to Reject H0 Correct Type II Error TYPE I Error: Null Hypothesis is true and you reject it. Probability = α. Significance level. We can choose this. TYPE II Error: Null Hypothesis is false and you fail to reject it. Probability = β . Power= 1 − β . This depends on the unknown population parameter. Have limited control over this. Utku Suleymanoglu (UMich) Hypothesis Testing 6 / 39 Null and Alternative Hypotheses Trial Analogy Reality Verdict INNOCENT (H0 ) GUILTY Reject (Verdict=Guilty) H0 Type I Error Correct Fail to Reject (Verdict= Not Guilty) H0 Correct Type II Error We set a high standard for convicting people. We assume innocence, then try to find evidence to reject this presumption. We do the same for null hypothesis as well: unless there is a lot of evidence, we do not reject it. Type I error: Innocent man gets the chair, Type II error: Murderer walks away. Society and statisticians try to minimize the probability of Type I error first, and demand a lot evidence to reject an H0 . Key thing: If we fail to reject H0 , we don’t say “we proved H0 ”, we just don’t have enough evidence against it. Analogy: if the defendant walks away, his innocence has not been proven, instead: his guilt has not been proven with enough evidence. Utku Suleymanoglu (UMich) Hypothesis Testing 7 / 39 Null and Alternative Hypotheses General Testing Procedure TEST PROCEDURE: 1 Formulate and state null and alternative hypothesis. 2 (Select a significance level: α) 3 Calculate a suitable test statistic using available sample statistics to use in conjuction with. . . 4 (Develop and) Use a decision rule to make a call about H0 . You don’t need to develop it everytime but this is how it works: (a) Assume the null hypothesis is valid. (b) Figure out the sampling distribution of the sample statistic under the assumption is null hypothesis correct. (c) Figure the distribution of the test statistic under the null. (d) Select a criteria that uses probability distribution of the test statistic to reject or fail to reject the null hypothesis. The criteria uses α as a tolerance level. 5 State your conclusion on the null hypothesis. Utku Suleymanoglu (UMich) Hypothesis Testing 8 / 39 Testing Hypothesis about the Population Mean: Case 1 σ known Case 1: σ known We start with unrealistic case where σ is known. This works exactly if population values have a normal distribution, and approximately if not. One-Tailed Tests A left-tailed test has the H0 and H1 : H0 :µ ≥ µ0 H1 :µ < µ0 A right-tailed test has the H0 and H1 : H0 :µ ≤ µ0 H 1 :µ > µ 0 Utku Suleymanoglu (UMich) Hypothesis Testing 9 / 39 Testing Hypothesis about the Population Mean: Case 1 σ known Test statistic for tests with known σ ’s will have the test statistic: z= x − µ0 ¯ √ σ/ n Now, we need to come up with a testing criteria. There are two equivalent ways of doing this: p-value approach Critical value (rejection region) approach These are best explained with an example. We will discuss the logic of hypothesis testing with this example. Important Note: We will discuss hypothesis testing regarding µ and p in different scenarios. The first scenario is for µ where σ is known. I will spend an extra amount of time on this case to explain to you the logic of hypothesis testing. This logic carries through everything we are going to do, so I will not repeat it again. Don’t mistake me spending a lot of time on the first case for other cases not being important. Utku Suleymanoglu (UMich) Hypothesis Testing 10 / 39 Testing Hypothesis about the Population Mean: Case 1 σ known Long Running Example Suppose you think the average lifespan of energy-saving light bulbs is less than 3 years. You collect a sample of 25 newly builty bulbs and measure their lifespan. You get x = 2.5. You ¯ (somehow) know standard deviation of their lifespan is σ = 1.5. Then we have the hypotheses: H0 :µ ≥ 3 H 1 :µ < 3 This is a left-tailed test. Relevant test statistic for this test (for all Case 1 cases, right or left-tailed or two-tailed) is: z= x − µ0 ¯ 2.5 − 3 = −1.66 √= σ/ n 1.5/5 We will see why we use this. Utku Suleymanoglu (UMich) Hypothesis Testing 11 / 39 Testing Hypothesis about the Population Mean: Case 1 σ known Decision rule will evaluate how likely it is to get a sample with x = 2.5 if you population ¯ mean of 3 years. ¯ We know X ∼ N (3, (1.5)2 ) 25 if