1. AP Statistics Tuesday, 09 February 2016 OBJECTIVE TSW explore Hypothesis Testing. Student to Ms. Havens: “Is either yesterday’s test or the previous test graded yet?” Ms. Havens to student: “No.” Student to Ms. Havens: “Do you know when they will be graded?” Ms. Havens to student: “No.” Ms. Havens is sorry to not be able to give student a good answer, but she will try to have them graded by the end of the week.

2. Hypothesis Tests One-Sample Means

3. Example: A government agency has received numerous complaints that a particular restaurant has been selling underweight hamburgers. The restaurant advertises that it’s patties are “a quarter pound” (4 ounces). How can I tell if they really are underweight? Take a sample & find x. expect unlikely But how do I know if this x is one that I expect to happen or is it one that is unlikely to happen? Hypothesis testing will help me decide!

4. What are hypothesis tests? Calculations that tell us if a value occurs by random chance or not. If it is statistically significant, is it... –a random occurrence due to variation? –a biased occurrence due to some other reason?

5. Nature of hypothesis tests - First begin by supposing the “effect” is NOT present Next, see if data provides evidence against the supposition Example:murder trial How does a murder trial work? First - assume that the person is innocent must Then – must have sufficient evidence to prove guilty Hmmmmm … Hypothesis tests use the same process!

6. Steps: 1)Assumptions 2)Hypothesis statements & define parameters 3)Calculations 4)Conclusion, in context Notice the steps are the same except we add hypothesis statements – which you will learn today

7. Assumptions for z-test (t-test): Have an SRS of context Distribution is (approximately) normal –Given –Large sample size –Graph data  is known (unknown) YEA YEA – These are the same assumptions as confidence intervals!!

8. Example 1: Bottles of a popular cola are supposed to contain 300 mL of cola. There is some variation from bottle to bottle. An inspector, who suspects that the bottler is under- filling, measures the contents of six randomly selected bottles. Are the assumptions met? 299.4 297.7 298.9 300.2 297 301 Have an SRS of bottles Sampling distribution is approximately normal because the boxplot is symmetrical  is unknown

9. Writing Hypothesis statements: Null hypothesis – is the statement being tested; this is a statement of “no effect” or “no difference” Alternative hypothesis – is the statement that we suspect is true H0:H0: Ha:Ha:

10. The form: Null hypothesis H 0 : parameter = hypothesized value Alternative hypothesis H a : parameter > hypothesized value H a : parameter < hypothesized value H a : parameter ≠ hypothesized value or

11. Example 2: A government agency has received numerous complaints that a particular restaurant has been selling underweight hamburgers. The restaurant advertises that it’s patties are “a quarter pound” (4 ounces). State the hypotheses : Where  is the true mean weight of hamburger patties H 0 :  = 4 H a :  < 4 You MUST indicate what μ represents!

12. Example 3: A car dealer advertises that his new subcompact models get 47 mpg. You suspect the mileage might be overrated. State the hypotheses : Where  is the true mean mpg H 0 :  = 47 H a :  < 47

13. AP Statistics Wednesday, 10 February 2016 OBJECTIVE TSW explore Hypothesis Testing. TEST: Sampling Distributions is graded. –You will get it back after lunch.

14. Example 4: Many older homes have electrical systems that use fuses rather than circuit breakers. A manufacturer of 40-A fuses wants to make sure that the mean amperage at which its fuses burn out is in fact 40. If the mean amperage is lower than 40, customers will complain because the fuses require replacement too often. If the amperage is higher than 40, the manufacturer might be liable for damage to an electrical system due to fuse malfunction. State the hypotheses : Where  is the true mean amperage of the fuses H 0 :  = 40 H a :  ≠ 40

15. Facts to remember about hypotheses: ALWAYS refer to populations (parameters) The null hypothesis for the “difference” between populations is usually equal to zero The null hypothesis for the correlation (rho) of two events is usually equal to zero. H 0 :  x-y = 0 H 0 :  = 0

16. Activity: For each pair of hypotheses, indicate which are not legitimate & explain why: Must use parameter (population) x is a statistic (sample)  is the population proportion! Must use same number as H 0 !  is parameter for population correlation coefficient – but H 0 MUST be “=“ ! Must be NOT equal!

