Lecture #19 Tuesday, October 25, 2016 Textbook: Sections 12.3 to 12.4 Statistics 200 Lecture #19 Tuesday, October 25, 2016 Textbook: Sections 12.3 to 12.4 Objectives: • Frame a 2x2 test of independence as a test of difference of two proportions • Apply ideas of forming hypotheses, calculating test statistics, and calculating p-values. • Use Minitab to find probabilities for normal, binomial, and chi- square distributions

Quick review of odds vs. risk/probability I came across this quotation recently: “It will take several days for serious scientific polls to come in, but the betting markets respond in real time. PaddyPower, the Irish bookie, puts the odds on Clinton at 2/11, which implies an 85% chance that Clinton will win. U.K.-based William Hill is offering 1/6 odds on Clinton, implying that she has an 86% chance of winning.” How are these percents calculated? Ans: Odds of 2/11 give a probability of 2/13 (of losing), which is a winning probability of 11/13=84.6%.

Recall this example: Are women more likely to have dogs? Has Dog No Dog Total Female 89 56.7% 68 43.3% 157 Male 66 50.8% 64 49.2% 130 155 132 287 Your class data

Recall this example: Are women more likely to have dogs? Has Dog No Dog Total Female 89 56.7% 68 43.3% 157 Male 66 50.8% 64 49.2% 130 155 132 287 Let’s reframe this problem: Examine the difference between two independent proportions, that is, pf–pm. Is it zero? Let’s run a statistical hypothesis test.

Clicker Quiz #12

Recall this example: Are women more likely to have dogs? Has Dog No Dog Total Female 89 56.7% 68 43.3% 157 Male 66 50.8% 64 49.2% 130 155 132 287 This is a two-sided alternative. H0: pf–pm = 0 Ha: pf–pm ≠ 0 Hypotheses: In this dataset,

Review: The sampling distribution of As long as both p-hat1 and p-hat2 are approximately normal… ...and the two samples are independent... Then the sampling distribution is approximately normal with mean p1–p2 and standard deviation

Recall the general test statistic formula: In our example, the parameter is pf–pm. Therefore: • The sample estimate is • The mean under H0 is 0 • The std dev. under H0 is Notice: Same value of p-hat in both fractions! That value is the combined sample proportion:

Recall the general test statistic formula: In our example, the parameter is pf–pm. Therefore: • The sample estimate is • The mean under H0 is 0 • The std dev. under H0 is Conclusion: The test statistic is

p-value definition We have a test statistic equal to 1.00. The p-value is the probability, if H0 is true, that our experiment would give a test statistic at least as extreme as the test statistic we observed. Also, the alternative is Ha: pf–pm ≠ 0. “At least as extreme” means in the direction determined by the alternative hypothesis. In this case, the p-value is P(Z ≥ 1.00 or Z ≤ –1.00). Therefore, the p-value is 0.317.

Chi-square statistic: 1.003 Recall result from Lecture 08 (Sept. 15): Are women more likely to have dogs? Has Dog No Dog Total Female 89 56.7% 68 43.3% 157 Male 66 50.8% 64 49.2% 130 155 132 287 Chi-square statistic: 1.003 P-value: 0.317 Note: There was a mistake in the Sept. 15 calculation!

Question from Midterm #2 Correct Answer: C Answered correctly by 72.1%

Finding p-values using Minitab We have a test statistic equal to 1.00. In this case, the p-value is P(Z ≥ 1.00 or Z ≤ –1.00). Also, the alternative is Ha: pf–pm ≠ 0. To find this probability in Minitab, select GraphProbability Distribution Plot: Then, click “view probability” and click OK.

Finding p-values using Minitab We have a test statistic equal to 1.00. In this case, the p-value is P(Z ≥ 1.00 or Z ≤ –1.00). Also, the alternative is Ha: pf–pm ≠ 0. Next, click “Shaded Area” and “Both Tails”.

