ISCAM-Chapter3 AC

Chance/Rossman, 2018 ISCAM III Investigation 3.7 209 Investigation 3.7: Is Yawning Contagious? The folks at MythBusters , a popular television program on the Discovery Channel, investigated whether yawning is contagious by recruiting fifty subjects at a local flea market and asking them to sit in one of three small rooms for a short period of time. For some of the subjects, the attendee yawned while leading them to the room (planting a yawn ³seed´), whereas for other subjects the attendee did not yawn. As time passed, the researchers watched (via a hidden camera) to see which subjects yawned. (a) Identify the explanatory variable (EV) and the response variable (RV) in this study. EV: RV: (b) Define the relevant parameter of interest, and state the null and alternative hypotheses for this study. Be sure to clearly define any symbols that you use. Parameter: H 0 : H a : In the study they found that 10 of 34 subjects who had been given a yawn seed actually yawned themselves, compared with 4 of 16 subjects who had not been given a yawn seed. (c) Create a two-way table summarizing the results, using the explanatory variable as the column variable . Totals Totals (d) Explain how you would carry out a simulation analysis to approximate a p-value for this study. [ Hint : How many cards? How many of each type? How many would you deal out? What would you record? How would you find the p-value?] weather them n' Seed or net Yawn or not 㱺 I The doffenme in prohahyuf yawing tickets - Bros ) two Gwp been a G - oaf nuked Teed > Tmsn 㱺 From > u seed nosed yawn 10 4 14 no yawn 24 12 36 34 16 50 ← d im - Em - 4g = 0.044 = .

* Differ . µ pnohahrj or = Diff .nu in ppnluhipnprhin II. a - a. " a thinner - nie i Ik n . . . . . " wth µ = So p - values Pl low mm ) M= 14 pl X 7101 n - - 34

Chance/Rossman, 2018 ISCAM III Investigation 3.7 211 (g) How many ways altogether are there to randomly assign these 50 subjects into one group of 34 (yawn-seed group) and the remaining group of 16 (no-yawn-seed group)? [ Hint : Recall what you saw earlier with the binomial distribution and counting the number of ways to obtain S successes and F failures in n trials. See the Technical Details in Investigation 1.1.] (h) Now consider the 14 successes and the 36 failures. How many ways are there to randomly select 10 of the successes? How many ways are there to randomly assign 24 of the failures to be in the yawn seed group? How should you combine these two numbers to calculate the total number of ways to obtain 10 successes and 24 failures in the yawn-seed group, the configuration that we observed in the study? Successes Failures Total: (i) To determine the exact probability that random assignment would produce exactly 10 successes and 24 failures into the group of 34 subjects, divide your calculation in (h) by your calculation in (g). (j) Explain why your answer to (i) is not yet the p-value for this study. Result: The probability of obtaining k successes in Group A, with n observations, when sampled from a two-way table with N observations, consisting of M successes and N ± M failures is: P(X = k ) = C( M , k ) × C( N ± M , n ± k ) / C( N , n ) where C( N , n ) = N !/[ n !( N ± n !)] is the number of ways to choose n items from a group of N items. X represents the number of successes randomly selected for group A. X is a hypergeometric random variable. Also note E(X) = n ( M / N ) and SD(X) = )] 1 ( /[ ) )( ( 2 ± ± ± N N n N M N nM . In this study, we had N = 50 subjects and we defined yawning to be success so M = 14. We also arbitrarily chose to focus on the yawn-seed group, so n = 34. This calculation works out the same if you had defined ³not yawning´ to be a success and/or if you had focused on the 16 people in the no - yawn-seed group. You just need to make sure you count consistently. O CoE O ( 58k go.nu . - 492*10 " x lool = ( Yf ) * 13241=125162700 pl # ok fzg/ I IM PIX 7107 Iplx - lulxplx -111 + plash ) -11714=13 ) tpl # 14 ) - -