17. P-values - The probability that the test statistic would have a value as extreme or more than what is actually observed In other words... is it far out in the tails of the distribution?

18. Level of significance - Is the amount of evidence necessary before we begin to doubt that the null hypothesis is true Is the probability that we will reject the null hypothesis, assuming that it is true Denoted by  –Can be any value –Usual values: 0.1, 0.05, 0.01 –Most common is 0.05

19. Statistically significant – as smallsmallerThe p-value is as small or smaller than the level of significance (  ) fail to rejectIf p > , “fail to reject” the null hypothesis at the  level. rejectIf p < , “reject” the null hypothesis at the  level.

20. Facts about p-values: ALWAYS make a decision about the null hypothesis! Large p-values show support for the null hypothesis, but never that it is true! Small p-values show support that the null is not true. Double the p-value for two-tail (≠) tests Never acceptNever accept the null hypothesis!

21. Never “accept” the null hypothesis!

22. At an  level of 0.05, would you reject or fail to reject H 0 for the given p-values? a)0.03 b)0.15 c)0.45 d)0.023 Reject Fail to reject

23. Calculating p-values For z-test statistic – –Use normalcdf(lb,ub) –[using standard normal curve] For t-test statistic – –Use tcdf(lb, ub, df)

24. Draw & shade a curve & calculate the p-value: 1) right-tail test t = 1.6; n = 20 2) left-tail testz = -2.4; n = 15 3) two-tail testt = 2.3; n = 25 p = 0.06305 p = 0.008198 p = 0.03045

25. Assignment WS Hypothesis Testing #1 –Due on Tuesday, 16 February 2016.

26. Writing Conclusions: 1)A statement of the decision being made (reject or fail to reject H 0 ) & why (linkage) 2)A statement of the results in context. (state in terms of H a ) AND

27. “Since the p-value ) , I reject (fail to reject) the H 0. There is (is not) sufficient evidence to suggest that H a.” Be sure to write H a in context (words)!

28. Example 5: Drinking water is considered unsafe if the mean concentration of lead is 15 ppb (parts per billion) or greater. Suppose a community randomly selects of 25 water samples and computes a t-test statistic of 2.1. Assume that lead concentrations are normally distributed. Write the hypotheses, calculate the p-value & write the appropriate conclusion for  = 0.05. H 0 :  = 15 H a :  > 15 Where  is the true mean concentration of lead in drinking water P-value = tcdf(2.1,10^99,24) =0.0232 t=2.1 Since the p-value < , I reject H 0. There is sufficient evidence to suggest that the mean concentration of lead in drinking water is greater than 15 ppb.

29. Example 6: A certain type of frozen dinners states that the dinner contains 240 calories. A random sample of 12 of these frozen dinners was selected from production to see if the caloric content was greater than stated on the box. The t-test statistic was calculated to be 1.9. (Assume calories vary normally.) Write the hypotheses, calculate the p-value & write the appropriate conclusion for  = 0.05. H 0 :  = 240 calories H a :  > 240 calories Where  is the true mean caloric content of the frozen dinners P-value = tcdf(1.9,10^99,11) =0.0420 t=1.9 Since the p-value < , I reject H 0. There is sufficient evidence to suggest that the true mean caloric content of these frozen dinners is greater than 240 calories.

30. ASSUMPTIONS SRS (given) Normal distribution (given)  unknown H o :  = 240 calories H a :  > 240 calories, where  is the true mean caloric content of frozen dinners p-value = tcdf(1.9, ∞, 11) = 0.04197 <  = 0.05 Since p ≤ , we reject H 0. There is evidence to suggest that the true mean caloric content of frozen dinners is greater than 240 calories. If  is not given, include it!

