ࡱ> lnkp xNbjbj =RF%**mmmmm8m"7778m:m:m:m:m:m:m=oq:mm+++:mmm77Om!!!+m7m78m!+8m!!xbh70Bhn \d6$mem0mdrararlhharmh "4!L`:m:m m++++ar* 3: Chapter 13 Notes Comparing Means In a study on the The Antihypertensive Effects of Fish Oil, researchers randomly divided 14 male volunteers with high blood pressure to one of two treatments. The first treatment was a 4-week diet that included fish oil and the second was a 4-week diet that included regular oil. The response variable was the reduction in diastolic blood pressure (mm of mercury). The results of this study are shown below. [Source: New England Journal of Medicine 320 (1989): 1037-1043; cited in The Statistical Sleuth, by Ramsey and Schafer, page 23.] Can we conclude that fish oil is better than regular oil for reducing diastolic blood pressure? Fish Oil: 8 12 10 14 2 0 0  EMBED Equation.DSMT4 = 6.57 Regular Oil: -6 0 1 2 -3 -4 2  EMBED Equation.DSMT4 = -1.14 Hypothesis Tests for the Difference of Two Means In chapters 11 and 12, we were comparing the mean of a population to some hypothesized value. In chapter 13, we are comparing the means of 2 populations (or 2 treatments) to each other, rather than to some hypothesized value. For example, we may want to compare the salaries of men and women in a certain industry or investigate if a new cholesterol drug is more effective than the currently used drug. In the first case, we would take a random sample of men and a random sample of women and compare their average salaries. In the second case, researchers would randomly assign the drugs to subjects in the study and compare the average reduction in cholesterol level for both drugs. When we are conducting a hypothesis test for 2 population or treatment means, we are usually testing the following null hypothesis: Before we continue, it is important that we consider how the samples were selected (or how the treatments were assigned). Two samples (or treatment groups) are said to be _____________________ if the selection of individuals that make up one sample (or treatment group) does not influence the selection of those in the other sample (other treatment group). However, when observations in the first sample (or treatment group) are matched in some meaningful way with observations in the second sample (second treatment group), the data is said to be ______________. For example, if we are assigning subjects to treatments in an experiment by drawing names out of a hat, the treatment groups are independent. However, if we use blocking (by finding the two most similar subjects and splitting up into each treatment group), then our data is paired (not independent). In this chapter (24) we will focus on the independent case. When we are comparing 2 means from independent samples, we usually do not perform a simulation to estimate the p-value. Instead we rely on our understanding of the sampling distribution of  EMBED Equation.DSMT4 : The 5-Step Testing Procedure: Using yesterdays example 1. At first glance, it appears that the true mean reduction using fish oil  EMBED Equation.DSMT4  is greater than the true mean reduction using regular oil ( EMBED Equation.DSMT4 ) since  EMBED Equation.DSMT4 . However, it is possible that in reality the oils have the same effect and the difference we observed was due to randomization variability. To decide I will conduct a 2 sample t-test for  EMBED Equation.DSMT4  ( EMBED Equation.DSMT4  = .05). 2. Ho:  EMBED Equation.DSMT4  = 0 Ha:  EMBED Equation.DSMT4  > 0 3. Conditions: a. The two treatments were randomly assigned to individuals? Given Note: This condition is to make sure the sample means are independent for experiments. If we are analyzing the results of a two independent samples, we would need to check two conditions for independence: Independent random samples of _____ and _____. Samples are less than 10% of populations. b. Both populations are approximately normal or large sample sizes?   Since both NPPs are roughly linear, it is OK to assume the populations are approximately normal. Note: If we use the simulation approach, there is no normality condition. 4.  EMBED Equation.DSMT4  with  EMBED Equation.DSMT4  p-value =  EMBED Equation.DSMT4 =  EMBED Equation.DSMT4 = P(t > 3.06) At this point, we have 2 options (neither of which is calculating the df by hand with the formula above): 1. Use the conservative estimate for df which is the smaller of  EMBED Equation.DSMT4 . However, this will give the test less power. 2. Using the Calculator: Enter the data, choose the proper alternative hypothesis ( EMBED Equation.DSMT4 ), say no for pooling, and calculate. This will use the ugly df formula and give more power. Using the conservative estimate: p-value = tcdf(3.06,99999,6) = .0111 Using the test menu, p-value = .0065 (df = 9.26) 5. Since p-value < alpha, we reject the null hypothesis and conclude that fish oil is better than regular oil for reducing blood pressure. Each person in a random sample of 228 male teenagers and a random sample of 306 female teenagers was asked how many hours he or she spent online in a typical week (Ipsos, 1-25-2006). Based on the summary statistics below, can we conclude there is a significant difference in the mean number of hours online per week for the two genders? Females: mean = 14.1 hours standard deviation = 11.8 hours Males: mean = 15.1 hours standard deviation = 11.4 hours Based on your decision, which type of error, Type I or Type II, could you have made? Explain. Confidence Intervals for the Difference of Two Means Review: A confidence interval gives a range of plausible values for a population parameter; in this case, the true difference of two means. To test if a new fertilizer is effective, 20 similar tomato plants were randomly assigned to