µ = 3 were true. ¯ Then the question is: If so what is the probability of getting an X < 2.5? Well we can calculate that! 2.5 − 3 0.3 ¯ P (X < 2.5) = P (Z < ) = P (Z < z ) = P (Z < −1.66) z test statistic We can calculate this probability using the z-table. It is 0.0485: the probability that you get a sample that produces an x which is lower than than our current estimate x = 2.5 if the ¯ ¯ null hypothesis were true. We will call this probability p-value. This is a small probability, H0 should probably be rejected. But what is small enough? We need a criteria. We will set “a small enough probability”: significance level and denote it with α. If you calculate a p-value which is less than α, you reject the null hypotesis. α is in our control: usually 0.1, 0.01 or 0.05. Remember α is also the probability of Type I error: we might be wrong! α is the probability associated the risk we are taking. This is the essence of p-value approach Utku Suleymanoglu (UMich) Hypothesis Testing 12 / 39 Testing Hypothesis about the Population Mean: Case 1 σ known Graphical recap: 0 Utku Suleymanoglu (UMich) Hypothesis Testing z 13 / 39 Testing Hypothesis about the Population Mean: Case 1 σ known p-value Approach for One-Tailed Tests We are working on the case: tests for µ where σ is known, but this logic generalizes to many cases. 1. After hypotheses statement and the calculation of test statistic (z for this case) : 2. Calculate Left-tailed tests: Calculate (left-tail) probability that the sample mean is less than x at hand if the ¯ null is true via: P (Z ≤ z ). Right-tailed tests: Calculate (right-tail) probability that the sample mean is more than x at hand if ¯ the null is true via: P (Z ≥ z ). 3. The probability you calculate is called the p-value. 4. Decision Rule: Compare the p-value with α. If p-value < α : Reject the H0 . You have enough evidence that H0 is false. If p-value > α : Fail to reject H0 . There is not enough evidence to reject the null hypothesis. In our example, we reject the null if α is set to be 0.05 or 0.1 but not if 0.01. p-value approach allows easy comparison of decision with different α’s. Notice: p-value is the smallest α choice where H0 is rejected. Utku Suleymanoglu (UMich) Hypothesis Testing 14 / 39 Testing Hypothesis about the Population Mean: Case 1 σ known Critical Value Approach for One-Tailed Tests Another equivalently valid approach to create to criteria for testing would be this: 1. After hypotheses statement and the calculation of test statistic (z for this case). 2. Set an α. Say, α = 0.05. 3. Figure out the zα such that P (Z ≥ zα ) = α: The z-value with upper-tail probability of α. 4. Make a decision about H0 by comparing the test statistic with a critical value. Critical value tells us which values are too far off from the null hypothesis value. Left-tailed tests: Reject H0 if z < −zα . Critical value= −zα Right-tailed tests: Reject H0 if z > zα . Critical value= zα 5. Choosing an α and finding the critical value creates a rejection region. If the test statistic is in this region, H0 is rejected. The key idea: Instead of comparing probability with a small enough probability (α), choose α first and establish a far enough test statistic. Utku Suleymanoglu (UMich) Hypothesis Testing 15 / 39 Testing Hypothesis about the Population Mean: Case 1 σ known Example Cont. For our example, critical value with α = 0.05 is −1.645 = −zα . Then any test static which is smaller than than −1.645 is in the rejection region. We had z = −1.66, so we reject the the null hypothesis. 0 z Exercise: Calculate the largest x value which will lead to the rejection of the null. ¯ Utku Suleymanoglu (UMich) Hypothesis Testing 16 / 39 Testing Hypothesis about the Population Mean: Case 1 σ known Right-Tailed Test Example Suppose now that you also believe that µ > 1.5. So you want to test the claim µ ≤ 1.5. Let’s do that. First, let’s properly state the hypotheses: H0 :µ ≤ 1.5 H1 :µ > 1.5 This is a right-tailed test. Under the assumption that null hypothesis is true, we need to evaluate the chances of x > 2.5. If small, the null hypothesis is not likely to be true. ¯ Next step: relevant test statistic is z = x −µ0 ¯ √ σ/ n = 2.5−1.5 0.3 = 3.33. Next: Critical value for α = 0.01 is z0.01 = 2.33. Next: Use the decision rule: z > 2.33 so we reject H0 . OR: p-value for