WEEK 6 - HOMEWORK 6: LANE CHAPTERS, 11, 12, AND 13; ILLOWSKY CHAPTER 9, 10 INTRODUCTION USING THE NORMAL, z-TABLES AND THE t-TABLES. Here are graphs showing the Normal (z-values) and the t-Distributions (t-with more than about 40 degrees of freedom looks like the Normal). The t or z-value separating the shaded and white sections can be either our calculated test statistic or the critical value that we compare the test statistic to (to see if we are in the \"unusual\" hence \"reject\" area). For now let's assume the indicated z and t values are the critical, decision making ones based on the level of significance (alpha) we have chosen. The values of are typically 1%, 5% or 10% . Graphs \"a\" and \"b\" represent ONE-TAILED tests, hence the area in the shaded region of graph \"b\" represents the 1%, 5% or 10% probabilities. We would use this critical t or z value for RIGHT TAILED tests, where the NULL hypothesis would be something like: The mean is less than or equal to 3.5 (m < 3.5) and the ALTERNATE hypothesis would be that the mean is greater than 3.5 (m > 3.5). The NULL MUST have an equality in it. To find the critical value of z, we go to the BODY OF THE TABLE and look for the zvalue that corresponds to 99% (0.9900), 95% (0.9500) or 90% (0.9000) depending on which alpha we chose. Then, if our calculated test statistic is a larger number than the critical value, we would REJECT Ho and accept the alternate hypothesis. If your NULL hypothesis were that the mean were greater than or equal to 3.5 (ALTERNATE is that the mean is less than 3.5) our critical value would be in the LEFT TAIL and we would want to determine if our test statistic is LESS than the critical value (a test statistic of -0.62 is less than a critical value of -0.61, hence we would REJECT Ho). Graphs \"c\" and \"d\" are for TWO-TAILED tests in which the NULL hypothesis is simply that the mean equals 3.5 ( no < or >) and the ALTERNATE is that the mean is simply NOT EQUAL to 3.5. In this case we must split the alpha between the two tails. So an alpha of 5% becomes 2.5% at each tail. So for the right shorter tail we need an area of 97.5% (0.9750) to the left and for the left shorter tail we need an area of simply 2.5% (0.2500). For an alpha of 1% this would be 0.5% (0.0500) at the left tail and 99.5% (0.9950) to the left to determine the right tail. You can figure this for an alpha of 10% split between both tails. Now, on to the homework. NON-TEXT PROBLEM #1: What are the CRITICAL z-values that correspond to alphas of 1%, 5% and 10% for the LEFT TAIL and the RIGHT TAIL (6 numbers required)? Now, split the alphas between both ends and give those pairs of values (right and left tails) for the these /2 (12 numbers required). That's it for this question. Read on for guidance on the remaining questions and concepts covered. NON-TEXT PROBLEM #2: Using the t-Table. It's different from the z-Table. The BODY of the t-Table contains the t-values, NOT the areas or probabilities. We find our calculated (or critical t-value) in the BODY of the Table along the row with our determined degrees of freedom. Then, we go up to the top, heading row and read the probability. For this problem, what are the critical t-test values with 20 df at the 1%, %5 and 10% levels of significance (one-tailed test - right tail)? Moving on. This week's concepts deal with population means and the sample means that are used to test them. We collect data samples and then we need to calculate the means and standard deviations of those samples. If we know the POPULATION STANDARD DEVIATIONS, we can use the z-test. MORE OFTEN we do NOT know that SD, so we substitute the sample standard deviations and use the t-TEST. BUT, then we