HLTH 501 Biostatistics Final Exam Due 12/11/12 at 2000 Hours 1 BMI is measured in a sample of patients, with a mean of 31.7 and a standard deviation of 5.9. How many subjects would be needed to ensure that a 95% confidence interval estimate of BMI had a margin of error not exceeding 2 units? 2 The mean BMI in patients free of diabetes was elsewhere reported as 28.2. The researcher producing the data in question 1 wonders if the BMI in hi Based on the data in Problem 1 is there evidence that the BMI is significantly higher that 28.2? Use a 5% level of significance. 3 Peak expiratory flow (PEF) is a measure of a patient's ability to expel air from the lungs. Patients with asthma or other respiratory conditions often h The mean PEF for children free of asthma is 306. An investigator wants to test whether children with chronic bronchitis have restricted PEF. A sample of 40 children with chronic bronchitis are studied and their mean PEF is 279 with a standard deviation of 71. Is there statistical evidence of a lower mean PEF in children with chronic bronchitis? Run the appropriate test at a=0.05. 4 Considering again the study in Problem 3, a different investigator conducts a second study to investigate whether there is a difference in mean PEF i Data on PEF are collected and summarized below. Based on the data, is there statistical evidence of a lower mean PEF in children with chronic bronchitis Number of Children Mean PEF Std Dev PEF Chronic Bronchitis 25 281 68 No Chronic Bronchitis 25 319 74 Group 5 A clinical trial is run to investigate the effectiveness of an experimental drug in reducing preterm delivery to a drug considered standard care and to p Pregnant women are enrolled and randomly assigned to receive either the experimental drug, the standard drug or placebo. Women are followed through delivery and classified as delivering preterm (< 37 weeks) or not. The data are shown below. Is there a statistically significant difference in the proportions of women delivering preterm among the three treatment groups? Run the test at a 5% level Preterm Delivery Experimental Standard Drug Drug Placebo Yes 23 35 No 6 17 83 77 65 Consider the data presented in Problem 5. Previous studies have shown that approximately 32% of women deliver prematurely without treatment. Is the proportion of women delivering prematurely significantly higher in the placebo group? Run the test at a 5% level of significance. 7 A clinical trial is run to assess the effects of different forms of regular exercise on HDL levels in persons between the ages of 18 and 29. Participants in the study are randomly assigned to one of three exercise groups - Weight training, Aerobic exercise or Stretching/Yoga - and instructed to f Their HDL levels are measured after 8 weeks and are summarized below. Is there a significant difference in mean HDL levels among the exercise groups? Run the test at a 5% level of significance. Exercise Group Mean Std Dev Weight Training 20 49.7 10.2 Aerobic Exercise 20 43.1 11.1 Stretching/Yoga 8 N 20 57 12.5 Consider again the data in Problem 7. Suppose that in the aerobic exercise group we also measured the number of hours of aerobic exercise per wee and the mean is 5.2 hours with a standard deviation of 2.1 hours. The equation describing the results is Y = 54.6 - 2.22x, where Y = HDL levels. Estimate the HDL level for a person who exercises 7 hours per week. a Estimate the HDL level for a person who does not exercise. b 9 The table below summarizes baseline characteristics on patients participating in a clinical trial. Characteristic Placebo (n=125) Experimental (n=125) P Mean (+ SD) Age 54 + 4.5 53 + 4.9 0.7856 % Female 39% 52% 0.0289 % Less than High School Education 24% 22% 0.0986 % Completing High School 37% 36% % Completing Some College 136 + 13.8 134 + 12.4 0.4736 Mean (+ SD) Total Cholesterol 214 + 24.9 210 + 23.1 0.8954 % Current Smokers 17% 15% 0.5741 % with Diabetes 10 42% Mean (+ SD) Systolic Blood Pressure a b c d 39% 8% 3% 0.0438 Are there any statistically significant differences in baseline characteristics between treatment groups? Justify your answer. Write the hypotheses and the test statistic used to compare ages between groups. (No calculations - just H 0, H1 and form of the test statistic) Write the hypotheses and the test statistic used to compare % females between groups. (No calculations - just H 0, H1 and form of the test sta Write the hypotheses and the test statistic used to compare educational levels between groups. (No calculations - just H 0, H1 and form of the A study is