Chap 11 Expect_The_Unexpected_A_First_Course_In_Biostatist..._----_(Statistics) (4)

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Chapter 11 Paired Samples In this chapter, we compare the means of two dependent populations using confidence intervals and hypothesis testing. Typical examples of dependent data sets are measurements made on the same individuals “before” and “after” a certain treatment: the weight before and after a diet program, the blood pressure before and after a physical exercise, etc. Other examples of dependent data sets are measurements made on the same individuals using two different treatments. In both cases, the observations come in pairs, and together they constitute a “paired sample”. 11.1 Confidence Intervals for μ D Let ( X 1 , Y 1 ) , ( X 2 , Y 2 ) , . . . , ( X n , Y n ) be the paired observations made on n individuals before and after a certain treatment (or when using two different treatments). We assume that X 1 , X 2 , . . . , X n is a random sample from a population whose mean is denoted by μ X , and Y 1 , Y 2 , . . . , Y n is a random sample from a population whose mean is denoted by μ Y . We would like to compare μ X and μ Y by calculating a confidence interval for the difference μ D = μ X - μ Y . If this interval contains mostly positive values, then we can say that μ X is larger than μ Y , with a certain confidence. On the other hand, if the interval contains mostly negative values, then μ X is smaller than μ Y , with a certain confidence. If the interval contains both positive and negative values, no conclusion can be drawn. We first calculate the differences D 1 = X 1 - Y 1 , D 2 = X 2 - Y 2 , . . . , D n = X n - Y n . These differences constitute a random sample from a population whose mean is μ D . This random sample is used for drawing statistical 191 Balan, R., & Lamothe, G. (2017). Expect the unexpected : A first course in biostatistics (second edition). World Scientific Publishing Company. Created from ottawa on 2023-12-03 18:32:54. Copyright © 2017. World Scientific Publishing Company. All rights reserved.

192 Expect the Unexpected: A First Course in Biostatistics conclusions about μ D . We denote by ¯ D the sample average and by S 2 D the sample variance of the difference data set, i.e. ¯ D = 1 n n X i =1 D i and S 2 d = 1 n - 1 n X i =1 ( D i - ¯ D ) 2 . We denote by ¯ d and s 2 d the observed values of ¯ D and S 2 d . Assuming that the differences D 1 , D 2 , . . . , D n are normally distributed, the 100(1 - α )% -confidence interval for μ D is given by formula (8.7): ¯ d ± t s d √ n where t is the value found in Table 18.4 such that P ( - t ≤ T ≤ t ) = 1 - α and T is a random variable with a T ( n - 1) distribution. Example 11.1. One of the standard ways of measuring the lung function is FEV 1 , the forced expiratory volume in one second (the total volume of air blown in one second). The FEV 1 is different in males and females and slows down with age, with a peak of 4.5 l at the age of 25. Smoking speeds up this decline. The effects of smoking on the decline of FEV 1 are studied in [51]. In this example, we want to show that smoking cessation improves the lung function in 3 months’ time. The following table gives the FEV 1 values for 9 males in their mid 30’s before quitting smoking (the x measurement), and 3 months after quitting (the y measurement), as well as the differences d = x - y : x i ( FEV 1 before) y i ( FEV 1 after) Difference d i = x i - y i 2 . 94 4 . 22 - 1 . 28 2 . 90 4 . 12 - 1 . 22 3 . 11 4 . 35 - 1 . 24 2 . 85 4 . 09 - 1 . 24 2 . 93 4 . 15 - 1 . 22 3 . 00 4 . 29 - 1 . 29 2 . 93 4 . 18 - 1 . 25 3 . 03 4 . 29 - 1 . 26 3 . 13 4 . 33 - 1 . 2 After quitting smoking, we notice an increase in the FEV 1 measurements for all subjects. Recall from Section 7.3, that a commonly used tool for assessing the normality of a data set is the QQ-plot. Figure 11.1 gives the QQ-plot of the difference data set. Since the plot shows a linear tendency, we may assume that the difference data set has a normal distribution. Balan, R., & Lamothe, G. (2017). Expect the unexpected : A first course in biostatistics (second edition). World Scientific Publishing Company. Created from ottawa on 2023-12-03 18:32:54. Copyright © 2017. World Scientific Publishing Company. All rights reserved.