z = 3.33 is smaller than 0.001, so we reject the null hypothesis for reasonable α. Exercise: Calculate the smallest x value that leads to rejection of the null. ¯ Utku Suleymanoglu (UMich) Hypothesis Testing 17 / 39 What determines test results? Determinants of The Test Results Before we go on and discuss two-tailed testing and other cases, let’s talk about a few things: Difference between hypothesized parameter and calculated statistic: x − µ0 . Generally ¯ speaking, if the claim is too far off from the hypothesized value, test statistic would be larger in absolute terms. The effect of this on test depend on the sign of the test statistic and the tail of the test. Here is an example: Remember we have x = 2.5 and σ = 1.5 and n = 25. Suppose now that you ¯ have to test H0 : µ ≥ 1 with the alternative hypothesis H1 : µ < 1. We can calculate the test statistic: z = 2.5−1 0.3 = 5. That is a pretty big z . But this is a left-tailed test. The p-value is calculated as the left tail probability. P (Z ≤ 5). It is almost 1, it is bigger than any α imaginable. So you fail to reject the null. Rejection region is (∞, −1.645) (if α = 0.1). Critical value is negative for left-tailed tests, so a positive test statistic cannot be in the rejection region. Even if the test results seems obvious, we still test it properly. And this is an example why a higher test statistic does not necessarily mean right away that you are more likely to reject the null. Generally, a large (in absolute value) test statistic in the direction of the tail of the test is more likely to reject the null hypothesis. Utku Suleymanoglu (UMich) Hypothesis Testing 18 / 39 What determines test results? Why Left/Right Tail? Why do we focus on the left-tail if the null hypothesis is µ ≥ µ0 ?. When testing the null H0 : µ ≥ 1: if x values that are larger than 1 are surely supporting the null. ¯ We saw this in the previous example. Got x = 100? It is fine, because the null is µ ≥ 1, they are in agreement, so we should not reject ¯ the null. We focus on the left-tail, because values of x less than 1 is still possible. Notice when ¯ x < 1 = µ0 ⇒ z < 0. As x gets smaller, they also get more and more imporabable. ¯ ¯ If µ = 1, x = 0.9 is possible, so is x = 0.1. As we move to lower x ’s, the probability shrinks. We ¯ ¯ ¯ can calculate this probability, and we set a criteria for a low probability. This is a criteria for: x is too small to have come from a µ which is 1 or larger. ¯ Notice: If an x value is too small if µ = 1, it will be even less improbable if µ were larger. ¯ This is why we pay attention to the left tail. Exercise: For H0 : µ ≥ 1, and SE (¯ ) = 0.3, α = 0.05, find the largest x where we reject H0 . x ¯ Exercise: To make sure you understand this, do the right-tailed version yourself. Utku Suleymanoglu (UMich) Hypothesis Testing 19 / 39 What determines test results? Precision of the sampling distribution: Standard error of the sample statistic determines how much variation we should expect in x from sample to sample. If it is high, it is more likely to have ¯ samples with x that deviates largely from the hypothesized value. ¯ Consider the formula for test statistic: z= x − µ0 ¯ σ √ n = x − µ0 ¯ SE (¯ ) x Remember from Chapter 7: as the standard error of the mean (denominator above) decreases, we say we measure x more precisely. The same x − µ0 difference might be too little and too large, ¯ ¯ depending on how precisely we measure x . ¯ As SE decreases (maybe we have a bigger sample), test statistic will increase magnitude. The effect of this is generally an increase in the chances of a null being rejected. Here is an example: Suppose we have a left-tailed test. Test statistic is z = −1, so the null is not rejected with α = 0.05. If, for some reason, SE(x ) were half of what it was before, the new would have been ¯ z = −2 and the null would be rejected. Intuition: when we have more precision, we need much less discrepancy between x and µ0 to ¯ reject the null. Notice: With a left-tailed test, if z > 0, a decrease in the SE cannot change the result. If z > 0, a left-tailed test will always result in “fail to reject”, regardless of the size of z . Confirm this for exercise. Think about the right-tailed as well. Utku Suleymanoglu (UMich) Hypothesis Testing 20 / 39 What determines test results? Signifance