designed to investigate whether there is a difference in response to various treatments in patients with rheumatoid arthritis. The outcome is patient's self-reported effect of treatment. The data are shown below. Is there a significant difference in effect of treatment? Run the test Symptoms No Effect Symptoms Improved Total Worsened Treatment 1 14 14 50 Treatment 2 14 15 21 50 Treatment 3 11 22 9 12 29 50 Using the data shown in Problem 10, suppose we focus on the proportions of patients who show improvement. Is there a statistically significant difference in the proportions of patients who show improvement between treatments 1 and 2? Run the test at a 5% level 12 An analysis is conducted to compare mean time to pain relief (measured in minutes) under four competing treatment regimens Summary statistics o Test the null hypothesis that there is no difference in mean time to pain relief between treatment groups. Treatment Sample Size Mean Time to Relief Sample Variance A 5 33.8 17.7 B 5 27 15.5 C 5 50.8 9.7 D 5 39.6 16.8 13 A small pilot study is conducted to investigate the effect of a nutritional supplement on total body weight. Six participants agree to take the nutritional sup To assess its effect on body weight, weights are measured before starting the supplementation and then after 6 weeks. The data are shown below. Is there a significant increase in body weight following supplementation? Run the test at a 5% level of significance. Weight after 6 D Weeks (difference) Subject Initial Weight 1 155 157 2 1.77 2 142 145 3 5.43 3 176 180 4 11.09 4 180 175 -5 32.15 5 210 209 -1 2.79 6 125 126 1 0.11 4 53.34 TOTAL 14 (D-0.67)2 A small pilot study is run to compare a new drug for chronic pain to one that is currently available. Participants are randomly assigned to receive eit currently available drug and report improvement in pain on a 5-point ordinal scale: 1=Pain is much worse, 2=Pain is slightly worse, 3= No change, 4=Pain Is there a significant difference in self-reported improvement in pain? Use the Mann-Whitney U test with a 5% level of significance. New Drug: 4 5 3 3 4 2 Standard Drug: 2 3 4 1 2 3 Answer True or False to each of the following 15 a The margin of error is always greater than or equal to the standard error. b If a test is run and p=0.0356, then we can reject H0 at a=0.01. 16 A randomized controlled trial is run to evaluate the effectiveness of a new drug for asthma in children. A total of 250 children are randomized to eith The mean age of children assigned to the new drug is 12.4 with a standard deviation of 3.6 years. The mean age of children assigned to the placebo is 13.0 Is there a statistically significant difference in ages of children assigned to the treatments? Run the appropriate test at a 5% level of significance. 17 \"Average adult Americans are about one inch taller, but nearly a whopping 25 pounds heavier than they were in 1960, according to a new report from The bad news, says CDC is that average BMI (body mass index, a weight-for-height formula used to measure obesity) has increased among adults from ap Boston is considered one of America's healthiest cities - is the weight gain since 1960 similar in Boston? A sample of n=25 adults suggested a mean increase of 17 pounds with a standard deviation of 8.6 pounds. Is Boston statistically significantly different in Run the appropriate test at a 5% level of significance. 18 In 2007, the CDC reported that approximately 6.6 per 1000 (0.66%) children were affected with autism spectrum disorder. A sample of 900 children with autism spectrum disorder. Is the proportion of children affected with autism spectrum disorder higher in Boston as compared to the national estimate Run the appropriate test at a 5% level of significance. 19 a b For each question below, provide a brief (1-2 sentences) response. When would a survival analysis model be used instead of a logistic regression model? What is the appropriate statistical test to assess whether there is an association between obesity status (normal weight, overweight, o Suppose each participant's obesity status (category) is known as is whether they develop CVD over the next 5 years or not. 20 An observational study is conducted to compare experiences of men and women between the ages of 50-59 years following coronary artery bypass s Participants undergo the surgery and are followed until the time of death, until they are lost to follow-up or up to 30 years, whichever comes first. The following table details the experiences of participating men and women. The data