Paired Samples 193 Fig. 11.1 QQ-plot of the differences between the FEV 1 levels The sample mean and the sample standard deviation for the before/after measurements and the differences are given below: Before After Difference Mean ¯ x = 2 . 980 ¯ y = 4 . 224 ¯ d = - 1 . 244 Standard Deviation s x = 0 . 0950 s y = 0 . 0949 s d = 0 . 029 Note that ¯ x - ¯ y = ¯ d , but s 2 x + s 2 y 6 = s 2 d . To find a 95% confidence interval for the average difference between the FEV 1 after quitting smoking and before quitting smoking, we use the value t = 2 . 306, which corresponds to a T random variable with a T (8) distribution, such that P ( T > 2 . 306) = 0 . 025. This interval is: - 1 . 244 ± 2 . 306 0 . 029 √ 9 = - 1 . 244 ± 0 . 022 = [ - 1 . 266; - 1 . 222] . Since the interval contains only negative values, we are 95% confident that the average difference μ D is negative, that is the average FEV 1 value ( μ X ) before quitting smoking is smaller than the average FEV 1 value ( μ Y ) after quitting smoking. Based on this data, we can say that smoking cessation induces an increase of the FEV 1 value. Example 11.2. A sample of 15 people participate in a study which com- pares the effectiveness of two drugs for reducing the level of the LDL (low density lipoprotein) blood cholesterol. After using the first drug for two weeks, the decrease in their cholesterol level is recorded as the x -measurement. After a pause of two months, the same individuals are administered another drug for two weeks, and the new decrease in their cholesterol level is recorded as the y -measurement. The following table Balan, R., & Lamothe, G. (2017). Expect the unexpected : A first course in biostatistics (second edition). World Scientific Publishing Company. Created from ottawa on 2023-12-03 18:32:54. Copyright © 2017. World Scientific Publishing Company. All rights reserved.

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194 Expect the Unexpected: A First Course in Biostatistics gives the ( x, y )-measurement pairs, together with the corresponding differ- ence d = x - y . The measurements are in mg/dl . First Drug ( x i ) Second Drug ( y i ) Difference d i = x i - y i 13.1 12.0 1.1 12.3 7.3 5.0 10.0 11.7 -1.7 17.7 12.5 5.2 19.4 18.6 0.8 10.1 12.3 -2.2 11.5 15.2 -3.7 12.6 16.3 -3.7 9.5 10.7 -1.2 12.1 9.8 2.3 18.0 15.3 2.7 7.5 6.4 1.1 6.9 8.5 -1.6 14.5 16.4 -1.9 8.6 7.8 0.8 It appears that the differences verify the assumption of normality. The sam- ple means and the sample standard deviations for the ( x, y ) measurements and the differences are given below: First Drug Second Drug Difference Mean ¯ x = 12 . 253 ¯ y = 12 . 053 ¯ d = 0 . 2 Standard Deviation s x = 3 . 8 s y = 3 . 711 s d = 2 . 809 A 90% confidence interval for μ D is 0 . 2 ± 1 . 761 2 . 809 √ 15 = 0 . 2 ± 1 . 277 = [ - 1 . 077; 1 . 477] . Here, the value t = 1 . 761 is chosen such that P ( T > t ) = 0 . 95, where T is a random variable with a T (14)-distribution. Since the interval contains both positive and negative values, we cannot conclude that the average decrease in the LDL cholesterol level caused by the first drug is larger (or smaller) than that caused by the second drug. 11.2 Hypothesis Testing for μ D As in the previous section, we denote by ( X 1 , Y 1 ) , ( X 2 , Y 2 ) , . . . , ( X n , Y n ) a sample of paired observations made on n individuals. We let D 1 = X 1 - Balan, R., & Lamothe, G. (2017). Expect the unexpected : A first course in biostatistics (second edition). World Scientific Publishing Company. Created from ottawa on 2023-12-03 18:32:54. Copyright © 2017. World Scientific Publishing Company. All rights reserved.

