1. 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

2. 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%.

3. 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

4. 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.

5. Clicker Quiz #12

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
This is a two-sided alternative.
H0: pf–pm = 0
Ha: pf–pm ≠ 0
Hypotheses:
In this dataset,

7. 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

8. 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:

9. 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

10. 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

11. 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!

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

13. 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.

14. 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”.

15. 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.

16. 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: =

17. Finding binomial probabilities using Minitab
Suppose that X is a binomial random variable with parameters 5 and (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).

18. Finding binomial probabilities using Minitab
Suppose that X is a binomial random variable with parameters 5 and (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).

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

20. 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

21. 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.

22. Finding chi-square probabilities using Minitab
We have a chi-square statistic of 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).

23. Finding chi-square probabilities using Minitab
We have a chi-square statistic of 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).

24. Finding chi-square probabilities using Minitab
We have a chi-square statistic of 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).

25. Finding chi-square probabilities using Minitab
We have a chi-square statistic of for a 2x2 table.
Step 3: The p-value is , 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).

26. 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?

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

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

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

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

31. 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