WEEK 6 - HOMEWORK 6: LANE CHAPTERS, 11, 12, AND 13; ILLOWSKY CHAPTERS 9, 10 INTRODUCTION TO HYPOTHESIS TESTING WHAT IS A HYPOTHESIS TEST? Here we are testing claims about the TRUE POPULATION'S STATISTICS based on SAMPLES we have taken. The most common statistic of interest is of course the POPULATION MEAN ( ). But, we can also test its VARIANCE and its STANDARD DEVIATION. (We can also compare TWO or more means to see if there are significant differences. We must have a basic hypothesis, referred to as the NULL Hypothesis (Ho) and an ALTERNATE Hypothesis (Ha). Our NULL ( and ALTERNATE) Hypotheses can take three forms: (1) Ho: < some number; Ha: > that number (< is \"less than or equal to\" and > is \"greater than or equal to\" ), (2) Ho: (3) Ho: = some number; > some number; Ha: Ha < that number , or that number ; ( means \"not equal to\") NOTE THAT Ho MUST HAVE THE \"EQUALS\" IN IT WHEREAS Ha NEVER DOES. (1) Is referred to as a \"ONE-TAILED TEST TO THE LEFT\" (2) Is a \"ONE-TAILED TEST TO THE RIGHT\" (3) Is a \"TWO-TAILED TEST\" NEXT, we need to decide what level of significance, i.e.(how sure we want to be about our hypothesis. This is where comes in again. Do we want to test at the 10%, 5% or 1% level of significance? Another wrinkle is that for the TWO- TAILED test, since our value could be greater OR less than some number, we use /2 for each extreme, so for 10% it's 5% (0.050) at each end (tail of the curve), for 5% it's 2.5% (0.0250) at each end, and for 1% it's 0.5% (0.0050) at the ends. You have heard about this kind of split before with confidence intervals, but think about it. Here is a graphical display of all this: As you can see, there is a CRITICAL z-VALUE for each of these test depending on the significance level alpha ( ) or /2. In HW4 questions 1 and 2, you found the critical z-values for alpha's of 1%, 5% and 10%, which would work for the onetailed tests. For the two tailed tests we need to split these alphas (/2) and find the critical z-values (at the positive and negative tails of the graph) So, for an of 1% (0.0100) it would be /2 or 0.005 in the left tail (negative z-value) = -2.575 and for the far right tail (0.005 in that tail) we would have to find the z-value for an area to the LEFT of 99.5% (0.9950) and this is +z = +2.575 Continuing on, for an of 5% for a two-tailed test the z-values for /2 would correspond to areas under the curve of 0.0250 at each end. The far left tail would have a negative z-value of -1.96 (see picture above) and the far right tail would have a positive z-value of +1.96 that in the Table represented an area of 97.5% (0.9750) to the LEFT. Lastly, for an alpha of 10%, hence an /2 at both ends of 5% (the two-tailed test), the negative z-value would be -1.645. The positive z-value marking the upper 5% (Table value from 95% to the left) is +1.645. SO, FOR YOUR USE IN ALL HYPOTHESIS TEST (AND WORKS FOR CONFIDENCE INTERVALS TOO) HERE IS A TABLE OF THE CRITICAL Z-VALUES FOR THE VARIOUS LEVELS OF SIGNIFICANCE (ALPHA's) MOST COMMONLY USED. ALPHA () -Z-value +Z-value ALPHA/2 -Z-value +Z-value (LEFT tail) (RIGHT tail) (/2) (LEFT tail) (RIGHT tail) 1% (0.0100) -2.33 +2.33 0.5% (0.0050) -2.575 +2.575 5% (0.0500) -1.645 +1.645 2.5% (0.0250) -1.96 +1.96 10% (0.1000) -1.28 +1.28 5% (0.0500) -1.645 +1.645 (For those in the NAVY the left and right colors are for PORT and STARBOARD and the history of these terms is interesting (\"google\" them). HYPOTHESIS TESTING: 1. WRITE OUT YOUR NULL (Ho) AND ALTERNATE (Ha) HYPOTHESES, MAKING SURE THE NULL INCLUDES AN \"EQUALS\". 