31. Formulas:  known: z = 

32. Formulas:  unknown: t = 

33. OBJECTIVE TSW explore the aspects of hypothesis testing. ASSIGNMENTS DUE TUESDAY –WS Hypothesis Testing #1 –WS Hypothesis Testing #2 –WS Hypothesis Testing #3 –WS Matched Pairs  due 02/19/16 LOOKING AHEAD –Tuesday, 02/16/2016: QUIZ: Hypothesis Testing –Wednesday, 02/17/2016: ASSESSMENT: Hypothesis Testing REVIEW: Hypothesis Testing –Friday, 02/19/2016: TEST: Hypothesis Testing AP Statistics Friday, 12 February 2016

34. Hypothesis Testing WS #1 1a) No, H 0 must be =b) No, must use parameter μ c) 2a) b) H 0 : μ = 30 ppmWhere μ is the true mean nitrate concentration in the H a : μ > 30 ppmwater c) d) H 0 : μ = $42,500 Where μ is the true mean household income of mall H a : μ > $42,500 shoppers e) 3a) Yes, the normal probability (quantile) plot is approximately linear so the distribution is approximately normal. b)

35. Hypothesis Testing WS #2 1) 2) reject H o 3) fail to reject H o 4) 5) 6) p-value =.0256 7) P-value =.092896 8) 9) p – value = tcdf(2.056, 1E99, 14)(2) =.02946(2) =.05892 Since the p-value > α, I fail to reject the null hypothesis. There is not sufficient evidence to suggest that the true mean diameter of the catheters is not equal to 2.00 mm.

36. Example 7: The Fritzi Cheese Company buys milk from several suppliers as the essential raw material for its cheese. Fritzi suspects that some producers are adding water to their milk to increase their profits. Excess water can be detected by determining the freezing point of milk. The freezing temperature of natural milk varies normally, with a mean of -0.545 degrees and a standard deviation of 0.008. Added water raises the freezing temperature toward 0 degrees, the freezing point of water (in Celsius). The laboratory manager measures the freezing temperature of five randomly selected lots of milk from one producer with a mean of -0.538 degrees. Is there sufficient evidence to suggest that this producer is adding water to his milk? (Full write-up.)

37. Assumptions: I have an SRS of milk from one producer The freezing temperature of milk is a normal distribution. (given)  is known SRS? Normal? How do you know? Do you know  ? H 0 : μ = -0.545 H a : μ > -0.545 where μ is the true mean freezing temperature of milk What are your hypothesis statements? Is there a key word? Plug values into formula. p-value = normalcdf(1.9566, 1E99) = 0.0252 Use normalcdf to calculate p-value. α =.05

38. Conclusion: Compare your p-value to α & make decision Since p-value < α, I reject the null hypothesis. Write conclusion in context in terms of H a. There is sufficient evidence to suggest that the true mean freezing temperature is greater than -0.545. This suggests that the producer is adding water to the milk.

39. Example 8: The Degree of Reading Power (DRP) is a test of the reading ability of children. Here are DRP scores for a random sample of 44 third-grade students in a suburban district: (data on note page) At the  = 0.1, is there sufficient evidence to suggest that this district’s third graders reading ability is different than the national mean of 34? (Full write-up.)

40. I have an SRS of third-graders Since the sample size is large, the sampling distribution is approximately normally distributed  is unknown SRS? Normal? How do you know? Do you know  ? What are your hypothesis statements? Is there a key word? Plug values into formula. p-value = 2*tcdf(0.6467, 1E99, 43) = 2*0.2606 = 0.5212 α = 0.1 H 0 : μ = 34where μ is the true mean reading H a : μ ≠ 34 ability of the district’s third-graders Use tcdf to calculate p-value.

41. Conclusion: Compare your p-value to α & make decision Since p-value > α, I fail to reject the null hypothesis. Write conclusion in context in terms of H a. There is not sufficient evidence to suggest that the true mean reading ability of the district’s third-graders is different than the national mean of 34.

42. Example 9: The Wall Street Journal (January 27, 1994) reported that based on sales in a chain of Midwestern grocery stores, President’s Choice Chocolate Chip Cookies were selling at a mean rate of $1323 per week. Suppose a random sample of 30 weeks in 1995 in the same stores showed that the cookies were selling at the average rate of $1208 with standard deviation of $275. Does this indicate that the sales of the cookies is different from the earlier figure? (Include assumptions.)