below are years of death or years of last contact for men and wome Estimate the survival functions for each treatment group (men, women) using the Kaplan-Meier approach Men Women Year of Death Year of Last Contact Year of Death Year of Last Contact 5 8 19 4 12 17 20 9 14 24 21 14 23 26 24 15 29 26 17 27 19 29 21 30 22 30 24 30 25 30 I in his patients free of diabetes is higher than this reported number. ften have restricted PEF. PEF in children with chronic bronchitis as compared to those without. chitis as compared to those without? Run the appropriate test at a=0.05. d to placebo. . level of significance. ent. d to follow the program for 8 weeks. r week tistic) est statistic) of the test statistic) e test at a 5% level of significance. level of significance. tics on the four treatments are shown below. al supplement. ve either the new drug or the =Pain improved slightly, 5=Pain much improved. o either the new drug or placebo (125 per group). s 13.0 with a standard deviation of 4.0 years. t from the Centers for Disease Control and Prevention (CDC). om approximately 25 in 1960 to 28 in 2002.\" nt in terms of weight gain since 1960? ildren from Boston are tested and 7 are diagnosed imate? ght, obese) and 5-year incident cardiovascular disease (CVD)? pass surgery. women. 1 BMI is measured in a sample of patients, with a mean of 31.7 and a standard deviation of 5.9. How many subjects would be needed to ensure that a 95% confidence interval estimate of BMI had a margin of error not exceeding 2 units? Solution: Given that margin of error E 2 sample 2 sample standard edviation Critical Z value at 5% level of significance * sample 2 Z0.05/2* 5.9 Z table at 5 level of significance Z sample 2 1.96* 5.9 sample 0.05/2 =Z0.025=1.96) 1.965.9 =5.782 2 2 Sample size 5.782 =33.43 33 Sample size 33 So 33 subjects would be needed at least to ensure that a 95% confidence interval estimate of BMI had a margin of error not exceeding 2 units. 2 The mean BMI in patients free of diabetes was elsewhere reported as 28.2. The researcher producing the data in question 1 wonders if the BMI in his patients free of diabetes is higher than this reported number. Based on the data in Problem 1 is there evidence that the BMI is significantly higher that 28.2? Use a 5% level of significance. Solution: Null hypothesis H0: 28.2 against the alternative H : 28.2 1 The test statistics will be Z test because n=33>30 and the test statistics Z= sample mean sample standard edviation Under the null hypothesis One-Sample Z Test of mu = 28.2 vs > 28.2 The assumed standard deviation = 5.9 N 33 Mean 31.70 SE Mean 1.03 95% Lower Bound 30.01 Z 3.41 P 0.000 Hence Z=3.41 and corresponding p value is 0.00 which is less than 0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that the BMI is significantly higher than 28.2. 3 Peak expiratory flow (PEF) is a measure of a patient's ability to expel air from the lungs. Patients with asthma or other respiratory conditions often have restricted PEF. The mean PEF for children free of asthma is 306. An investigator wants to test whether children with chronic bronchitis have restricted PEF. A sample of 40 children with chronic bronchitis is studied and their mean PEF is 279 with a standard deviation of 71. Is there statistical evidence of a lower mean PEF in children with chronic bronchitis? Run the appropriate test at a=0.05. Solution: Null hypothesis H0: 306 against alternative hypothesis H : 1 < 306 Since the sample size n=40>30 therefore test will be Z test and test statistics is sample standard edviation / sample sample mean Sample mean=279 and sample standard deviation =71. Under the null hypothesis the test result is: One-Sample Z Test of mu = 306 vs < 306 Z= The assumed standard deviation = 71 N 40 Mean 279.0 SE Mean 11.2 95% Upper Bound 297.5 Z -2.41 P 0.008 From the result we see Z=-2.41 and corresponding p value is 0.008< 0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that < 306 which means there is a statistical evidence of a lower mean PEF in children with chronic bronchitis. Considering again the study in Problem 3, a different investigator conducts a second study to investigate whether X there is a difference in mean PEF in children with chronic 4 bronchitis as compared to those without. Data on PEF are collected and summarized below. Based on the data, is there statistical evidence of a lower mean PEF in children with chronic bronchitis as compared to those without? Run the appropriate test at a=0.05. St M d Nu ea D mbe Gro n e r of up P v Chil E P dren F E F Chr onic 28 6 Bro 