be planted in either regular soil or soil with fertilizer. After 3 months, the total weight of the tomatoes from each plant was measured. These figures (in pounds) are below. Construct and interpret a 95% confidence interval for the difference in mean production for the two conditions. Does the interval give evidence that soil with fertilizer is more effective than regular soil for growing tomato plants? regular soil: 7, 9, 5, 8, 7, 6, 7, 10, 8, 6 soil w/ fertilizer: 9, 9, 11, 7, 12, 8, 10, 10, 13, 10 4 step procedure: 1. We are trying to estimate the true difference in the mean weight of tomatoes grown in regular soil and mean weight of tomatoes grown in fertilized soil ( EMBED Equation.DSMT4 ). Our best guess is  EMBED Equation.DSMT4  = -2.6, but because of randomization variability this is probably incorrect. Instead, we will calculate a 95% 2 sample t-interval for  EMBED Equation.DSMT4 . 2. Conditions: a. Treatments randomly assigned? Given b. Populations approximately normal?   Since both NPPs are roughly linear, it is reasonable to assume the populations are both approximately normal. 3. CI =  EMBED Equation.DSMT4  =  EMBED Equation.DSMT4  =  EMBED Equation.DSMT4  Note: the df calculations are the same as for the 2 sample t test. The interval above is using the conservative df. You may also use the Test Menu to calculate the more exact 2 sample t interval, however you should know how to do it by hand if necessary (e.g. on a MC question). 4. I am 95% confident that the interval from -4.27 to -0.93 captures the true difference in mean weight for tomatoes grown in regular soil and for tomatoes grown in fertilized soil. Since the interval of plausible values is entirely below 0 (0 = no difference), we can conclude that the mean weight is greater for fertilized plants ( EMBED Equation.DSMT4 =.025). A recent study (Pediatrics [2004]: 112-118) investigated the effect of fast food consumption on other dietary variables. For a sample of 663 teens who reported that they did not eat fast food during a typical day, the mean caloric intake was 2258 with a SD of 1519. For a sample of 413 teens who reported that they did eat fast food on a typical day, the mean caloric intake was 2637 with a SD of 1138. Construct and interpret a 99% CI for the true difference in mean caloric intake for these two types of teens. Comparing Means with Paired Data Recall that two samples are said to be independent if the selection of one group has no effect on the selection of the other group. However, in many cases, using independent samples is not the best way to investigate a difference between two populations or treatments. For example, in our experiment to determine if caffeine affects pulse rate, we could have assigned the treatments completely at random. However, since other factors affect pulse rate (such as gender, resting pulse rate, etc.), the variability in pulse rate caused by the caffeine may be obscured by the variability created by the other factors. To even out the effects of these other factors, we paired subjects by gender and initial pulse rate. When we analyze the results, we look at the difference in each pair. After subtracting, the variability created by these extraneous factors has been removed and any remaining variability between the treatments groups can be explained by the caffeine or random chance. Paired data occurs in many ways, especially in experiments: measurements on a single individual before and after a treatment measurements of two individuals who are paired based on some similar characteristic that might otherwise confound the results (i.e. blocks of size 2 = matched pairs) measurements of naturally occurring pairs (e.g. twins, right hand/left hand) Paired data can also occur in observational studies: measurements at two different locations in a river (e.g. to measure pollution levels) measurements of husbands and wives measurements on the inside and outside of a house (e.g. to measure insulation) Paired samples often provide more information than independent samples because extraneous variables are screened out. In other words, if the observations are paired in a meaningful way, a paired design will have more power than a 2 independent samples design. To compare two SAT schools (A and B), 14 subjects were recruited and paired by PSAT score. One person from each pair was randomly assigned to go to school A and the other was assigned to school B. After 3 months, they all took the same SAT test (results below). Is there a significant difference between the schools? Pair #ABDifference119501940217101680316301590415401500515001490614801500714201370 The five steps: For this test, we are analyzing the differences in each pair, not the individual observations. Thus, a matched pairs t test is really a special case of the one sample t test that we learned about in chapter 23. 1. At first glance it seems that the true mean difference in SAT scores at the two schools ( EMBED Equation.DSMT4 ) is not 0 since  EMBED Equation.DSMT4 = 22.9. However, it is possible that the schools are in reality equally effective and that we got an sample mean difference this large due to randomization variability. To decide, I will conduct a matched pairs t-test for EMBED Equation.DSMT4  ( EMBED Equation.DSMT4  = .05). 2. Ho:  EMBED Equation.DSMT4  = 0 Ha:  EMBED Equation.DSMT4  EMBED Equation.DSMT4  0 3. Conditions: a. The treatments were randomly assigned to members of the pair? Given. Note: for surveys we need to check different conditions for independence: the differences can be viewed as a random sample from population of differences The sample size is less than 10% of the population size b. The population of differences is approximately normal or large number of differences?  