level: α is our choice as researchers. Think about the p-value approach. You compare your calculated p-value with different α’s. If p = 0.04, you reject the null with α = 0.05, but not if α = 0.01. To reject a null with α = 0.01 or α = 0.001, you need a really small p-value. So as α decreases, you ask for more and more evidence against the null hypothesis to be able to reject it. As α decreases, rejection region gets smaller. A small α choice means you have a small probability of rejecting a true hypothesis (Type I error, executing the innocent). But a small α is also asking a lot of evidence and not rejecting H0 most of the time. So maybe you are also not rejecting some false hypotheses: probability of committing Type II error increases. So there is a trade-off between Type I and Type II error probabilities. This is why we don’t set α = 0.000001. Generally speaking an α = 0.05 is norm. If one needs to be more conservative about rejecting the null (asking for a little bit more evidence), α = 0.01 is the choice. An α = 0.1 is also ok. There is no clear-cut reason to choose one over the others. But we usually don’t use α = 0.2. And nevery a high α like α = 0.8. The nice thing about providing p-values is that you allow the readers to pick their own α’s and arrive at their own conclusions quickly. Utku Suleymanoglu (UMich) Hypothesis Testing 21 / 39 What determines test results? σ is known: Two Tailed Tests Now we will discuss a slightly different type of test. The difference is in the null and alternative hypothesis tests: H0 :µ = µ0 H1 :µ = µ0 These type of tests judge the claim that unknown population parameter is exactly equal to some number. In economics, two-tailed tests are performed a lot to test things like: Whether a production technology have constant returns to scale. (population parameter=1) Whether a job training program has any effect on wages whatsoever. (population parameter=0) We will come back to the latter one again when we do regression analysis. Utku Suleymanoglu (UMich) Hypothesis Testing 22 / 39 What determines test results? Example Test procedure is very similar to the one-tailed tests with a few but important differences. Suppose with your lightbulb sample (remember x = 2.5, n = 25 and σ = 1.5.) Now suppose that ¯ there is a claim that says the mean life expectancy of lightbulbs is 2.6 years: H0 :µ = 2.6 H1 :µ = 2.6 The test statistic is going to be identical with one-tailed tests: z= 2.5 − 2.6 x − µ0 ¯ = −0.33 √= σ/ n 0.3 The test statistic calculates the relative position of 2.5 with respect to hypothesized value for µ: 2.6. You can see it is fairly close as measured by z = −0.33. Given that normal distribution is ¯ bell-shaped, we know x = 2.5 draw from the distribution of X is quite probable if µ = 2.6, so we ¯ should not reject the H0 . Key thing: Because of the equality in the null, what we consider unlikely (under the assumption that the null hypothesis is true) can be on either tail. We will build our rejection regions on both tails. Utku Suleymanoglu (UMich) Hypothesis Testing 23 / 39 What determines test results? This time lets start with the Critical Value Approach: After stating the hypotheses and calculating the z-statistic, the decision rule is going to be: x −µ ¯ Reject H0 if | σ/√0 | > zα/2 n In other words, reject the null if the test statistic is outside the interval (−zα/2 , zα/2 ) where critical value zα/2 is the z-value for upper tail probability α/2. For our example, we have z = 2.5−2.6 √ 1/ 25 = 0.1/0.3 = −0.33. Let’s pick α = 0.05, then α/2 = 0.025 and zα/2 = 1.96. Because z = 0.33 lies inside the interval (−1.96, 1.96), we do not reject the null hypothesis. We don’t have enough evidence to assert that µ = 2.6 is not the case. Let’s see what we are doing on a picture. Utku Suleymanoglu (UMich) Hypothesis Testing 24 / 39 What determines test results? Graphical Explanation 0 Utku Suleymanoglu (UMich) Hypothesis Testing z 25 / 39 What determines test results? p-value Approach for Two-Tailed Tests We can also calculate a p-value for the test statistic z , to compare with different α’s to get a conclusion. The p-value can be calculated as the area outside the interval (−z , z ), or simply (due to symmetricity) as 2 × P (Z > z = 0.33) In our example we get a p-value 2 × P (Z > z = 0.33) = 2 × 0.3707 = 0.7414. This