25 1 8 nchi tis No Chr onic 31 7 25 Bro 9 4 nchi tis Solution: Null hypothesis H0: 1 2 against the alternative H1: 1<2 where 1, 2 are the population mean of children with chronic bronchitis and without bronchitis. since sample size n =25+25=50>30 therefore the test will be two sample independent Z test and test x 1 x 2( 1 2) 2 2 s1 s2 statistics is Z= where s1 and s2 are the standard deviation of the two sample and ( + ) n1 n2 n1 and n2 be the sample size of the two sample. Under the null hypothesis the test results as follows: Two-Sample Z-Test Sample 1 2 N 25 25 Mean 281.0 319.0 SE Mean 14 15 StDev 68.0 74.0 Difference = mu (1) - mu (2) Estimate for difference: -38.0 95% CI for difference: (-78.4, 2.4) Z-Test of difference = 0 (vs not =): Z-Value = -1.89 P-Value = 0.02938 From the result Z=-1.89 and the corresponding p value =0.02938<0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that 1<2 which means there is a statistical evidence of a lower mean PEF in children with chronic bronchitis as compared to those without. 5 A clinical trial is run to investigate the effectiveness of an experimental drug in reducing preterm delivery to a drug considered standard care and to placebo. . Pregnant women are enrolled and randomly assigned to receive either the experimental drug, the standard drug or placebo. Women are followed through delivery and classified as delivering preterm (< 37 weeks) or not. The data are shown below. Is there a statistically significant difference in the proportions of women delivering preterm among the three treatment groups? Run the test at a 5% level of significance. Preterm Experimental Standard Drug Placebo Delivery Drug Yes 17 23 35 No 83 77 65 Solution: Solution: Chi-Square Test Preterm Delivery Observed Frequencies Column variable Experimental Standard Placebo Total Drug Yes No Total Preterm Delivery Yes No Total Drug 17 23 45 83 77 66 100 100 111 Expected Frequencies Column variable Experimental Standard Drug Drug Placebo 27.33119 27.33119 30.33762 72.66881 72.66881 80.66238 100 100 111 Data Level of Significance Number of Rows Number of Columns Degrees of Freedom 85 226 311 Total 85 226 311 0.05 2 3 2 Results Critical Value Chi-Square Test Statistic p-Value 5.991465 16.07015 0.000324 Reject the null hypothesis Since P-value is less than 0.05, so we can conclude that there is a statistically significant difference in the proportions of women delivering preterm among the three treatment groups Solution: Solution: Z Test of Hypothesis for the Proportion Data Null Hypothesis = 0.32 Level of Significance 0.05 Number of Successes 45 Sample Size 111 Intermediate Calculations 0.40540540 Sample Proportion 5 0.04427595 Standard Error 7 Z Test Statistic 1.92893415 2 Upper-Tail Test Upper Critical Value p-Value Reject the null hypothesis 1.64485362 7 0.02686951 9 Since P-value is less than 0.05, so we can conclude that the proportion of women delivering prematurely is significantly higher in the placebo group. Solution: 1 BMI is measured in a sample of patients, with a mean of 31.7 and a standard deviation of 5.9. How many subjects would be needed to ensure that a 95% confidence interval estimate of BMI had a margin of error not exceeding 2 units? Solution: Given that margin of error E 2 sample 2 sample standard edviation Critical Z value at 5% level of significance * sample 2 Z0.05/2* 5.9 Z table at 5 level of significance Z sample 2 1.96* 5.9 sample 0.05/2 =Z0.025=1.96) 1.965.9 =5.782 2 2 Sample size 5.782 =33.43 33 Sample size 33 So 33 subjects would be needed at least to ensure that a 95% confidence interval estimate of BMI had a margin of error not exceeding 2 units. 2 The mean BMI in patients free of diabetes was elsewhere reported as 28.2. The researcher producing the data in question 1 wonders if the BMI in his patients free of diabetes is higher than this reported number. Based on the data in Problem 1 is there evidence that the BMI is significantly higher that 28.2? Use a 5% level of significance. Solution: Null hypothesis H0: 28.2 against the alternative H : 28.2 1 The test statistics will be Z test because n=33>30 and the test statistics Z= sample mean sample standard edviation Under the null hypothesis One-Sample Z Test of mu = 28.2 vs > 28.2 The assumed standard deviation = 5.9 N 33 Mean 31.70 SE Mean 1.03 95% Lower Bound 30.01 Z 3.41 P 0.000 Hence Z=3.41 and corresponding p value is 0.00 which is less than 0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that the BMI is significantly higher than 28.2. 