Chance/Rossman, 2018 ISCAM III Investigation 3.7 212 We will continue to define the p-value to be the probability of obtaining results at least as extreme as those observed in the actual study. Because we expected more yawners in the yawn-seed group, the p- value is the probability of randomly assigning at least 10 of the yawners in the yawn-seed group. So far you have found P(X = 10) = C(14, 10) × C(36, 24) / C(50, 34) = 0.2545. (k) Calculate P(X = 11), P(X = 12), P(X = 13), and P(X = 14) using the hypergeometric probability formula. P(X = 11) P(X = 12) P(X = 13) P(X = 14) Why do we stop at 14? (l) Sum all five probabilities together (including P(X = 10)) to determine the exact p-value for the yawning study. How does this p-value compare to the empirical p-value from the applet simulation? Write a one or two sentence interpretation of this p-value. Exact p-value: Comparison: Interpretation: Definition: Using the hypergeometric probabilities to determine a p-value in this fashion for a two- way table is called Fisher¶s Exact Test , named after R. A. Fisher. (m) Calculate this hypergeometric probability using technology (see Technology Detour on next page). (n) Set up and carry out the calculation to determine the exact p-value where you define the success to be ³not yawning´ and the group of interest to be the yawn seed group. (o) Set up and carry out the calculation to determine the exact p-value, where you focus on the number that did not yawn in the no-yawn-seed group. Show that you obtain the same exact p-value as before. 0.1165 = ipb " " - b. 0702 0.0198 = = = 0.0015 0.5128 V. sina.MN T - * R - Cohen - Iscamhypwprubfk > 24 , f- 50,1=36 ,n=34 17124 Leg ) hairtail . - TRUE ) p( nor More ) , f- SO , a- 16 , f- 36 Iscamhyperptrobl Koh ,N=5UM=3f,a= 16 lower tail - - FALSE I

p - van can the written as to " ) 1311 = 0.5/28

Chance/Rossman, 2018 ISCAM III Investigation 3.7 213 Technology Detour ʹ Calculating Hypergeometric Probabilities ;Fisher’s Exact TestͿ In R , the iscamhyperprob function takes the following inputs: x k , the observed value of interest (or the difference in conditional proportions, assumed if value is less than one, including negative) x total , the total number of observations in the two-way table x succ , the overall number of successes in the table x n , the number of observations in ³group A´ x lower.tail , a Boolean which is TRUE or FALSE For example: iscamhyperprob(k=10, total=50, succ=14, n=34, lower.tail=FALSE) Analyzing Two-way Tables applet x Check the box for ShRZ FLVheU¶V E[acW TeVW in the lower left corner. x A check box will appear for determining the two-sided p-value Discussion : You should see that there are several equivalent ways to set up the probability calculation. Make sure it is clear how you define success/failure and which group you are considering ³group A.´ This will help you determine the numerical values for N , M , and n in the calculation. Below is a graph of the Hypergeometric distribution with N = 50, M = 14, and n = 34. Using probability rules, you can show that the expected value of this distribution is n N M u ) / ( = (14/50) × 34 = 9.52 yawners in yawn seed group and the standard deviation of the probability distribution is the square root of n × ( M/N ) × ( N ± n )/ N × ( N ± M )/( N ± 1) = 1.496 yawners. (p) Compare this graph and the mean and standard deviation values to your simulation results.

Chance/Rossman, 2018 ISCAM III Investigation 3.7 215 Practice Problem 3.7A (a) For the Mythbusters¶ study (p-value > 0.5), is it reasonable to conclude from this study that we have strong evidence that yawning is not contagious? Explain. (b) Explain, in this context, what is meant in the Study Conclusions box by ³the researchers could try the study again with a larger sample size to increase the power of their test´ and why that is a reasonable recommendation here. (c) To calculate the p-value here, why are we using the hypergeometric distribution instead of the binomial distribution? Practice Problem 3.7B Reconsider the Dolphin Therapy study (Investigation 3.6). Dolphin Therapy Control Group Total Showed substantial improvement 10 3 13 Did not show substantial improvement 5 12 17 Total 15 15 30 Continue to focus on the number of improvers randomly assigned to the dolphin group, and represent this value by X. (a) When the null hypothesis is true, the random variable X has a hypergeometric distribution. Specify the values of N , M , and n . (b) Calculating the exact p-value involves finding P(X __________). [ Hint : Fill in the blank with an inequality symbol and a number.] (c) Calculate this exact p-value, either by hand or with technology. Comment on whether this p-value is similar to the approximate one from your simulation results. (Be sure it¶s clear how you calculated this value.) (d) Suppose that the dolphin study had involved twice as many subjects, again with half randomly assigned to each group, and with the same proportion of improvers in each group. Determine the exact p-value in this case, and comment on whether/how it changes from the p-value with the real data. Explain why this makes sense.