Paired Samples 195 Y 1 , D 2 = X 2 - Y 2 , . . . , D n - X n - Y n be the corresponding differences. We assume that the differences D 1 , D 2 , . . . , D n are normally distributed. We want to compare the average difference μ D = μ X - μ Y with the value 0, by using a test of hypothesis. More precisely, we confront the null hypothesis H 0 : μ D = 0 with one of the alternatives H 1 : μ D > 0 or H 1 : μ D < 0. By Theorem 8.2, if H 0 is true, then the test statistic T 0 = ¯ D - 0 S d / √ n has a T ( n - 1) distribution. We consider separately the two cases: Case (1) H 0 : μ D = 0 versus H 1 : μ D > 0 In this case, we want to reject H 0 and gain evidence that the average difference μ D is positive, i.e. μ X is larger than μ Y . Our decision is based on the p -value. To perform the test, we calculate the observed value of the test statistic: t 0 = ¯ d - 0 s d / √ n . (11.1) If this (positive) ratio is large, then it is unlikely that H 0 is true, and there is some evidence in favor of H 1 . The fact that the ratio is large corresponds to a small p -value, which is calculated as: p -value = P ( T > t 0 ) , where T is a random variable with a T ( n - 1)-distribution. If the p -value is small (usually smaller than 0.05), we reject H 0 and conclude that there is some evidence for H 1 . Otherwise, we do not reject H 0 . Example 11.3. In the study [50], a group of 15 rats were inoculated in each hind leg with a 50 μl injection of a colon tumor cell suspension (DHD/K12/TRb). The cell inoculation grew into a solid tumor at the in- jection site, which was used to model colon cancer. 6 weeks after the tumor inoculations, a drug called doxorubicin (Dox) was administered weekly to the rats. Each rat had one of the two tumors exposed to low-frequency ultrasound for an hour every week. At the end of the treatment time, the tumor volumes were measured in both legs. The table below gives for each rat the volume of the tumor in the leg which received only Dox (the x mea- surement) and the volume of the tumor in the leg which received Dox and ultrasound treatment (the y measurement). In all 15 rats, it was observed that the volume of the insonated (i.e. ultrasound treated) tumor is smaller than the volume of the noninsonated tumor. Balan, R., & Lamothe, G. (2017). Expect the unexpected : A first course in biostatistics (second edition). World Scientific Publishing Company. Created from ottawa on 2023-12-03 18:32:54. Copyright © 2017. World Scientific Publishing Company. All rights reserved.

196 Expect the Unexpected: A First Course in Biostatistics Noninsonated Insonated Difference Tumor ( x i ) Tumor ( y i ) d i = x i - y i 17.5 13.9 3.6 19.9 18.5 1.4 20.7 16.4 4.3 17.7 14.3 3.4 21.5 12.5 9.0 18.7 14.4 4.3 16.5 11.7 4.8 22.1 17.4 4.7 18.6 10.8 7.8 20.5 13.2 7.3 17.6 15.4 2.2 15.7 10.7 5.0 20.5 19.6 0.9 18.3 16.3 2.0 19.7 15.6 4.1 We assume that the difference data set has a normal distribution, an as- sumption which is supported by the QQ-plot. We want to confront the hypothesis H 0 : μ D = 0 (which says that there is no difference between the volumes of the two tumors), with the alternative hypothesis H 1 : μ D > 0 (which says that noninsonated tumors have larger volumes, on average). The sample mean and the sample standard deviation of the ( x, y ) measurements are: ¯ x = 19 . 033, s x = 1 . 858, ¯ y = 14 . 713 , s y = 2 . 682. These statistics are not needed for the analysis. What we need are the sample mean and the sample standard deviation of the differences: ¯ d = 4 . 32 and s d = 2 . 318. The observed value of the test statistic is t 0 = ¯ d - 0 s d / √ n = 4 . 32 - 0 2 . 318 / √ 15 = 7 . 218 . From Table 18.4, we see that P ( T > 2 . 997) = 0 . 005 where T is a random variable with a T (14) distribution. Since the observed value 7.218 is larger than 2.997, we infer that p -value = P ( T > 7 . 218) < 0 . 005 . The p -value being very small, we reject H 0 and conclude that the nonin- sonated tumors have larger volumes, on average, than the insonated tumors. Balan, R., & Lamothe, G. (2017). Expect the unexpected : A first course in biostatistics (second edition). World Scientific Publishing Company. Created from ottawa on 2023-12-03 18:32:54. Copyright © 2017. World Scientific Publishing Company. All rights reserved.