2. DECIDE ON THE LEVEL OF SIGNIFICANCE () YOU WANT (OR ARE TOLD TO USE-BUT IF NOT TOLD ASSUME 5%) 3. SOMETIMES THE LEVEL IS REFERREED TO AS 95% (FOR 5%), 90% FOR (10%) AND 99% FOR (1%). THIS IS JUST TO CONFUSE YOU. 4. USING THE FORMULAS PROVIDED (IN THE TEXTS) CALCULATE THE TEST STATISTIC, FOR NOW A z-VALUE OR tVALUE 5. COMPARE THE CALCULATED z OR t TEST STATISTIC TO THE CRITICAL VALUE (YOU ONLY HAVE THE CRITICAL zVALUES IN THE TABLE ABOVE) AND IF THE TEST STATISTIC IS A SMALLER NUMBER THAN THE NEGATIVE CRITICAL VALUE IT IS IN THE \"UNUSUAL\" (RED) AREA, HENCE IT IS \"UNLIKELY\" THE NULL HYPOTHESIS (Ho) IS CORRECT, SO IN THIS CASE WE REJECT Ho AND ACCEPT Ha. 6. ALSO, IF THE CALCULATED TEST STATISTIC IS GREATER THAN THE POSITIVE CRITICAL VALUE WE AGAIN REJECT Ho. FOR EXAMPLE: ONE-TAILED TEST TO THE LEFT WITH ALPHA = 5% : OUR CRITICAL LEFT TAIL NEGATIVE z-VALUE IS -1.645. WE CALCUALTE OUR TEST STATISTIC AND GET A -1.650. DO WE ACCEPT OR REJECT Ho ? WE REJECT Ho BECAUSE -1.650 IS LESS THAN -1.645 (FURTHER LEFT IN THE RARE AREA TAIL) SIMILARLY, FOR A ONE-TAILED TEST TO THE RIGHT WITH ALPHA = 5%: OUR CRITICAL RIGHT TAIL POSITIVE z-VALUE IS +1.645. IF OUR CALCULATE TEST STATISTIC IS 1.650 WE AGAIN REJECT Ho SINCE +1.650 IS FURTHER RIGHT IN THE (RED) \"RARE\" AREA THAT THE THRESHOLD z-VALUE OF +1.645. HYPOTHESIS TESTING USING CRITICAL t-VALUES AND t-TEST STATISTICS: IF WE ARE NOT GIVEN THE POPULATION STANDARD DEVIATON WE HAVE TO USE THE t-VALUES (t-TEST) RATHER THAN THE z-VALUES (z-TEST). IT WOULD BE DIFFICULT TO SET UP A TABLE OF THE CRITICAL t-VALUES (LIKE WE DID FOR THE CRITICAL z-VALUES ABOVE) SINCE THE t-VALUES VARY WITH THE SAMPLE SIZE (N) THAT DETERMINES THE DEGREES OF FREEDOM (df = N-1) WE ENTER THE t-TABLE WITH. NOW, ON TO THE HOMEWORK FOR WEEK 6: REVIEW: Using the t-Table (NOT the z-Table). The BODY of the t-Table actually contains the t-values, NOT the areas or probabilities like the z-Tables. As review, we locate our desired probability along the TOP row of the t-Table based on the alpha selected. NOTE that these are the UPPER-TAIL probabilities - area shown in blue. We then go down that column (under the probability) until we reach the row with the calculated degrees of freedom (df = N-1). The number at that intersection is our desired t-Value. #1. WHAT are the critical t-Values with 20 df (NOT 14 as the example circles show) at the 1%, %5 and 10% levels of significance (one-tailed test - right tail). The Ho in this case would be the MEAN is GREATER THAN OR EQUAL TO ___ and the Ha: MEAN < _____ THIS WEEK'S CONCEPT: HYPOTHESIS TESTING DEALING WITH POPULATION MEANS AND THE SAMPLE MEANS AND STANDARD DEVIATIONS THAT ARE USED TO TEST THOSE POPULATION MEANS. HOWEVER, WE MUST ADJUST THE SAMPLE'S SD FOR \"SAMPLING ERROR\" and