Since the NPP of the differences is roughly linear, it is reasonable to assume the population of differences is approximately normal. 4. Test Statistic:  EMBED Equation.DSMT4  with df =  EMBED Equation.DSMT4 -1 Note: You can use the 1 sample t-test in the Test menu to do the calculations, but you should also know how to do this by hand. 5. Since p-value < alpha, we reject Ho and conclude that there is a mean difference in the SAT scores at the two schools. What if we did not know that the data were paired and did a 2 independent samples test? We wouldnt see the difference in the schools: t = 0.31, p = .7636, df = 12 Compare the boxplots of A vs. B. There are 3 major sources of variability: The variability between the schools (the difference in the centers of the boxplots), the variability of the individual students (the widths of the boxplots), and randomization variability. When we pair the subjects and subtract their scores, the variability of the individual students is eliminated (mostly) since we are comparing students of roughly the same ability. The only variability left is due to the difference in schools or due to the randomization and the hypothesis test allows us to decide if the difference is due to randomization variability. Removing this source of variability increases the powerwe have a better chance to see if there really is a difference in the schools! Confidence Intervals for Paired Data New homes often come with energy guarantees. That is, the builder guarantees that the cost of heating and cooling the house will be below a certain cost. When there is a complaint, the builder will estimate the insulating power of the house by taking temperature measurements inside and outside the house at randomly selected locations (such as inside and outside of a particular window). Use a 95% confidence interval to estimate the average difference in temperature inside and outside of the following house. Location #Inside ( EMBED Equation.DSMT4 )Outside ( EMBED Equation.DSMT4 )172.148.6273.551.1372.650.7474.550.9561.249.3670.549.7770.950.1 Comparing Two Proportions Is Yawning Contagious? According to the popular show Mythbusters, the answer is yes. In the March 9, 2005 episode the Mythbusters team presented the results of an experiment involving 50 subjects. All of the subjects were placed in a booth for an extended period of time and monitored by a hidden camera. Two-thirds of the subjects were given a yawn seed by one of the experimenters; that is, the experimenter yawned in the subjects presence prior to leaving the booth. The remaining subjects were given no yawn seed. What were the results? Of the 16 subjects who had no yawn seed, 4 yawned (25%). Of the 34 subjects given a yawn seed, 10 yawned (29.4%). Adam Savage and Jamie Hyneman, the co-hosts of Mythbusters, used this 4.4% difference to conclude that yawning is contagious. What do you think? (from Daren Starnes) In most cases we do not calculate p-values with a simulation. Instead, we use the sampling distribution of  EMBED Equation.DSMT4 . Just like inferences for  EMBED Equation.DSMT4  depend on the distribution of  EMBED Equation.DSMT4 , inferences for  EMBED Equation.DSMT4  depend on the distribution of  EMBED Equation.DSMT4 : 1.  EMBED Equation.DSMT4  2.  EMBED Equation.DSMT4  3. If both  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  are large then the distribution of  EMBED Equation.DSMT4  is approximately normal check:  EMBED Equation.DSMT4  are all at least 10. The five steps: 1. At first glance, it appears that the proportion of people who yawn when given a yawn seed ( EMBED Equation.DSMT4 ) is greater than the proportion people who yawn when not given a yawn seed ( EMBED Equation.DSMT4 ) since  EMBED Equation.DSMT4  > 0 However, it is possible that the true proportions are the same and we got a difference this large due to randomization variability. To decide, I will conduct a 2 sample z-test for  EMBED Equation.DSMT4 ( EMBED Equation.DSMT4  = .05). 2. Ho:  EMBED Equation.DSMT4 = 0 and Ha:  EMBED Equation.DSMT4 >0 3. Conditions: Treatments Randomly Assigned? Not stated, must assume. Note: If we are comparing 2 independent random samples, we must check that they are both random samples from the populations of interest and that the samples are less than 10% of the populations. The sample sizes are large?  EMBED Equation.DSMT4  = 10  EMBED Equation.DSMT4  = 24  EMBED Equation.DSMT4 = 4  EMBED Equation.DSMT4  = 12 Not all are greater than 10, so we will proceed with caution. Note: We use  instead of p, since we do not know the value of p. These four quantities are simply the number of successes and failures for each sample. 4. Calculations: When we are performing a test for Ho:  EMBED Equation.DSMT4  ( EMBED Equation.DSMT4 ), we are assuming that the population proportions are equal. However, we do not know the numerical value of  EMBED Equation.DSMT4  or  EMBED Equation.DSMT4 . To estimate this value, we pool (combine) both estimates to create a pooled estimate:  EMBED Equation.DSMT4  Thus, the standard error is:  EMBED Equation.DSMT4  and the test statistic is:  EMBED Equation.DSMT4  Note: The 2 prop z test in the test menu automatically pools the two proportions. 5. Since the p-value > alpha (.37 > .05) we fail to reject  EMBED Equation.DSMT4  and cannot conclude that yawning is contagious. A 1997 article in USA Today reported that in a survey of 150 males ages 20-24, 72 live with their parents and in a similar survey of 150 females ages 20-24, 51 live with their parents. Does this give evidence that 20-24 year old males are more likely to live with their parents than females of the same age?     PAGE  PAGE 1 !"  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