need to adjust the SAMPLE standard deviations (s) for \"error\" by dividing \"s\" by the square root of its sample size. If we are comparing two population means this involves two sample SD's and two sample sizes (n's). Often the s/ n is changed to s2 / n. You should note that the s2 is actually the VARIANCE. Remember too that the variance is just the square of the distance of each data point from the mean. The \"squaring\" gets rid of negative values as you should recall. A typical problem statement would be: We believe that the average (mean) heights of high school freshmen is the same as the mean heights of high school juniors. The NULL hypothesis is that these two means are EQUAL: Ho: f = j Since we have no population standard deviations we must use our freshmen and junior sample data to generate sample means and standard deviations. We must also use the t-test for our hypothesis testing. Illowsky pages 530-531 have the formulas for t-test statistic calculation and the degrees of freedom (df) calculation. Both are a little involved, but you NEED TO USE THEM (NOT software) to calculate these necessary values. In the equation used to generate the t test statistic the formula subtracts f from j which, if they are equal, this subtraction equals zero. This will give you the necessary t-value to compare to the critical t-value and make the accept/reject decision. HOWEVER, REMEMBER THAT THE CRITICAL VALUE OF t DEPENDS ON THE DEGREES OF FREEDOM CALCULATED FROM THE ILLOWSKY EQUATIONS. Round off your calculated df to the nearest whole number and then YOU CAN AND NEED TO USE A TABLE TO FIND THESE VALUES. (Search for a t-test critical values table but here are a couple of links that might help). t-Tables and z-Tables are in the Course Resources section of our course and you can also find you own on-line such as the following. CRITICAL t-VALUES RIGHT TAIL: https://www.stat.tamu.edu/~lzhou/stat302/TTable.pdf MORE CRITICAL t-VALUES: table.pdf http://www.sjsu.edu/faculty/gerstman/StatPrimer/t- LANE CHAPTERS, 11, 12, AND 13; LANE (CHAPTER 11) PROBLEM #4: THINK CAREFULLY ABOUT THIS ONE. THE NULL HYPOTHESIS SHOULD BE AN EQUALITY. ADD THE ALTERNATE HYPOTHESIS. ANY TWO-TAILED SITUATIONS HERE? 4. State the null hypothesis for: a. An experiment testing whether echinacea decreases the length of colds. b. A correlational study on the relationship between brain size and intelligence. c. An investigation of whether a self-proclaimed psychic can predict the outcome of a coin flip. d. A study comparing a drug with a placebo on the amount of pain relief. (A onetailed test was used.) LANE (CHAPTER 11) PROBLEMS #18 - #24 (LATTER ONES ARE T/F). 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. THE T/F QUESTIONS REINFORCE THESE CONCEPTS. #19 IS A LITTLE DIFFERENT AND REQUIRES JUST A SIMPLE SENTENCE ANSWER. 11. Distinguish between probability values and significance level 18. You choose an alpha level of .01 and then analyze your data. a. What is the probability that you will make a Type I error given that the null hypothesis is true? 0.01 b. What is the probability that you will make a Type I error given that the null hypothesis is false? 