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Paired Samples 197 Case (2) H 0 : μ D = 0 versus H 1 : μ D < 0 In this case, we want to reject H 0 and gain evidence that the average difference μ D is negative, i.e. μ X is smaller than μ Y . To perform the test, we calculate the same ratio t 0 given by (11.1) as in Case (1). If the absolute value of this (negative) ratio is large, then it is unlikely that H 0 is true, and there is some evidence in favor of H 1 . The p -value is calculated as: p -value = P ( T < t 0 ) , where T is a random variable with a T ( n - 1)-distribution. If the p -value is small, we reject H 0 and conclude that there is some evidence for H 1 . Otherwise, we do not reject H 0 . Example 11.4. Exercise therapy has been shown to influence human car- tilage properties (see [4]). Shortly after exercise, it was noticed an elevation of serum levels of cartilage oligomeric matrix protein (COMP). The follow- ing table gives the serum COMP levels for 12 patients before 60 minutes of exercise (the x -measurement) and right after the exercise period (the y -measurement). Before Exercise ( x i ) After Exercise ( y i ) Difference d i = x i - y i 6.32 6.48 - 0 . 16 7.85 8.27 - 0 . 42 12.87 13.26 - 0 . 39 11.27 11.84 - 0 . 57 7.89 8.23 - 0 . 34 15.56 15.87 - 0 . 31 16.34 16.60 - 0 . 26 7.83 8.17 - 0 . 34 9.23 9.61 - 0 . 38 10.22 10.38 - 0 . 16 14.67 14.91 - 0 . 24 15.30 15.61 - 0 . 31 From this data, we notice an increase in the serum COMP levels due to the exercise. The assumption of normality of the differences appears to be verified. The sample means and sample standard deviations for the ( x, y ) measurements and the differences are given below: Before After Difference Mean ¯ x = 11 . 28 ¯ y = 11 . 60 ¯ d = - 0 . 323 Standard Deviation s x = 3 . 56 s y = 3 . 55 s d = 0 . 114 Balan, R., & Lamothe, G. (2017). Expect the unexpected : A first course in biostatistics (second edition). World Scientific Publishing Company. Created from ottawa on 2023-12-03 18:32:54. Copyright © 2017. World Scientific Publishing Company. All rights reserved.

198 Expect the Unexpected: A First Course in Biostatistics We want to test H 0 : μ D = 0 against H 1 : μ D < 0. The observed value of the test statistic is t 0 = ¯ d - 0 s d / √ n = - 0 . 323 - 0 0 . 114 / √ 12 = - 9 . 82 . From Table 18.4, we see that P ( T < - 3 . 106) = 0 . 005 where T is a random variable with a T (11) distribution. Since the observed value - 9 . 82 is smaller than - 3 . 106, we infer that p -value = P ( T < - 9 . 82) < 0 . 005 . The p -value being very small, we reject H 0 and conclude that the serum COMP levels increase after exercise (on average). Technology Component using R : Assume that the paired measurements are assigned to the numerical vectors x and y , respectively. • To calculate the difference d=x-y, we use: d=x-y • To produce the QQ-plot for d, together with the fitted line, we use: qqnorm(d) abline(mean(d),sd(d)) Remark: This procedure gives the plot of the pairs ( z i , y i ) for i = 1 , . . . , n for the variable d, together with the fitted line y = ˆ μ + ˆ σz , where ˆ μ = ¯ d and ˆ σ = s d . It is used for verifying the assumption that the differences are normally distributed. We say that the differences are normally distributed if the plot seems to be linear. • To test the hypothesis H 0 : μ X = μ Y against μ X 6 = μ Y and calcu- late a 95% confidence interval for μ D = μ X - μ Y (assuming that the differences are normally distributed), we use: t.test(x,y,paired=TRUE) Remark: To change the confidence level to 98% (or any other value), we use: t.test(x,y,conf.lev=0.98,paired=TRUE) • To test the hypothesis H 0 : μ X = μ Y against μ X > μ Y (assuming that the differences are normally distributed), we use: Balan, R., & Lamothe, G. (2017). Expect the unexpected : A first course in biostatistics (second edition). World Scientific Publishing Company. Created from ottawa on 2023-12-03 18:32:54. Copyright © 2017. World Scientific Publishing Company. All rights reserved.