value is bigger than any reasonable α so reach at the same conclusion as before. We fail to reject the null hypothesis. Let’s go back one slide and see this on a picture. Utku Suleymanoglu (UMich) Hypothesis Testing 26 / 39 What determines test results? Hypothesis Testing Fundamentals Recap Before we go on to different cases, let’s repeat the general idea of hypothesis testing: We have an hypothetical value for a population parameter (µ = µ0 ) as a claim and we want to test this. We have a sample and a point estimate x = 2, let’s say. ¯ We know the sampling distribution of x assuming the claim is true from chapter 6. ¯ Then we can evaluate the probability of x or a similar draw from this sampling distribution. ¯ To do that we need to transform our normal random variable to standard normal, this gives us z-statistic. Then we can either calculate the probability associated with the z-statistic and see if it is small or big (p-value approach) compare it to some critical z-value so that we can assess how far off it is from the claimed value. (critical value approach) Either way, based on the assumption the claim is true, we assess the correctness of the claim by comparing it to what we observe in the data. If two are “different enough”, we say the claim is (probably) not correct. Notice ALL these boils down to a simple step-by-step procedure, so that when you need to do a test, you don’t have to think about and explain what you are doing, but just follow the procedure and get a result. But it is good to enough the reasoning and the mathematics behind the procedure. Utku Suleymanoglu (UMich) Hypothesis Testing 27 / 39 σ not Known Case 2: σ not known, population normal When σ is not known, we can use s , sample standard deviation, instead. Just like we did before. . . for CI’s. But remember, we need a modification to make this work. We need to use t-distribution instead of standard normal distribution. p-value approach is hard to perform with t-distribution without the help of a computer, so we will just use the critical value approach in the notes and exam. In the section and problem sets, you might use p-value approach. Let’s do a one tailed example first . . . Utku Suleymanoglu (UMich) Hypothesis Testing 28 / 39 σ not Known One-Tailed t-Tests For one tailed tests involving hypotheses: Left tailed: H0 :µ ≥ µ0 H1 :µ < µ0 Right tailed: H0 :µ ≤ µ0 H1 :µ > µ0 We reject the null the hypothesis if test statistic: t= x − µ0 ¯ √ s/ n is such that t < −tα,n−1 for left-tailed tests t > tα,n−1 for right-tailed tests where tα is the t value with probability α in the upper tail. You look this up via the t-table. Utku Suleymanoglu (UMich) Hypothesis Testing 29 / 39 σ not Known Example Suppose you are interested in the labor supply of elderly. You have a data set that consists of 90 people aged 65. They are measured to have a weekly labor supply (hours worked) of 30 hours. The standard deviation is calculated to be s = 10. You believe that in the population mean hours worked per week is less than 32 hours/week. Perform a test to support your idea. Step 1: The null and alternative hypotheses are: H0 :µ ≥ 32 H1 :µ < 32 Utku Suleymanoglu (UMich) Hypothesis Testing 30 / 39 σ not Known Step 2: Calculate the relevant test statistic: t= −2 x − µ0 ¯ 30 − 32 = = −1.897 √ √= s/ n 1.054 10/ 90 Step 3: Find the critical value: Let’s pick α = 0.05. Then −tα,n−1 = t0.05,89 ≈ −1.662 Step 4: Use the decision rule for left-tailed: Reject H0 if t < −tα,n−1 . t = −1.897 < −1.662, so reject the null. Finally, state your conclusion: We have enough evidence to reject the claim that µ ≥ 32. So our claim that it is less than 32 is supported by the data. Notice that with a lower α choice, the result could have been different. Utku Suleymanoglu (UMich) Hypothesis Testing 31 / 39 σ not Known Two-Tailed Tests: σ Unknown We will again use t-distribution instead of standard normal. First state the null and alternative hypotheses. H0 :µ = µ0 H1 :µ = µ0 We reject the null the hypothesis if test statistic (same as one-tailed tests): t= x − µ0 ¯ √ s/ n is such that |t | > tα/2,n−1 where tα/2 is the critical value with probability α/2 in the upper tail. So if test statistic falls outside the interval (−tα/2 , tα/2 ), the null is rejected. Utku Suleymanoglu (UMich) Hypothesis Testing 32 / 39 σ not Known