0 24. True/false: A researcher risks making a Type I error any time the null hypothesis is rejected. LANE (CHAPTER 12) PROBLEM #13: STRAIGHT FORWARD 13. You are conducting a study to see if students do better when they study all at once or in intervals. One group of 12 participants took a test after studying for one hour continuously. The other group of 12 participants took a test after studying for three twenty minute sessions. The first group had a mean score of 75 and a variance of 120. The second group had a mean score of 86 and a variance of 100. a. What is the calculated t value? Are the mean test scores of these two groups significantly different at the .05 level? b. What would the t value be if there were only 6 participants in each group? Would the scores be significant at the .05 level? a) t(22)= 2.57, p=0.0175 significant b) t(10)= 1.82 p=0.099, non-sig a. Standard error = 120 = 12 = 4.28 100= H : 1 - 2 = 0 Ha : 1 - 2 0 Df = 21.82 t(22) = 2.57 P ( t < -2.57 or t > 2.57) = 0.0175 < 0.05, reject Ho The test scores - two groups are significantly different at the 0.05 level. b. Standard error = 6.055 Degree of freedom = 9.918 t(10) = 0.099> 0.05, fail to reject Ho the mean test scores of these two groups are significantly different at the 0.05 level. LANE (CHAPTER 13) PROBLEMS #1 AND #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 ] 1. Define power in your own words. 2. List 3 measures one can take to increase the power of an experiment. Explain why your measures result in greater power. ILLOWSKY CHAPTER 9 ILLOWSKY (CHAPTER 9) PROBLEM #77: STATE THE NULL AND ALTERNATE HYPOTHESIS. WILL YOU USE THE NORMAL (z) OR t-TEST? WHAT IS THE CRITICAL VALUE OF THIS z OR t (From your Critical Value list)?, WHAT IS THE CALCULATED z OR t VALUE (the TEST statistic) ? COMPARE THE CALCULATED TEST STATISTIC TO ITS CRITICAL VALUE. DO YOU ACCEPT OR REJECT THE NULL HYPOTHESIS. WHY? YOU CAN ALSO USE THE P-VALUE WHICH COMPARES THE CALCULATED PROBABILITY TO THE CHOSEN LEVEL OF SIGNIFICANCE (ALPHA). IF THE PVALUE IS LARGER THAN ALPHA WE FAIL TO REJECT THE NULL HYPOTHESIS, FOR EXAMPLE, A P-VALUE OF 5.2% IS LESS RARE THAN A LEVEL OF SIGNIFICANCE () OF 5%. IT IS SIMPLY NOT THAT UNUSUAL A RESULT. THIS p-VALUE CAN BE APPROXIMATED USING THE TABLES (BUT PRECISE ANSWERS REQUIRE SOFTWARE). WE WILL STICK WITH COMPARING THE CALCULATED TEST STATISTIC TO THE CRITICAL VALUE FOR THE LEVEL OF SIGNIFICANCE CHOSEN. SEE IF YOU CAN APPROXIMATE THE PROBABILITY (P-VALUE) USING THE TABLE. FOLLOW UP WITH A SOFTWARE CALCULATION OF THIS P-VALUE IF YOU HAVE TIME AND INTEREST. THE TEST STATISTIC APPROACH AND THE P-VALUE APPROACH ALWAYS GIVE THE SAME RESULTS. CONFIDENCE INTERVALS CAN ALSO BE USED, BUT THEY CAN GIVE DIFFERENT RESULTS. 77. An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level? H0 : 4.5 Ha : > 4.5 Let X = Average time to finish an undergraduate degree. Student's-t distribution z = 3.5 p-value = 0.0005 <0.01, reject the null hypothesis Conclusion: > 4.5 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. YOU CAN USE THE LINKS PROVIDED EARLIER TO GET THE CRITICAL VALUE OF t ONCE YOU HAVE CALCULATED THE df. 80. At Rachel's 11th birthday party, eight girls were timed to see how long (in seconds) they could hold their breath in a relaxed position. After a two-minute rest, they timed themselves while jumping. The girls thought that the mean difference between their jumping and relaxed times would be zero. Test their hypothesis. Relaxed time (seconds) Jumping time (seconds) 26 21 47 40 30 28 22 21 23 25 45 43 37 35 29 32 Table 10.24 Xd is the average difference between jumping and relaxed times. t7 test statistic: -1.51 p-value: 0.1755 > 0.05, do not reject the null hypothesis. At the 5% level of significance, insufficient evidence to answer that the average difference is not zero. PROBLEM #87: THIS IS A MULTIPLE CHOICE QUESTION, WRITE OUT YOUR CHOICE (NOT JUST A LETTER) AND THEN EXPLAIN YOUR CHOICE BASED ON YOUR READINGS IN BOTH TEXTS. GIVE PAGE REFERENCES WHERE POSSIBLE. LASTLY, EXPLAIN WHY EACH OF THE OTHER CHOICES IS INCORRECT. 87. The exact distribution for the hypothesis test is: a. the normal distribution b. the Student's t-distribution c. the uniform distribution d. the exponential distribution