3 Peak expiratory flow (PEF) is a measure of a patient's ability to expel air from the lungs. Patients with asthma or other respiratory conditions often have restricted PEF. The mean PEF for children free of asthma is 306. An investigator wants to test whether children with chronic bronchitis have restricted PEF. A sample of 40 children with chronic bronchitis is studied and their mean PEF is 279 with a standard deviation of 71. Is there statistical evidence of a lower mean PEF in children with chronic bronchitis? Run the appropriate test at a=0.05. Solution: Null hypothesis H0: 306 against alternative hypothesis H : 1 < 306 Since the sample size n=40>30 therefore test will be Z test and test statistics is sample standard edviation / sample sample mean Sample mean=279 and sample standard deviation =71. Under the null hypothesis the test result is: One-Sample Z Test of mu = 306 vs < 306 Z= The assumed standard deviation = 71 N 40 Mean 279.0 SE Mean 11.2 95% Upper Bound 297.5 Z -2.41 P 0.008 From the result we see Z=-2.41 and corresponding p value is 0.008< 0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that < 306 which means there is a statistical evidence of a lower mean PEF in children with chronic bronchitis. Considering again the study in Problem 3, a different investigator conducts a second study to investigate whether X there is a difference in mean PEF in children with chronic 4 bronchitis as compared to those without. Data on PEF are collected and summarized below. Based on the data, is there statistical evidence of a lower mean PEF in children with chronic bronchitis as compared to those without? Run the appropriate test at a=0.05. St M d Nu ea D mbe Gro n e r of up P v Chil E P dren F E F Chr onic 28 6 Bro 25 1 8 nchi tis No Chr onic 31 7 25 Bro 9 4 nchi tis Solution: Null hypothesis H0: 1 2 against the alternative H1: 1<2 where 1, 2 are the population mean of children with chronic bronchitis and without bronchitis. since sample size n =25+25=50>30 therefore the test will be two sample independent Z test and test x 1 x 2( 1 2) 2 2 s1 s2 statistics is Z= where s1 and s2 are the standard deviation of the two sample and ( + ) n1 n2 n1 and n2 be the sample size of the two sample. Under the null hypothesis the test results as follows: Two-Sample Z-Test Sample 1 2 N 25 25 Mean 281.0 319.0 SE Mean 14 15 StDev 68.0 74.0 Difference = mu (1) - mu (2) Estimate for difference: -38.0 95% CI for difference: (-78.4, 2.4) Z-Test of difference = 0 (vs not =): Z-Value = -1.89 P-Value = 0.02938 From the result Z=-1.89 and the corresponding p value =0.02938<0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that 1<2 which means there is a statistical evidence of a lower mean PEF in children with chronic bronchitis as compared to those without. 5 A clinical trial is run to investigate the effectiveness of an experimental drug in reducing preterm delivery to a drug considered standard care and to placebo. . Pregnant women are enrolled and randomly assigned to receive either the experimental drug, the standard drug or placebo. Women are followed through delivery and classified as delivering preterm (< 37 weeks) or not. The data are shown below. Is there a statistically significant difference in the proportions of women delivering preterm among the three treatment groups? Run the test at a 5% level of significance. Preterm Experimental Standard Drug Placebo Delivery Drug Yes 17 23 35 No 83 77 65 Solution: Solution: Chi-Square Test Preterm Delivery Observed Frequencies Column variable Experimental Standard Placebo Total Drug Yes No Total Preterm Delivery Yes No Total Drug 17 23 45 83 77 66 100 100 111 Expected Frequencies Column variable Experimental Standard Drug Drug Placebo 27.33119 27.33119 30.33762 72.66881 72.66881 80.66238 100 100 111 Data Level of Significance Number of Rows Number of Columns Degrees of Freedom 85 226 311 Total 85 226 311 0.05 2 3 2 Results Critical Value Chi-Square Test Statistic p-Value 5.991465 16.07015 0.000324 Reject the null hypothesis Since P-value is less than 0.05, so we can conclude that there is a statistically significant difference in the proportions of women delivering preterm among the three treatment groups Solution: Solution: Z Test of Hypothesis for the Proportion Data Null Hypothesis = 0.32 Level of Significance 0.05 Number of Successes 45 Sample Size 111 Intermediate Calculations 0.40540540 Sample Proportion 5 0.04427595 Standard Error 7 Z Test Statistic 1.92893415 2 Upper-Tail Test Upper Critical Value p-Value Reject the null hypothesis 1.64485362 7 0.02686951 9 Since P-value is less than 