Example Now let’s say the data set has 900 people aged 65. They are measured to have a weekly labor supply (hours worked) of 30 hours. The standard deviation is calculated to be s = 10. Now, somebody claims that mean hours worked is exactly 29.5. Test this claim. Step 1: The null and alternative hypotheses are: H0 :µ = 29.5 H1 :µ = 29.5 Utku Suleymanoglu (UMich) Hypothesis Testing 33 / 39 σ not Known Step 2: Calculate the relevant test statistic: t= 30 − 29.5 0.5 x − µ0 ¯ = = 1.515 √= s/ n 10/30 0.33 Step 3: Find the critical value: Let’s pick α = 0.05. Then tα/2,n−1 = t0.025,899 ≈ 1.96 (Table 2 in Appendix B) (n is too large, use infinity as df, note that this is the same critical value as z). Step 4: Use the decision rule: Reject H0 if |t | > tα/2,n−1 . t = 1.515 is outside the rejection region (−1.96, 1.96). We failed to reject the null. As a final step, state your conclusion: We do NOT have enough evidence to reject the claim that µ = 29.5. Big picture note: this does not mean that we know that µ = 29.5. We would not have rejected the null with 29.6 or 30 either! Utku Suleymanoglu (UMich) Hypothesis Testing 34 / 39 σ not Known MIDTERM TOPICS END HERE! This means the hypothesis tests that you will need to perform in the second midterm II will have normal distribution as the distribution of the population values. The remainder of the chapter will be included in the last midterm. Utku Suleymanoglu (UMich) Hypothesis Testing 35 / 39 σ not Known Case 3: σ not Known, Population Not Normal The t-test relies on the assumption that population values are distributed normally. If not, we can still test hypothesis thanks to CLT, but these are going to be approximations which are close to true values in large samples. To make the procudure work: you switch back to standard normal distribution even though σ is not known. ¯ −µ0 So the test statistic is zt = x /√n , and z-test rules apply. I gave this test statistic a slightly s different name so that you don’t confuse it with the case 1 statistic. PLEASE BE CAREFUL: We have the same formula for test statistic for cases 2 and 3. The difference is Case 2: Population values are normally distributed, so test statistic has t-distribution. Hence called “t”. And you use the t-table. Case 3: Population values are from unknown distribution, so test statistic is approximately normal. So it is called “z” and z-table is the relevant one. If one knows that population has normal distribution, “t-test” must be performed (especially with small n). This yields more accurate test results. Utku Suleymanoglu (UMich) Hypothesis Testing 36 / 39 Hypothesis about Population Proportions Hypothesis Tests about Population Proportion We have learned Sampling distribution of p ¯ Building confidence intervals for p using p ¯ These worked only in large samples (CLT). Using the same ideas, we can perform hypothesis testing for population proportions. The test statistic you will need will be a z-statistic: z= p − p0 ¯ p0 (1−p0 ) n where p0 is the value in the null hypothesis. Note that denominator uses p0 , not p . ¯ And critical values looked up from the standard normal table. Utku Suleymanoglu (UMich) Hypothesis Testing 37 / 39 Hypothesis about Population Proportions Example Suppose you work at a polling company and you are interested the proportion of Americans who support Obamacare. In particular, you are interested in whether the majority of people support it. Your company interviewed 10, 000 people, 5, 200 of which said they support Obamacare. Perform a test to support the Democratic claim that the majority of Americans support Obamacare with 0.01 significance level. [this is fictitious data] We have a large enough sample, so we can do hypothesis testing. The null and the alternative: H0 :p ≤ 0.50 H1 :p > 0.50 Utku Suleymanoglu (UMich) Hypothesis Testing 38 / 39 Hypothesis about Population Proportions p = 0.52 so the test statistic is: ¯ z= 0.52 − 0.50 0.5(1−0.5) 10000 = 0.02 = 40 0.005 Find the critical value zα : z0.01 = F (2.33) = 0.99 Because this is right-tailed test, decision rule says “reject H0 if z > zα ”. So we reject the null hypothesis that says a minority supports Obamacare. The fictitious data supports the Democratic claim. Utku Suleymanoglu (UMich) Hypothesis Testing 39 / 39 ...
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