this is done by dividing the sample SD (abbreviated \"s\") by the square root of its sample size N. So, the \"adjusted\" s = original sample s/ n. This can also be 2 2 calculated as s / n. NOTE that the s is actually the VARIANCE. Remember too from Week 1 that the variance is just the square of the distance of each data point from the true mean. The \"squaring\" gets rid of negative values as you should recall. IT'S ALL ABOUT THESE DATA POINT DISTANCES FROM THE TRUE MEAN. FOR THESE HYPOTHESIS TESTS, If we KNOW the POPULATION'S STANDARD DEVIATION, we use the z-TEST. LOOK AT ILLOWSKY'S SAMPLE PROBLEM 9.9 ON TEST PAGE 380 AS YOU READ ON ILLOWSKY USES THE P-VALUE TO TEST HYPOTHESES AND COMPARES THEIR CALCULATED P-VALUE TO THE ALPHA LEVEL OF SIGNIFICANCE WE CHOSE. IF THE P-VALUE IS LESS THAN THE ALPHA WE REJECT Ho. THIS IS FINE, BUT IT REQUIRES SOFTWARE TO GET THE PRECISE PROBABILITY (P-VALUE). WE WILL BE USING THE TEST STATISTIC APPROACH AND ACTUALLY READING THE TABLES. (YOU WILL SEE THAT YOU CAN STILL GET A CLOSE APPROXIMATION OF THE P-VALUE FROM THE TABLE AS WELL.). ILLOWSKY HAS AN EXAMPLE USING TEST STATISTICS ON PAGE 486 (EXAMPLE 9.11) BOTH OF THESE HYPOTHESIS TESTING APPROACHES: P-VALUE AND TEST STATISTIC WILL SUPPORT THE SAME CONCLUSION (ACCEPT OR REJECT Ho). CONFIDENCE INTERVALS CAN ALSO BE USED TO TEST HYPOTHESIS, BUT THIS APPROACH CAN (BUT NOT ALWAYS) SUPPORT DIFFERENT CONCLUSIONS. ILLOWSKY 9.9 (page 480) Baker claims his bread height is at least 15 cm. Customers say it isn't that high. This is a one-tailed test and we will use the customer's statement as our NULL Hypothesis Ho: mean bread height is less than or equal to 15 cm so Ha: mean bread height is greater than 15 cm OR Ho: < 15 cm and Ha: > 15 cm (NOTE that the Ho has the \"equals\" in it as it must) This phrasing makes it a one-tailed test to the RIGHT. If the customers had said the bread simply was NOT 15 cm, we could have ended up with a two tailed test with Ho: mean equals 15 cm and Ha: mean is simply not equal to 15 cm (height is higher or lower than 15cm). Let's keep it simple. He bakes 10 loaves that have an average height of 17cm (this is a SAMPLE) The Baker has produced enough loaves of bread (his POPULATION of bread) over his career to have calculated a standard deviation = 0.5 and we adjust this for \"error\" as well so it is 0.5/N and N = 10 so = 0.5/10 = 0.16 NOW, since we have the standard deviation we can use the z-Test statistic. The 10 loaf sample mean is 17 cm, this is our x-value we are using to test the accuracy of the Population mean of 15 cm. We simply standardize that x-value: ztest statistic = (x - mean) /adjusted SD = (17 - 15) / 0.16 = 12.5 This is the z-TEST value we now go to our +z-Table with. You will quickly see that a z-value of +12.5 is WAY off the chart with more than 99.999% to the LEFT ( and less than 0.00000000. . .1 to the right as the PROBABILITY. Normally you would compare this calculated z-Test statistic to our critical values, but you can see that at whatever level of significance we chose (1%, 5% or 10%) we are MUCH further to the right (in the \"reject Ho) area than any of them. So, we REJECT Ho and