Solved Questions for Final Exam - Mathematics and Statistics | MATH 243, Exams of Probability and Statistics

Download Solved Questions for Final Exam - Mathematics and Statistics | MATH 243 and more Exams Probability and Statistics in PDF only on Docsity! Math 243: Lecture File 10 N. Christopher Phillips 30 April 2009 N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 1 / 35 A refinement: Significance level One should really decide ahead of time how small a P-value to ask for. The number is called the significance level and denoted α. We have succeeded if, after running our test, it turns out that P ≤ α. Caution: α is not just another name for P. Instead, it is a criterion we choose before the test, preferably even before the experiment, like what the treatments are. Caution: “significant” does not mean important here. Rather, it means the evidence for something is strong. For not particularly implausible alternative hypotheses, common values for α are α = 0.05 and α = 0.01. N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 2 / 35 Recall Example 7 You have bought a large quantity of 7 mm ball bearings from Wang’s Widgets Inc., and you are preparing to sue Wang’s Widgets Inc. because, you claim, the diameter of the balls in the ball bearings is not as advertised. We compare the mean diameter of the balls in the ball bearings with 7 mm, which is what the diameter is supposed to be. Let µ be the mean diameter, in mm, of the balls in 7 mm ball bearings from Wang’s Widgets Inc. H0: µ = 7. Ha: µ 6= 7. Our lawyers are careful, and ask that we run a test at significance α = 0.01. N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 3 / 35 Example 7 (continued) Let µ be the mean diameter, in mm, of the balls in 7 mm ball bearings from Wang’s Widgets Inc. Suppose the standard deviation of their diameters is known to be 0.1 (in mm). H0: µ = 7. Ha: µ 6= 7. We are asking for significance level α = 0.01. We choose a simple random sample of size 100, and get x = 7.03. So z = x − µ0 σ/ √ n = 7.03− 7 0.1/ √ 100 = 3. We got P ≈ 0.003. Is it true that P ≤ α? N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 4 / 35 Example 7 (continued) Let µ be the mean diameter, in mm, of the balls in 7 mm ball bearings from Wang’s Widgets Inc. H0: µ = 7. Ha: µ 6= 7. We are asking for significance level α = 0.01, and we got P ≈ 0.003. Is it true that P ≤ α? Yes, P ≤ α. Accordingly, in formal terms, we reject the null hypothesis. In terms related to the statement of the problem, we conclude that there is strong evidence, at the significance level α = 0.01, that the mean diameter of the balls in 7 mm ball bearings from Wang’s Widgets Inc. is not really 7 mm. N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 5 / 35 Example 7 (continued) Let µ be the mean diameter, in mm, of the balls in 7 mm ball bearings from Wang’s Widgets Inc. H0: µ = 7. Ha: µ 6= 7. We are asking for significance level α = 0.01, and we got P ≤ α. Accordingly, in formal terms, we reject the null hypothesis. In terms related to the statement of the problem, we conclude that there is strong evidence, at the significance level α = 0.01, that the mean diameter of the balls in 7 mm ball bearings from Wang’s Widgets Inc. is not really 7 mm. In a homework, quiz, or exam problem, you must state the outcome both in formal terms (“We reject the null hypothesis”) and in terms related to the statement of the problem (“We conclude that there is strong evidence . . . ”). N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 6 / 35 Example 7 (continued) As before: You have bought a large quantity of 7 mm ball bearings from Wang’s Widgets Inc., and you are preparing to sue Wang’s Widgets Inc. because, you claim, the diameter of the balls in the ball bearings is not as advertised. We compare the mean diameter of the balls in the ball bearings with 7 mm, which is what the diameter is supposed to be. Let µ be the mean diameter, in mm, of the balls in 7 mm ball bearings from Wang’s Widgets Inc. H0: µ = 7. Ha: µ 6= 7. Our lawyers are careful, and ask that we run a test at significance α = 0.01. N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 7 / 35 Example 7 (continued) Let µ be the mean diameter, in mm, of the balls in 7 mm ball bearings from Wang’s Widgets Inc. Suppose the standard deviation of their diameters is known to be 0.1 (in mm). H0: µ = 7. Ha: µ 6= 7. We are asking for significance level α = 0.01. We choose a simple random sample of size 100. Suppose that we instead get x = 6.98. So z = x − µ0 σ/ √ n = 6.98− 7 0.1/ √ 100 = −2. We got P ≈ 0.0456. Is it true that P ≤ α? N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 8 / 35 Example: IQ scores The IQ scores of a simple random sample of 31 girls in the 7th grade in the Gorman school district are: 72, 74, 86, 89, 91, 93, 96, 98, 100, 102, 103, 103, 103, 104, 105, 107, 108, 111, 111, 112, 112, 112, 114, 114, 114, 118, 119, 120, 128, 130, 132. Assume the standard deviation of the IQ scores of 7th grade girls in the Gorman school district is σ = 15, the same as for the general population. Test at the significance level α = 0.02 whether the mean IQ of 7th grade girls in the Gorman school district is greater than 100. We