0.05, so we can conclude that the proportion of women delivering prematurely is significantly higher in the placebo group. Solution: Solution: Null hypothesis H0: 1 2=3 against the alternative H1: any of them are not equal. Since the sample size n=25+25=50>30 therefore the test will be two sample independent Z test and test x 1 x 2( 1 2) s 12 s 2 2 statistics is Z= where s1 and s2 are the standard deviation of the two sample and ( + ) n1 n2 n1 and n2 be the sample size of the two sample. Under the null hypothesis the test results as follows: Solution: a) Y=54.6-2.22*7 = 39.06 b) Y=54.6-2.22*0 = 54.6 Solution: Solution: We assume that censoring of data is done after the death occurs. Terminology: tj = Time of jth death dj = No. of death at 'tj' nj = No. of people alive just before jth death. j = Probability of death at tj. MEN j tj dj nj j = dj / nj 1 - j S(t) 1 5 1 15 0.0667 0.9333 0.9333 2 12 1 13 0.0769 0.9231 0.8615 3 14 1 12 0.0833 0.9167 0.7897 4 23 1 10 0.1000 0.9000 0.7108 5 29 1 5 0.2000 0.8000 0.5686 1 0 29 WOMEN j tj dj nj j = d j / nj 1 - j S(t) 1 19 1 10 0.1000 0.9000 0.9000 2 20 1 8 0.1250 0.8750 0.7875 3 21 1 7 0.1429 0.8571 0.6750 4 24 1 4 0.2500 0.7500 0.5063 1 0 < t < 19 0.9 19 < t < 20 0.7875 20 < t < 21 Survival Function = S(t) 0.675 21 < t < 24 0.5063 t > 24 1 BMI is measured in a sample of patients, with a mean of 31.7 and a standard deviation of 5.9. How many subjects would be needed to ensure that a 95% confidence interval estimate of BMI had a margin of error not exceeding 2 units? Solution: Given that margin of error E 2 sample 2 sample standard edviation Critical Z value at 5% level of significance * sample 2 Z0.05/2* 5.9 Z table at 5 level of significance Z sample 2 1.96* 5.9 sample 0.05/2 =Z0.025=1.96) 1.965.9 =5.782 2 2 Sample size 5.782 =33.43 33 Sample size 33 So 33 subjects would be needed at least to ensure that a 95% confidence interval estimate of BMI had a margin of error not exceeding 2 units. 2 The mean BMI in patients free of diabetes was elsewhere reported as 28.2. The researcher producing the data in question 1 wonders if the BMI in his patients free of diabetes is higher than this reported number. Based on the data in Problem 1 is there evidence that the BMI is significantly higher that 28.2? Use a 5% level of significance. Solution: Null hypothesis H0: 28.2 against the alternative H : 28.2 1 The test statistics will be Z test because n=33>30 and the test statistics Z= sample mean sample standard edviation Under the null hypothesis One-Sample Z Test of mu = 28.2 vs > 28.2 The assumed standard deviation = 5.9 N 33 Mean 31.70 SE Mean 1.03 95% Lower Bound 30.01 Z 3.41 P 0.000 Hence Z=3.41 and corresponding p value is 0.00 which is less than 0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that the BMI is significantly higher than 28.2. 3 Peak expiratory flow (PEF) is a measure of a patient's ability to expel air from the lungs. Patients with asthma or other respiratory conditions often have restricted PEF. The mean PEF for children free of asthma is 306. An investigator wants to test whether children with chronic bronchitis have restricted PEF. A sample of 40 children with chronic bronchitis is studied and their mean PEF is 279 with a standard deviation of 71. Is there statistical evidence of a lower mean PEF in children with chronic bronchitis? Run the appropriate test at a=0.05. Solution: Null hypothesis H0: 306 against alternative hypothesis H : 1 < 306 Since the sample size n=40>30 therefore test will be Z test and test statistics is sample standard edviation / sample sample mean Sample mean=279 and sample standard deviation =71. Under the null hypothesis the test result is: One-Sample Z Test of mu = 306 vs < 306 Z= The assumed standard deviation = 71 N 40 Mean 279.0 SE Mean 11.2 95% Upper Bound 297.5 Z -2.41 P 0.008 From the result we see Z=-2.41 and corresponding p value is 0.008< 0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that < 306 which means there is a statistical evidence of a lower mean PEF in children with chronic bronchitis. Considering again the study in Problem 3, a different investigator conducts a second study to investigate whether X there is a difference in mean PEF in children with chronic 4 bronchitis as compared to those without. Data on PEF are collected and summarized below. Based on the data, is there statistical evidence of a lower mean PEF in children with chronic bronchitis as compared to those without? Run the appropriate test at a=0.05. St M d Nu ea D mbe Gro n e r of up P v Chil E P dren F E F Chr onic 28 