43. Assumptions Have an SRS of weeks Distribution of sales is approximately normal due to large sample size s unknown H 0 : μ = 1323 where μ is the true mean cookie sales per week H a : μ ≠ 1323 Since p-value < α of 0.05, I reject the null hypothesis. There is sufficient evidence to suggest that the sales of cookies are different from the earlier figure.

44. Example 9 (Continued): President’s Choice Chocolate Chip Cookies were selling at a mean rate of $1323 per week. Suppose a random sample of 30 weeks in 1995 in the same stores showed that the cookies were selling at the average rate of $1208 with standard deviation of $275. Compute a 95% confidence interval for the mean weekly sales rate. (Just compute the interval.) CI = ($1105.30, $1310.70) Based on this interval, is the mean weekly sales rate statistically different from the reported $1323? The sales rate is statistically different, since the reported mean of $1323 is not in the interval.

45. Assignment WS Hypothesis Testing #3 –Due on Tuesday, 16 February 2016. WS Matched Pairs –Due on Friday, 19 February 2016.

46. Matched Pairs Test A special type of t- inference

47. Matched Pairs – Two Forms Form #1 Pair individuals by certain characteristics Randomly select treatment for individual A Individual B is assigned to other treatment Assignment of B is dependent on assignment of A Form #2 Individual persons or items receive both treatments Order of treatments is randomly assigned; or, before & after measurements are taken The two measures are dependent on the individual

48. Is this an example of matched pairs? 1)A college wants to see if there’s a difference in time it took last year’s class to find a job after graduation and the time it took the class from five years ago to find work after graduation. Researchers take a random sample from both classes and measure the number of days between graduation and first day of employment. No, there is no pairing of individuals, you have two independent samples

49. Is this an example of matched pairs? 2) In a taste test, a researcher asks people in a random sample to taste a certain brand of spring water and rate it. Another random sample of people is asked to taste a different brand of water and rate it. The researcher wants to compare these samples. No, there is no pairing of individuals, you have two independent samples – If you would have the same people taste both brands in random order, then it would be an example of matched pairs.

50. Is this an example of matched pairs? 3)A pharmaceutical company wants to test its new weight-loss drug. Before giving the drug to a random sample, company researchers take a weight measurement on each person. After a month of using the drug, each person’s weight is measured again. Yes, you have two measurements that are dependent on each individual.

51. A whale-watching company noticed that many customers wanted to know whether it was better to book an excursion in the morning or the afternoon. To test this question, the company collected the following data on 15 randomly selected days over the past month. (Note: days were not consecutive.) Day 123456789101112131415 Morning 897910131082577687 After- noon 810989118104789669 First, you must find the differences for each day. Since you have two values for each day, they are dependent on the day – making this data matched pairs You may subtract either way – just be careful when writing H a

52. Day 123456789101112131415 Morning 897910131082577687 After- noon 810989118104789669 Differe nces 0 -2-2 1122-2 -2-2 -1 -2-2 02 -2-2 Assumptions: Have an SRS of days for whale-watching  unknown Since the normal probability plot is approximately linear, the distribution of differences is approximately normal. You need to state assumptions using the differences! Notice that, even with the granularity in this plot, it still displays a nice linear relationship! I subtracted: Morning – afternoon You could subtract the other way!

53. Differenc es 0-21122 -202 Is there sufficient evidence that more whales are sighted in the afternoon? Be careful writing your H a ! Think about how you subtracted: M-A If afternoon is more should the differences be + or -? Don’t look at numbers!!!! H 0 : μ D = 0 H a : μ D < 0 Where μ D is the true mean difference in whale sightings from morning minus afternoon Notice we used μ D for differences & it equals 0 since the null should be that there is NO difference. If you subtract afternoon – morning; then H a : μ D >0

54. Differenc es 0-21122 -202 Finishing the hypothesis test: Since p-value > a, I fail to reject H 0. There is insufficient evidence to suggest that more whales are sighted in the afternoon than in the morning. Notice that if you subtracted A-M, then your test statistic t = + 0.945, but p- value would be the same In your calculator, perform a t-test using the differences (L3)

55. Assignment WS Matched Pairs –Due on Friday, 19 February 2016 (TEST day).