use the one sample z procedure. Let µ be the mean IQ of 7th grade girls in the Gorman school district. Hypotheses: H0: µ = 100. Ha: µ > 100. N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 17 / 35 Sample size 31, population standard deviation σ = 15, significance level α = 0.02. Hypotheses: H0: µ = 100. Ha: µ > 100. Find x . I got with a calculator x ≈ 105.839 (as before). Compute the test statistic. z = x − µ0 σ/ √ n ≈ 105.839− 100 15/ √ 31 ≈ 2.167 (as before). Find the P-value. My calculator gave a P-value P ≈ 0.01511 for the one-sided test. This is different. It is half as much as before. (We could have used Table A, or at least gotten a range using Table C.) N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 18 / 35 P-value (continued) 95 100 105 110 0.02 0.04 0.06 0.08 0.10 0.12 0.14 The curve is N ( 100, 15/sqrt31 ) , which is the density function for the sampling distribution. The red region is above the values of x which are as extreme, or more extreme, than the actually observed value x = 105.839. It has area P ≈ 0.01511. N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 19 / 35 Example: IQ scores (continued) Sample size 31, population standard deviation σ = 15, significance level α = 0.02. Hypotheses: H0: µ = 100. Ha: µ > 100. We got P ≈ 0.01511. Since P ≤ α, we do reject the null hypothesis. We conclude that there is strong evidence, at the significance level α = 0.02, that the mean IQ of 7th grade girls in the Gorman school district is greater than 100. N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 20 / 35 Example: IQ scores (continued) A two-sided test was not significant, but a one-sided test, using the same data, was significant. (In real life, this situation is not very common.) You must decide before doing your test whether to do a one-sided test or a two-sided test. Preferably, decide before collecting any data. N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 21 / 35 The logic behind hypothesis testing Suppose we have a large collection of coins, some of which are ordinary fair coins and some of which have tails on both sides. I choose one of these coins, flip it 10 times, and report the result: it came up tails every time. For a fair coin, the probability that 10 tosses come out all tails is 1/1024, or a bit less than 0.1%. Since this outcome is unlikely for a fair coin, we interpret it as evidence that the coin has tails on both sides. N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 22 / 35 The coin example is “hard”: if even one toss is heads, it can’t have tails on both sides. For a “soft” example, which is in some ways more realistic, see the discussion of free throw percentage at the beginning of Chapter 14. The coin example has the advantage of allowing some explicit probability calculations, which allow one to give explicit examples for “how small a P-value is convincing” (see Chapter 15). N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 23 / 35 An example We test the null hypothesis, for example µ = 64, against an alternative hypothesis, say µ > 64. We choose a simple random sample, and compute x . The P-value is the probability that, if the null hypothesis is true, a simple random sample gives a value of x which is as extreme (here, is as large) as the one we got. N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 24 / 35 An example (continued) H0: µ = 64. Ha: µ > 64. Example: I know σ = 2.7, and a simple random sample of size 9 gives x = 66.7. We have z = x − µ0 σ/ √ 9 = 3. Looking up in Table A, we find that the probability of having z ≥ 3 is 1− 0.9987 = 0.0013, so the P-value is 0.13%. This means that, if the null hypothesis is true, the probability of getting x ≥ 66.7 is 0.13%. N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 25 / 35 Be careful with the interpretation! This does not mean that there is a 99.87% probability that the null hypothesis is wrong! How convincing this evidence is depends, among other things, on how plausible the alternative hypothesis is. If it is (subjectively) unlikely, a smaller P-value is needed to persuade people to believe it. N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 26 / 35 If half the coins have tails on both sides Example with coins, where the probability calculations are simple (although I will only report the results here). Suppose we choose a coin at random from a bag in which half the coins are fair and half have tails on both sides. We flip the coin 10 times, and get tails every time. This outcome has a P-value of 1/1024 ≈ 0.000977, or a bit less than 0.1%. N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 27 / 35 The actual probability that the coin has tails on both sides is 1024/1025 ≈ 0.9990. That is, if we do this experiment a very large number of times, then in about 99.90% of all the cases in which we get 10 tails, the coin in fact has tails on both sides. The alternative hypothesis is quite plausible: there is a 50% chance that a randomly chosen coin has tails on both sides. Since the alternative hypothesis is plausible, the P-value of about 0.001 is strong evidence that the null hypothesis is false. N. Christopher Phillips () Math 243: Lecture File 10 30 April 2009 28 / 35