Paired Samples 199 t.test(x,y,alternative="greater",paired=TRUE) Remark: This procedure produces also a one-sided confidence interval for μ D which is not discussed in this book. • To test the hypothesis H 0 : μ X = μ Y against μ X < μ Y (assuming that the differences are normally distributed), we use: t.test(x,y,alternative="less",paired=TRUE) Remark: This procedure produces also a one-sided confidence interval for μ D which is not discussed in this book. 11.3 Problems Problem 11.1. Exposure to volatile organic compounds (VOC) which have been identified in indoor air is suspected as a cause for headaches and respiratory symptoms. Indoor plants have not only a positive psychological effects on humans, but may also improve the air quality. Certain species of indoor plants were found to be effective removers of VOCs. The following data gives the benzene level (in ppm) in 10 test chambers measured at the beginning of the study and after 3 days, using the plant Epipremnum aureum (Devils Ivy). Initial Benzene Level ( x i ) Benzene Level after 3 Days ( y i ) 28.4 27.4 27.3 26.3 25.5 25.6 29.4 24.5 30.2 28.7 31.3 29.6 28.6 27.5 28.4 28.4 26.5 23.2 27.3 24.3 Is there any evidence that this species of indoor plants is effective in re- moving the benzene from the indoor air? Justify your answer using a 95% confidence interval and a test of hypothesis. (Verify first that the differences d i = x i - y i , i = 1 , . . . , 10 satisfy the normality assumption.) Problem 11.2. Fibristal (ulipristal acetate) is a drug which was approved by Health Canada in 2014 for the treatment of the signs and symptoms Balan, R., & Lamothe, G. (2017). Expect the unexpected : A first course in biostatistics (second edition). World Scientific Publishing Company. Created from ottawa on 2023-12-03 18:32:54. Copyright © 2017. World Scientific Publishing Company. All rights reserved.

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200 Expect the Unexpected: A First Course in Biostatistics of uterine fibroids in adult women of reproductive age who are eligible for surgery. The following data gives the size of the fibroids (in cm) for n = 20 women before and after using fibristal for three months: woman before ( x i ) after ( y i ) woman before ( x i ) after ( y i ) 1 6.5 6.6 11 7.5 6.9 2 5.6 6.0 12 8.3 7.7 3 7.8 7.9 13 6.1 6.4 4 9.8 9.1 14 5.2 3.9 5 9.9 9.6 15 5.5 5.8 6 5.6 5.1 16 8.7 8.8 7 6.2 5.3 17 6.6 5.8 8 4.9 4.6 18 7.8 7.5 9 5.8 5.4 19 6.2 5.8 10 8.3 7.1 20 7.7 7.0 Let μ D = μ X - μ Y be the difference between the average fibroid size before treatment ( μ X ) and after treatment ( μ Y ). Test the hypotheses H 0 : μ D = 0 against μ D > 0 at level α = 0 . 05. Is there enough evidence that a three-month treatment with fibristal is efficient in reducing the fibroid size? (Verify first that the differences d i = x i - y i , i = 1 , . . . , 20 are normally distributed.) Problem 11.3. The purpose of the study [35] was to determine whether twelve months of intense exercise training can induce an increase in left ventricular stroke volume in patients with coronary artery disease. 11 male patients were studied. Before training, the mean stroke volume was 66 ml/beat and the standard deviation 11 ml/beat. After training, the mean stroke volume was 81 ml/beat and the standard deviation 13 ml/beat. The standard deviation of the differences between the stroke volume before the exercise program and after the program was 5.4 ml/beat. Do these findings suggest that prolonged and intense training induces an increase in stroke volume in patients with coronary artery disease? Justify your answer using a 95% confidence interval for the mean difference between the stroke volume before the exercising program, and after the program. Problem 11.4. Exotic predators are sometimes introduced into agricul- tural ecosystems to aid in biological control of crop pests, see [72]. We consider a laboratory experiment to study the effect of mantis excrement on the behavior of wolf spiders. Each wolf spider was observed in an indi- vidual container for one hour. In the container, there is a filter paper with Balan, R., & Lamothe, G. (2017). Expect the unexpected : A first course in biostatistics (second edition). World Scientific Publishing Company. Created from ottawa on 2023-12-03 18:32:54. Copyright © 2017. World Scientific Publishing Company. All rights reserved.