Chance/Rossman, 2018 ISCAM III Investigation 3.8 216 Investigation 3.8: CPR vs. Chest Compressions For many years, if a person experienced a heart attack and a bystander called 911, the dispatcher instructed the bystander in how to administer chest compression plus mouth-to-mouth ventilation (a combination known as CPR) until the emergency team arrived. Some researchers believe that giving instruction in chest compression alone (CC) would be a more effective approach. In the 1990s, a randomized comparative experiment was conducted in Seattle involving 518 cases (Hallstrom, Cobb, Johnson, & Copass, New England Journal of Medicine , 2000): In 278 cases, the dispatcher gave instructions in standard CPR to the bystander, and in the remaining 240 cases the dispatcher gave instructions in CC alone. A total of 64 patients survived to discharge from the hospital: 29 in the CPR group and 35 in the CC group. (a) Identify the observational units, explanatory variable, and response variable. Is this an observational study or an experiment? Observational units: Explanatory: Response: Type of study: Observational Experimental (b) Construct a two-way table to summarize the results of this study. Remember to put the explanatory variable in the columns. (c) Calculate the difference in the conditional proportions who survived (CC ± CPR). Does this seem to be a noteworthy difference to you? (d) Use technology to carry out Fisher¶s Exact Test (by calculating the corresponding hypergeometric probability) to assess the strength of evidence that the probability of survival is higher with CC alone as compared to standard CPR. Write out how to calculate this probability, report the p-value, and interpret what it is the probability of. p-value = P(X ) = where X follows a hypergeometric distribution with N = , M = , and n = Interpretation:

Chance/Rossman, 2018 ISCAM III Investigation 3.8 218 (j) Suppose you had defined the parameter by subtracting in the other direction (e.g., CPR ± CC instead of CC ± CPR). How would that change: (i) the observed statistic? (ii) the test statistic? (iii) the alternative hypothesis? (iv) the p-value? (v) confidence interval? Practice Problem 3.8 (a) Researchers in the CPR study also examined other response variables. For example, the 911 dispatcher¶s instructions were completely delivered in 62% of episodes assigned to chest compression plus mouth-to-mouth compared to 81% of the episodes assigned to chest compression alone. (i) Calculate the difference in conditional proportions and compare it to the original study. (ii) Without calculating, do you suspect the p-value for comparing this new response variable between the two groups will be larger or smaller or about the same as the p-value you determined above? Explain your reasoning. (b) The above study was operationally identical to that of another study and the results of the two studies were combined. Of the 399 combined patients randomly assigned to standard CPR, 44 survived to discharge from the hospital. Of the 351 combined patients randomly assigned to chest compression alone, 47 survived to discharge. (i) Calculate the difference in conditional proportions and compare it to the original study. (ii) Without calculating, do you suspect the p-value for this comparison will be larger, smaller, or the same as the p-value you determined? Explain your reasoning.