Finding p-values using Minitab We have a test statistic equal to 1.00. In this case, the p-value is P(Z ≥ 1.00 or Z ≤ –1.00). Also, the alternative is Ha: pf–pm ≠ 0. Next, click “Shaded Area” and “Both Tails”. Then, click to define shaded area by “X value” and type 1.00 for the value.

Finding p-values using Minitab We have a test statistic equal to 1.00. In this case, the p-value is P(Z ≥ 1.00 or Z ≤ –1.00). Also, the alternative is Ha: pf–pm ≠ 0. Here is the result: You have to add the two probabilities together to get the p-value: 0.1587 + 0.1587 = 0.3164

Finding binomial probabilities using Minitab Suppose that X is a binomial random variable with parameters 5 and 0.25. (Do you know what this means?) Step 1: Select “Probability Distribution Plot” and enter the parameters. Use Minitab to find the exact value of P(X ≥ 3).

Finding binomial probabilities using Minitab Suppose that X is a binomial random variable with parameters 5 and 0.25. (Do you know what this means?) Step 2: Under “Shaded area”, select right tail and type 3. (Why not 2?) Use Minitab to find the exact value of P(X > 2).

Finding binomial probabilities using Minitab Suppose that X is a binomial random variable with parameters 5 and 0.25. (Do you know what this means?) Step 3: The answer is shown below: 0.0135 Use Minitab to find the exact value of P(X > 2).

Which midterm #3 date do you prefer? Friday, Nov. 11 Monday, Nov. 14 Keep these things in mind: • Nov. 14 is three weeks from yesterday. • Thanksgiving break begins on Monday, Nov. 21. • Starting Nov. 28, we have two weeks left of class

Chi-square statistic: 1.003 Recall result from Lecture 08 (Sept. 15): Are women more likely to have dogs? Has Dog No Dog Total Female 89 56.7% 68 43.3% 157 Male 66 50.8% 64 49.2% 130 155 132 287 Chi-square statistic: 1.003 P-value: ??? Let’s use Minitab to find the p-value directly from the chi-square statistic.

Finding chi-square probabilities using Minitab We have a chi-square statistic of 1.003 for a 2x2 table. Use Minitab to find the corresponding p-value for a test of independence (recall that the null says that the variables ARE independent).

Finding chi-square probabilities using Minitab We have a chi-square statistic of 1.003 for a 2x2 table. Step 1: Choose the chi-square distribution and enter 1 degree of freedom. Use Minitab to find the corresponding p-value for a test of independence (recall that the null says that the variables ARE independent).

Finding chi-square probabilities using Minitab We have a chi-square statistic of 1.003 for a 2x2 table. Step 2: Enter the value of the statistic and be sure to click Right Tail (always for chi-square!) Use Minitab to find the corresponding p-value for a test of independence (recall that the null says that the variables ARE independent).

Finding chi-square probabilities using Minitab We have a chi-square statistic of 1.003 for a 2x2 table. Step 3: The p-value is 0.3166, as shown in the plot below. Use Minitab to find the corresponding p-value for a test of independence (recall that the null says that the variables ARE independent).

Why do we enter 1 “degree of freedom”? Has Dog No Dog Total Female 89 68 157 Male 66 64 130 155 132 287 Ans: Only one value is free to vary if we know the totals. The other values are then automatic and not free. How many d.f. in a 2x3 table? How about a 3x3 table?

Question from Midterm #2 Correct Answer: A Answered correctly by 42.0%

Question from Midterm #2 Correct Answer: A Answered correctly by 47.4%

Question from Midterm #2 Correct Answer: B Answered correctly by 64.4%

Question from Midterm #2 Correct Answer: C Answered correctly by 54.2%

If you understand today’s lecture… 12.60, 12.63, 12.65, 12.66 Objectives: • Frame a 2x2 test of independence as a test of difference of two proportions • Apply ideas of forming hypotheses, calculating test statistics, and calculating p-values. • Use Minitab to find probabilities for normal, binomial, and chi- square distributions