6 Bro 25 1 8 nchi tis No Chr onic 31 7 25 Bro 9 4 nchi tis Solution: Null hypothesis H0: 1 2 against the alternative H1: 1<2 where 1, 2 are the population mean of children with chronic bronchitis and without bronchitis. since sample size n =25+25=50>30 therefore the test will be two sample independent Z test and test x 1 x 2( 1 2) 2 2 s1 s2 statistics is Z= where s1 and s2 are the standard deviation of the two sample and ( + ) n1 n2 n1 and n2 be the sample size of the two sample. Under the null hypothesis the test results as follows: Two-Sample Z-Test Sample 1 2 N 25 25 Mean 281.0 319.0 SE Mean 14 15 StDev 68.0 74.0 Difference = mu (1) - mu (2) Estimate for difference: -38.0 95% CI for difference: (-78.4, 2.4) Z-Test of difference = 0 (vs not =): Z-Value = -1.89 P-Value = 0.02938 From the result Z=-1.89 and the corresponding p value =0.02938<0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that 1<2 which means there is a statistical evidence of a lower mean PEF in children with chronic bronchitis as compared to those without. 5 A clinical trial is run to investigate the effectiveness of an experimental drug in reducing preterm delivery to a drug considered standard care and to placebo. . Pregnant women are enrolled and randomly assigned to receive either the experimental drug, the standard drug or placebo. Women are followed through delivery and classified as delivering preterm (< 37 weeks) or not. The data are shown below. Is there a statistically significant difference in the proportions of women delivering preterm among the three treatment groups? Run the test at a 5% level of significance. Preterm Experimental Standard Drug Placebo Delivery Drug Yes 17 23 35 No 83 77 65 Solution: Solution: Chi-Square Test Preterm Delivery Observed Frequencies Column variable Experimental Standard Placebo Total Drug Yes No Total Preterm Delivery Yes No Total Drug 17 23 45 83 77 66 100 100 111 Expected Frequencies Column variable Experimental Standard Drug Drug Placebo 27.33119 27.33119 30.33762 72.66881 72.66881 80.66238 100 100 111 Data Level of Significance Number of Rows Number of Columns Degrees of Freedom 85 226 311 Total 85 226 311 0.05 2 3 2 Results Critical Value Chi-Square Test Statistic p-Value 5.991465 16.07015 0.000324 Reject the null hypothesis Since P-value is less than 0.05, so we can conclude that there is a statistically significant difference in the proportions of women delivering preterm among the three treatment groups Solution: Solution: Z Test of Hypothesis for the Proportion Data Null Hypothesis = 0.32 Level of Significance 0.05 Number of Successes 45 Sample Size 111 Intermediate Calculations 0.40540540 Sample Proportion 5 0.04427595 Standard Error 7 Z Test Statistic 1.92893415 2 Upper-Tail Test Upper Critical Value p-Value Reject the null hypothesis 1.64485362 7 0.02686951 9 Since P-value is less than 0.05, so we can conclude that the proportion of women delivering prematurely is significantly higher in the placebo group. Solution: Solution: Null hypothesis H0: 1 2=3 against the alternative H1: any of them are not equal. Since the sample size n=25+25=50>30 therefore the test will be two sample independent Z test and test x 1 x 2( 1 2) s 12 s 2 2 statistics is Z= where s1 and s2 are the standard deviation of the two sample and ( + ) n1 n2 n1 and n2 be the sample size of the two sample. Under the null hypothesis the test results as follows: Solution: a) Y=54.6-2.22*7 = 39.06 b) Y=54.6-2.22*0 = 54.6 Solution: Solution: Chi-square critical value with (row-1)(column -1) = (3-1)93-1) = 4 degrees of freedom = 9.49 Calculate chi-square: Find all expected frequencies: (use the method shown above) Treatment 1.....symptoms worsened=15, no effect=13.67, symptoms improved=21.33; total 50 Treatment 2.....symptoms worsened=15, no effect=13.67, symptoms improved=21.33 ; total 50 Treatment 3.....symptoms worsened=15, no effect=13.67, symptoms improved=21.33 ; total 50 --------Total ....... ........ ......... ........ . .45 ...... .... .41 .......... ....... ........ ...