Chance/Rossman, 2018 ISCAM III Investigation 3.9 219 SECTION 4: OTHER STATISTICS Investigation 3.9: Peanut Allergies Peanut allergies have increased in prevalence in the last decade, but can they be prevented? Even among infants with a high risk of allergy? Is it better to avoid the problematic food or to encourage early introduction? Du Toit et al. ( New England Journal of Medicine , Feb. 2015) randomly assigned U.K. infants (4-11 months old) with pre-existing sensitivity to peanut extract to either consume 6 g of peanut protein per week or to avoid peanuts until 60 months of age. The table below shows the results for infants who were not initially sensitized to peanuts and whether or not the child had developed a peanut allergy at 60 months. Peanut avoidance Peanut consumption Total Peanut allergy 11 2 13 No allergy 172 193 365 Total 183 195 378 (a) Calculate the proportion of children developing a peanut allergy in each group. Does this appear to be a large difference to you? (b) Use Fisher¶s Exact Test to investigate whether these data provide convincing evidence that the probability of developing a peanut allergy is larger among children who avoid peanuts for the first 60 months. [ Hint : State the hypotheses in symbols and in words. Define the random variable and outcomes of interest in computing your p-value.] Do you consider this strong evidence that the peanut consumption effectively deters development of a peanut allergy in this population? (c) Would you feel any differently about the magnitude of the difference in proportions if the conditional proportions developing a peanut allergy had been 0.500 and 0.55? Explain. Discussion : When the baseline rate (probability) of success is small, an alternative statistic to consider rather than the difference in the conditional proportions (which will also have to be small by the nature of the data) is the ratio of the conditional proportions. First used with medical studies where ³success´ is often defined to be an unpleasant event (e.g., death), this ratio was termed the relative risk . - -0 - - T I - - Rauf ' 4183=01101 Fan = 4450£03 xn hypergeometric III } ! I 9%4=724 - - -

Chance/Rossman, 2018 ISCAM III Investigation 3.9 221 But can we apply a mathematical model to this distribution? (i) Why does it make sense that the mean of the simulated relative risks is close to the value 1? [ Hint : Remember the assumption behind your simulation analysis.] (j) You should notice skewness in the distribution of relative risk values. Explain why it is not surprising for the distribution of this statistic to be skewed to the right (especially with smaller sample sizes). Note: If the number of successes equals zero, the applet adds 0.5 to each cell of the table before calculating the relative risk. (k) In fact, this distribution is usually well modeled by a log normal distribution. To verify this, check the ln relative risk box (in the lower left corner) to take the natural log of each relative risk value and display a new histogram of these transformed values. Describe the shape of this distribution. Is the distribution of the lnrelrisk well modeled by a normal distribution? (l) What is the mean of the simulated lnrelrisk values? Why does this value make sense? (m) What is the standard deviation of the lnrelrisk values? (n) Calculate the observed value of ln( p Ö 1 / p Ö 2 ) for this study (but don¶t round up). Where does this value fall (near in the middle or in the tail) of this simulated distribution of lnrelrisk values? Has the empirical p-value changed? (o) If you found the empirical p-value using ln( p Ö 1 / p Ö 2 ), it would be identical to the empirical p-value found in (h). Why? What did change about the distribution? [ Hint : What percentage of the simulated ln rel risk values are more extreme than ln(2.1), how does this compare to (h)?] Hae . Real Risk > I H-H-ocitrue.ee I Hoo . Taro - - Icons 㱺 Real Risk - - foin Al • Ha : Taro > Theon . In C O - 0601/0.0103 ) = In ( 5.8611 = 1.77 In ( 2nd ) = 0.741937