accept Ha: This Baker's bread has an average height of greater than 15 cm (we don't say equal to or greater than since the Ho included the equals and rejected all of it). NO POPULATION STANDARD DEVIATION MORE OFTEN we do NOT know the POPULATION'S SD, so we substitute the SAMPLE'S STANDARD DEVIATION and use the t-TEST. LOOK AT ILLOWSKY EXAMPLE 9.16 ON PAGE 488 THAT USES THE t-TEST STATISTIC. THEY STILL USE THE P-VALUE BUT WE WILL CALCULATE THE TEST STATISTIC This example deals with scores on a Statistics test. Student believe the mean score is 65 and the instructor believes it is higher. So, let's go with Ho: mean < 65 and Ha: > 65 We don't have POPULATION mean of SD. So we use a sample. This is a \"single population mean\" one-tailed test to the right (since the Ha: has the \">\") The chosen level of significance is 5%, but we can't use the critical z-values we tabulated earlier. We need to find the critical t-Value. Ten student tests were sampled so N= 10 and the df = 10 - 1 = 9 The sample of the 10 tests had a mean (X-bar as it's referred to) of 67. And we need to calculate the sample's standard deviation (s) which ends up s = 3.20 and then correct it for sampling error = s/10 = 3.197/3.162 = 1.011 Now, our t-Test statistic = (X-bar - mean)/ adjusted SD = (67 - 65)/1.01 = 1.978 WHAT IS THE CRITICAL t-VALUE WITH AN ALPHA OF 5% AND A df = 9 ? Go to the T-Table and find 5% (0.05) along the TOP row. Go down that COLUMN till you come to the row for a DF of 9 and read: 1.833. THIS IS OUR CRITICAL t-VALUE FOR THIS PROBLEM. COMPARE THE CALCULATED t-TEST STATISTIC 1.978 TO THE CRITICAL VALUE 1.833 AND WE SEE THAT THE TEST STATISTIC IS GREATER THAN (FURTHER TO THE RIGHT) OF THE CRITICAL VALUE, HENCE WE REJECT Ho. THE MEAN IS GREATER THAN 65. ILLOWSKY USED SOFTWARE BUT LET'S GO INTO THE t-TABLE WITH OUR t-TEST STATISTIC CALUCLATED VALUE OF 1.978 AT A df OF 9. FIND IT? NOW, LOOK UP TO THE TOP ROW OF THE TABLE AND WHAT PROBABILITIES ARE WE BETWEEN? WE ARE LESS THAN 0.0500 BUT MORE THAN 0.025. PER ILLOWSKY, THE ACTUAL P-VALUE IS 0.0396. SEE HOW YOU CAN GET AN APPROXIMATE P-VALUE JUST FROM THE TABLE? HOWEVER, WE WILL USE THE TEST STATISTIC APPROACH. YOU CAN CERTAINLY PRACTICE USING THE P-VALUE SOFTWARE, BUT SUBMIT ANSWERS USING TEST STATISTICS. WE CAN ALSO TEST PROPORTIONS RATHER THAN MEANS. FOREXAMPLE WE BELIEVE THAT 65% OF STUDENTS GET A \"C\" OR BEETER IN STATSTICS. THERE IS NO MEAN PROVIDED. WE GET WHAT WE NEED FROM A SAMPLE. CHECK ILLOWSKY EXAMPLE 9.17 PAGE 489) ONE LAST HYPOTHESIS INVOLVES COMPARING TWO MEANS AS IN OUR DRINKING OR NOT DRINKING COFFEE BEFORE A LONG LECTURE. THE Ho: would be that the means are equal expressed as Ho: M1 -M2 = 0 and the Ha: could be a < or > depending on how it's stated. Illowsky pages 530-531 have the formulas for t-test statistic calculation and the degrees of freedom (df) calculation FOR THESE MEAN COMPARISON HYPOTHESIS TESTS. Both are a little involved, but you NEED TO USE THEM (NOT software) to calculate these necessary values. Round off your calculated df to the nearest whole number and then YOU CAN AND NEED TO USE A TABLE TO FIND THESE VALUES. HYPOTHESIS TESTING CAN MAKE