Paired Samples 201 mantis excrement and also filter paper without mantis excrement. Walk- ing speeds for 15 wolf spiders on both filters were measured (in cm/s) and are displayed below. Assume that the difference between the two walking speeds is normally distributed. Mantis Excrement Mantis Excrement Wolf Spider Without With Wolf Spider Without With 1 2.3 1.2 9 3.3 3.5 2 2.4 2.7 10 3.0 2.8 3 2.9 2.3 11 2.3 2.1 4 2.3 1.2 12 3.4 3.4 5 2.9 2.6 13 2.6 2.3 6 3.0 2.9 14 2.5 1.7 7 3.3 2.9 15 2.9 1.9 8 2.9 2.3 (a) Is there significant evidence that the mean walking speeds are different? Use level α = 0 . 05. (b) Using a 95% confidence interval for the average difference between the two walking speeds, describe the effect that the mantis excrement has on the walking speed of the wolf spider. Problem 11.5. Almost two-thirds of iron in the body is found in hemoglobin, the protein in the red blood cells that carries oxygen to the tissues. Iron deficiency could lead to anemia, a condition characterized by less than the normal quantity of hemoglobin in the blood. The following data gives the hemoglobin values for 7 female patients at risk for anemia, before and after they followed a 3-month dietary iron intake program: Before After 10.4 12.3 11.5 13.6 9.6 13.7 8.7 10.3 11.5 14.3 11.8 13.9 10.7 12.8 (a) Was the program efficient in increasing the hemoglobin level? Justify your answer using a test of hypothesis for H 0 : μ 1 = μ 2 against H 1 : μ 1 < μ 2 , where μ 1 and μ 2 are the average hemoglobin levels before, respectively Balan, R., & Lamothe, G. (2017). Expect the unexpected : A first course in biostatistics (second edition). World Scientific Publishing Company. Created from ottawa on 2023-12-03 18:32:54. Copyright © 2017. World Scientific Publishing Company. All rights reserved.

202 Expect the Unexpected: A First Course in Biostatistics after the program. Use the level α = 0 . 005. (b) What is the conclusion of a test of level α = 0 . 005, if we assume (incorrectly) that the hemoglobin level after the program is independent of the hemoglobin level before the program? Assume that the two populations are normally distributed with equal variances. Problem 11.6. 16 professional marathon runners participated in a 3- month training program which included one hour of swimming three times a week. The best personal time (BPT) in minutes for a 5 km run of these athletes was recorded, before and after the training program. The data is summarized in the following table: BPT Before BPT After Difference ( x ) ( y ) ( d = x - y ) Mean 30.8 29.1 1.7 Standard Deviation 5.2 4.1 1.6 Using this data, can we say that integrating swimming into the training practice of professional runners improves their BPT? Use a test of hypothe- ses of level α = 0 . 005. Assume that variables X , Y and D are normally distributed, and the variables X and Y have the same variances. Problem 11.7. Two different methods were used to measure the cisternal milk volume (in kg) for 10 cows. Cow Method 1 Method 2 1 1.39 1.46 2 1.54 1.56 3 1.62 1.61 4 1.70 1.74 5 1.71 1.76 6 1.73 1.76 7 1.73 1.84 8 1.81 1.90 9 1.85 1.95 10 1.91 2.02 Do the methods give significantly different measurements on average? Use the significance level α = 0 . 05. Problem 11.8. Nine patients are evaluated for pain on a scale of 0 to 10, after using a control medication for pain relief (0 = no pain, 10 = Balan, R., & Lamothe, G. (2017). Expect the unexpected : A first course in biostatistics (second edition). World Scientific Publishing Company. Created from ottawa on 2023-12-03 18:32:54. Copyright © 2017. World Scientific Publishing Company. All rights reserved.

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Paired Samples 203 severe pain). One week later, the same patients are evaluated again after being given a new medication for pain relief. The following results are obtained: Control New Difference ( x 1 ) ( x 2 ) ( d = x 1 - x 2 ) Mean ¯ x 1 = 4 . 224 ¯ x 2 = 2 . 98 ¯ d = 1 . 244 Standard Deviation s 1 = 0 . 05 s 2 = 0 . 01 s d = 0 . 03 Construct a 95% confidence interval for the difference between the average pain level using the control medication ( μ 1 ) and the average pain level using the new medication ( μ 2 ). Using this confidence interval, can we say that the new treatment is effective in pain reduction? Problem 11.9. An experiment was designed to test the effects of a growth hormone (GH) on the daily milk production. The study involved ten pairs of identical twin dairy cows. For each pair of twins, only one cow was given the growth hormone, the other being considered a control. The table below gives the milk production (in kg per day) for each set of twins. Twin Set Control GH 1 9.86 9.69 2 12.10 12.38 3 13.33 14.24 4 13.69 14.09 5 9.04 9.05 6 9.72 10.67 7 9.89 11.48 8 10.22 11.14 9 9.46 9.54 10 9.02 9.05 Is there any evidence that the growth hormone increases the milk produc- tion? Justify your answer using a 98% confidence interval and a test of hypothesis at level α = 0 . 025. (Verify first that the normality assumption is satisfied.) Problem 11.10. We wish to compare the effect of two preparations of a virus on tobacco plants using a study which involves 8 plants. For each plant, half a leaf is inoculated with preparation 1 and the other half is inoculated with preparation 2. The number of lesions are measured and Balan, R., & Lamothe, G. (2017). Expect the unexpected : A first course in biostatistics (second edition). World Scientific Publishing Company. Created from ottawa on 2023-12-03 18:32:54. Copyright © 2017. World Scientific Publishing Company. All rights reserved.