Chance/Rossman, 2018 ISCAM III Investigation 3.9 222 Theoretical Result: It can be shown that the standard error of the ln relative risk is approximated by D B B C A A p p SE ² ± ² ² ± ¸ ¸ ¹ · ¨ ¨ © § 1 1 1 1 Ö Ö ln 2 1 where A , B , C , and D are the observed counts in the 2 × 2 table of data, with A and B representing the number of ³successes´ in the two groups. Having this formula allows us to determine the variability from sample to sample without conducting the simulation first. (p) Calculate the value of this standard error of the ln(rel risk) for this study. Interpret this value and compare it to the standard deviation from your simulated lnrelrisk values. (q) You may find this approximation is in the ballpark but not all that close. What assumption is made by the simulation that is not made by this formula? What if you made the same assumption in this formula? [ Hint : Think pooled p Ö .] (r) Now that you have a statistic (ln rel risk) that has a sampling distribution that is approximately normal, what general formula can we use to determine a confidence interval for the parameter? (s) Calculate the midpoint, 95% margin-of-error, and 95% confidence interval endpoints using the observed value of ln(rel risk) as the statistic and using the standard error calculated in (p). (t) What parameter does the confidence interval in (s) estimate? (u) Exponentiate the endpoints of this interval to obtain a confidence interval for the ratio of the probabilities of developing a peanut allergy between these two treatments. Interpret this interval. (v) Is zero in this interval? Do we care? What value is of interest instead? A B C D A+C B+D - SE = 1k¥ I 0.762 Skip O Lyn - statistic I z* SE / In 15.864 1.768 I 1.96 ( 0.762 ) s ( o . 025 , 3.291 In 1 Real Risk ) 95 X CI for Real Risk 60025 ! ' fo . ogseseg , = ( l - 32 , 26 . I )

Chance/Rossman, 2018 ISCAM III Investigation 3.9 224 Practice Problem 3.9A (a) For the peanut allergy study, find a 95% confidence interval for the probability of not developing a peanut allergy comparing the consumption treatment to the avoidance treatment. (b) Provide a one-sentence interpretation of the interval in (a). (c) Is the interval the same or closely related to the one you found in Investigation 3.9? Does one interval provided strong evidence of a treatment effect? Which interval would you report to new parents? (d) The article reports that ³the power to detect a difference in risk of 30 percentage points was 80.0%.´ Explain what this means in your own words. Practice Problem 3.9B A multicenter, randomized, double-blind trial involved patients aged 36-65 years who had knee injuries consistent with a degenerative medial meniscus tear (Shivonen et al., New England Journal of Medicine , 2013). Patients received either the most common orthopedic procedure (arthroscopic partial meniscectomy, n 1 = 70) or sham surgery that simulated the sounds, sensations, and timing of the real surgery ( n 2 = 76). After 12 months, 54 of those in the treatment group, reported satisfaction, compared to 53 in the sham surgery. (a) Calculate and interpret a confidence interval for the ratio of the probabilities (relative risk) of satisfaction for these two procedures. (b) What does your interval in (a) indicate about whether those receiving the orthopedic surgery are significantly more likely that those receiving a sham surgery to report satisfaction after 12 months? Explain your reasoning. Summary of Inference for ͞Relative Risk͟ Statistic: ratio of conditional proportions (typically set up to be larger than one) = 1 Ö p / 2 Ö p Hypotheses : H 0 : S 1 / S 2 = 1; H a : S 1 / S 2 <, >, or 1 p-value : Fisher¶s Exact Test or normal approximation on ln( 1 Ö p / 2 Ö p ) Confidence interval for S 1 / S 2 : exponentiate endpoints of » ¼ º « ¬ ª ² ± ² ² ± r D B B C A A z p p 1 1 1 1 * ) Ö / Ö ln( 2 1 Note: The confidence interval for the relative risk will not necessarily be symmetric around the statistic.

Chance/Rossman, 2018 ISCAM III Investigation 3.9 225 Technology Detour ʹ Simulating Random Assignment (two-way tables) We can select observations from a hypergeometric distribution for the cell 1 counts and then compute the cell 2 counts and the number of failures based on the fixed row and column totals. With this information you can compute the difference in conditional proportions, relative risk, etc. We show how to calculate p Ö unvac below, the rest is up to you. Also keep in mind you can use ³log´ to calculate the natural logs of values. Also recall how you created a Boolean expression in Investigation 3.1 to find the p-value from the simulated results. In R > VacInfCount=rhyper(10000, 210, 4985, 2584) x 210 is the number of successes ( M ) x 4985 is the number of failures ( N ± M) x 2584 is the sample size ( n ) > UnvacInfCount = 210-VacInfCount > Unvacphat = UnvacInfCount/2584