MISTAKES. THESE ARE REFERRED TO AS TYPE I AND TYPE II ERRORS. WE COULD REJECT Ho WHEN IT IS ACTUALLY TRUE OR ACCEPT Ho WHEN IT IS ACTUALLY FALSE. BIG MISTAKES LEGALLY. ACTION Do NOT reject Ho Reject H0 Ho IS ACTUALLY TRUE FALSE Correct Outcome Type II Error Type I Error Correct Outcome CONTINUING ON WITH THE HW PROBLEMS (DON'T OVERLOOK QUESTION #1 WAY EARLIER), THESE ARE PROBLEMS FROM OUR ACTUAL TEXTS: #2. LANE (CHAPTER 11) PROBLEM #18 YOU CAN BE RIGHT WHEN YOU THINK YOUR ARE WRONG AND WRONG WHEN YOU FEEL YOU ARE RIGHT. THIS IS WHAT THE TYPE I AND II ERRORS ARE ALL ABOUT. #3. LANE (CHAPTER 12) PROBLEM #13: LANE PROVIDES A GOOD EXAMPLE OF THIS TYPE OF PROBLEM BEGINNING N FPAGE 408 #4. LANE (CHAPTER 13) PROBLEM #2: IT'S ABOUT THE DEFINITIONS. [ PROBLEM #4 USED TO BE ASSIGNED, BUT IT IS VERY CONFUSING - CHECK IT OUT IF YOU HAVE TIME. THE RANK ORDER ANSWER IS C, A, B, D ] #5. ILLOWSKY (CHAPTER 9) PROBLEM #81: SHOW ALL STEPS FROM STATING THE HYPOTHISIS TO DRAWING THE CONCLUSION. MAKE SURE TO PROVIDE YOU CALCULATED TEST STATISTIC AND THE CRITICAL VALUE. CALCULATE BY HAND THE TEST STATISTIC AND COMPARE IT TO THE CRITICAL VALUE TO DRAW YOU CONCLUSION. THE SAME RESULTS. YOU CAN USE SOFTWARE TO GENERATE A P-VALUE FOR COMPARISON VALUE ONLY NO CREDIT FOR JUST A SOFTWARE GENERATED P-VALUE. #6. ILLOWSKY (CHAPTER 10) PROBLEM # 80: STRAIGHT FORWARD. USE THE TEST STATISTIC APPROACH USING THE EQUATIONS ON PAGES 530-531 IN ILLOWSKI FOR THE t-TEST STATISTIC AND THE DEGREES OF FREEDOM (THE LATTER TO GET THE CRITICAL t-VALUE. LANE ALSO HAS PERHAPS A SIMPLER VERSION OF THESE EQUATIONS. SHOW ALL STEPS. #7. FOR OUR COFFEE DRINKING AND STAYING AWAKE DISCUSSION: IS THERE A SIGNIFICANT DIFFERENCE AT THE 5% LEVEL BETWEEN THE STAYING AWAKE TIMES OF THESE TWO GROUPS OF STUDENTS IN A LECTURE HALL? WORK BY HAND AND SHOW THE RESULTS OF EACH STEP AS WITH #5 ABOVE. THAT'S ENOUGH FOR THIS WEEK. Wk 5 Discussion Topic: Condence lnte rval or the lack of it....... You are trying to decide whether to take Stat 200 as an online or facetoface class, but you don't know if you can stay awake through an inclass. 2hour (120 minute} lecture that starts at 3:00 AM. It is critical that you are able to since the instructor has said that all of the quiz and nal exam questions will come from material from those lectures. Recording them is ridiculous since who could listen to this stuff twice?! As a test you decide to audit a session of a current fo Stat 200 class to see how it goes. You position yourself bythe door and ask EVERY student {the entire POPULATION) 3 questions: 1] Are you hungover? 2] How many hours of sleep did you get last night? 3] Did you drink any caffeinated beverage (coffee, tea, Red Bull) just before coming to class? To your surprise (?) none said they were hungover,all said they had gotten at least 7 hours of sleep the previous night, and 15 of the 30 said they had drunk a caffeinated beverage within the last hour. Consequently, you feel that the only applicable variable affecting staying awake for the next two hours is caffeine. Here are the times {in minutes] that each of these 15 coffee drinking students and each of the 15 uncaffeinated students was able to stay awake: r-smmrstm'aEEDEE: cmsmmni BASED ON THESE DATA: 1) IS THERE A SIGNIFICANT DIFFERENCE BETWEEN THESE TWO GROUPS IN TERMS OF AWAKE TIME? WHAT IF YOU HAD (L'INL'IIr BEEN ABLE TO SURVEY 14 STUDENTS [A SAMPLE RATHER THAN A POPU LATIDN}? 