204 Expect the Unexpected: A First Course in Biostatistics denoted as x 1 and x 2 , respectively. The data are given in the following table. Plant x 1 x 2 Plant x 1 x 2 1 30 20 11 18 16 2 8 5 12 14 13 3 14 17 13 12 4 4 14 11 14 13 13 5 19 17 15 16 14 6 17 15 16 16 14 7 3 5 17 18 12 8 19 15 18 21 15 9 12 12 19 12 10 10 21 14 20 17 12 (a) Do the preparations have a different effect on the tobacco plants? Justify your answer using a test of hypothesis for H 0 : μ 1 = μ 2 against H 1 : μ 1 6 = μ 2 , where μ 1 is the mean number of lesions when preparation 1 is used and μ 2 is the mean number of lesions when preparation 2 is used. Use the level α = 0 . 05. (b) What is the conclusion of a test of level α = 0 . 05, if we assume (incor- rectly) that the observations from the same plant are independent? Assume that the two populations are normally distributed with equal variances. Problem 11.11. A new surgical procedure is compared to the old method. Fifteen surgeons performed the operation on two patients, who are similar in terms of relevant factors such as age and gender. The table below gives the duration (in minutes) for each surgery. Old New Old New Surgeon Method Method Surgeon Method Method 1 33.1 31.4 9 19.9 21.7 2 46.6 40.5 10 46.0 36.6 3 21.5 29.9 11 49.0 54.7 4 34.3 45.0 12 45.5 39.5 5 19.1 12.5 13 60.3 61.2 6 38.9 44.0 14 37.4 34.0 7 56.2 57.2 15 25.3 19.1 8 67.1 65.6 Is there any evidence that the new surgical procedure will reduce the dura- Balan, R., & Lamothe, G. (2017). Expect the unexpected : A first course in biostatistics (second edition). World Scientific Publishing Company. Created from ottawa on 2023-12-03 18:32:54. Copyright © 2017. World Scientific Publishing Company. All rights reserved.

Paired Samples 205 tion of the surgery? Justify your answer with a test of hypothesis at level α = 0 . 05. (Verify first that the normality assumption is satisfied.) Did you know? Haemophilia is a rare hereditary disease, characterized by an impaired ability of blood coagulation. Bleeding is a general symptom of the disease, but one of the difficulties in recognizing the presence of the disease is that bleeding can be internal. The disease is more likely to occur in males, but females can transmit it to their offsprings. Haemophilia is sometimes called “the royal disease”, since it occurred frequently among the European royal families. It is thought that Queen Victoria inherited the gene and passed the mutation through her children, across the European continent, to the royal families of Spain, Germany, and Russia. Alexei Nikolaevich, the only son of Russia’s last tsar Nicholas II, was a descen- dant of Queen Victoria through his mother Empress Alexandra, and suffered from haemophilia. According to some sources, the mystic healer Rasputin succeeded in treating the tsar’s son, by simply advising against the tradi- tional medical treatment with aspirin. Balan, R., & Lamothe, G. (2017). Expect the unexpected : A first course in biostatistics (second edition). World Scientific Publishing Company. Created from ottawa on 2023-12-03 18:32:54. Copyright © 2017. World Scientific Publishing Company. All rights reserved.

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This page intentionally left blank This page intentionally left blank This page intentionally left blank This page intentionally left blank Balan, R., & Lamothe, G. (2017). Expect the unexpected : A first course in biostatistics (second edition). World Scientific Publishing Company. Created from ottawa on 2023-12-03 18:32:54. Copyright © 2017. World Scientific Publishing Company. All rights reserved.

Chap 11 Expect_The_Unexpected_A_First_Course_In_Biostatist..._----_(Statistics) (4)

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