2) You decide that If you can stay awake for at least 90 of the 120 minutes the lecture lasts, you will get enough out of it to pass the course (this may be a bad criterion, but it's the one you have decided to use). DOES IT MATTER IF YOU DRINK CAFFEINE OR NOTTD BE ABLE TO STAY AWAKE 90 MINUTES? WHAT IS YOUR STATISTICAL BASIS FOR THIS CHOICE? 3) WHAT IS THE 95% CONFIDENCE INTERVAL FOR BOTH GROUPS? 4) Bottom line: Is CAFFEINE the key to success in a f2f Stat 200 class? {Would It be necessary in our online course?) 5) FINALLY, what are some of the flaws in this observational analysis? Howr might it be better designed? OK,taI-ce your time. but hurryr up. 1} There is not a signicant difference between the two groups of students. Although, there are more students who stayed awake who drank caffeine than students who did not. I would not have received as accurate results of students falling asleep who drank caffeine verses students who did not drink caITeine. It is important to survey all students as it will give you the exact numbers. 2} To decide whether or not I rely on drinking a caffeinated beverage before class to keep me awake for 90 to 120 minutes, I will nd the mean of each group of swdenrs. The mean time of the students who stayed awake and a drank caeinated beverage before class is 80.6 minutes. The mean time of the students who stayed awake and did not drink a caffeinated beverage before class was 89 minutes. These calculations show that there is a difference of 3A minutes where the students who did not drink caffeinated beverages stayed up Longer on average than those who did. Based off of these calculations I will decide that it is not necessary for me to get a caffeinated beverage before class. Although I Love my coffee and I would get one because I enjoy the taste in the morning. 3} The 95% condence interval for the group who drank caffeine before class is Elk-18. The 95% condence for the group of students who did not drink caffeine before class is 35+-15. 4} After calculating the 95% condence, I can say that caffeine is not a necessity when it comes to relying on it to stay a wake in a Hf classroom. For an online class like ours, this question would have to be asked to every student in the class to get data for the calculations. Figuring out this statistically would be tough because an online class is broken up during the week and the work Load can be a little bit at a time. This would allow the student to get me work done in thirty minutes or less and not need to drink or rely on caffeine because there is no two hour Long session. 5} The observational analysis could be more specic on how many milligrams of caffeine was in each of me caffeinated beverages that each student drank before class. Not incLuding a question about what they ate for breakfast before class is also a bug factor and can play a role in students lacking energy and falling asleep