--------.64 .........150 Chi-squared: (O(i)-E(i) )^2 / E(i) = (22-15)^2/15+(14-13.67)^2/13.67+(14-21.3... (9-15)^2/15+(12-13.67)^2/13.67+(29-21.33... =11.356 11.356 > 9.49 Reject H0 , we have significant evidence, =0.05, to show that treatment and response are not independent. Solution: Solution: Apparently it's absolute weight loss (after - before), given your reported s.d. You haven't said whether you're testing for weight loss or weight gain, so I'll assume that you're interested in any weight change and use a two-tailed z test. At the 0.05 level of significance, z = +/- 1.96. The 95% confidence interval about zero (no change) is 0 +/- 1.96 (3.27 / sqrt(6) ) = +/- 2.61 The mean change = +0.67, which is well within the bounds of the CI, so this is not a significant result at the 0.05 level of significance. Alternatively, p = 1 - { CDF(0.67 / (3.27 / sqrt(6)) ) - CDF(-0.67 / (3.27 / sqrt(6)) ) } = 1 - (0.6915 - 0.3085) = 0.617 so that the result is only significant at the 0.05 level of significance. Solution: Step 1. Set up hypotheses and determine level of significance. H0: The populations of self-reported improvement in pain who took new drug and those who took standard drug are equal H1: The populations of self-reported improvement in pain who took new drug and those who took standard drug are not equal. =0.05 Step 2. Select the appropriate test statistic. The test statistic for the Mann Whitney U Test is U. Step 3. Set up decision rule. The critical value for this two-sided test with n 1=n2=6 and =0.05 is found in Table and is 5. The decision rule is Reject H0 if U < 5. Step 4. Compute the test statistic. Total Sample (Ordered Smallest to Largest) New Standard Ranks New Drug Standard Drug New Standard 4 2 1 1 5 3 2 3 3 4 2 3 3 1 2 3 4 2 3 6.5 2 3 3 6.5 3 6.5 3 6.5 4 10 4 10 4 5 10 12 R1=39 U1 n1n 2 n1 (n 1 1) 6(7) R 1 6(6) 39 18 2 2 U 2 n1n 2 R2=30 n 2 (n 2 1) 6(7) R 2 6(6) 30 27 2 2 The test statistic is U = 27. Step 5. Conclusion. Do not reject H0 because 27 > 5. We do not have statistically significant evidence at =0.05 to show that populations of self-reported improvement in pain who took new drug and those who took standard drug are not equal. Solution: A: False B: False Solution: Hypotheses of interest : Ho: d (mean difference) = 1 - 2 = 0 against Ha : d 0. where 1, 2 are mean age of children assigned to the new drug and placebo respectively. Test: Paired T test. Test Statistics : t = [ (x1-bar - x2-bar ) - ( 1 - 2 ) ] / { [s1^2 + s2^2 ] / n} Here t = [ (12.4 - 13 ) - 0 ] / [[3.62 + 42] / 125] = -1.25 Here our test statistics is student's t statistic. The degrees of freedom is the number of independent estimates of variance = 124 Tabulated value of t for 124 degrees of freedom at 5% level of significance for two-tailed test is +/- 1.97. Decision: Since the modulus of calculated value of t ( |-1.25| = 1.25 ) is less than the modulus of tabulated value of t (|-1.97| = 1.97 ), so it is inside the acceptance region of the test and thus is not significant. Hence we are failed to reject the null hypothesis at 5% level of significance. Conclusion: At 5% level of significance we can conclude that there is not a statistically significant difference in the ages of children assigned to the treatments. Solution: Solution: We assume that censoring of data is done after the death occurs. Terminology: tj = Time of jth death dj = No. of death at 'tj' nj = No. of people alive just before jth death. j = Probability of death at tj. MEN j tj dj nj j = dj / nj 1 - j S(t) 1 5 1 15 0.0667 0.9333 0.9333 2 12 1 13 0.0769 0.9231 0.8615 3 14 1 12 0.0833 0.9167 0.7897 4 23 1 10 0.1000 0.9000 0.7108 5 29 1 5 0.2000 0.8000 0.5686 1 0 29 WOMEN j tj dj nj j = d j / nj 1 - j S(t) 1 19 1 10 0.1000 0.9000 0.9000 2 20 1 8 0.1250 0.8750 0.7875 3 21 1 7 0.1429 0.8571 0.6750 4 24 1 4 0.2500 0.7500 0.5063 1 0 < t < 19 0.9 19 < t < 20 0.7875 20 < t < 21 Survival Function = S(t) 0.675 21 < t < 24 0.5063 t > 24 1 BMI is measured in a sample of patients, with a mean of 31.7 and a standard deviation of 5.9. How many subjects would be needed to ensure that a 95% confidence interval estimate of BMI had a margin of error not exceeding 2 units? Solution: Given that margin of error E 2 sample 2 sample standard edviation Critical Z value at 5% level of significance * sample 2 Z0.05/2* 5.9 Z table at 5 level of significance Z sample 2 1.96* 5.9 sample 0.05/2 =Z0.025=1.96) 1.965.9 =5.782 2 2 Sample size 5.782 =33.43 33 Sample size 33 So 33 subjects would be needed at least to ensure that a 95% confidence interval estimate of BMI had a margin of error not exceeding 2 units. 2 The mean BMI in patients free of diabetes was elsewhere reported as 28.2. The researcher producing the data in question 1 wonders if the BMI in his patients free of diabetes is higher than this reported number. Based on the data in Problem 1 is there evidence that the BMI is significantly higher that 28.2? Use a 5% level of significance. Solution: Null hypothesis H0: 28.2 against the alternative H : 28.2 1 The test statistics will be Z test because n=33>30 and the test statistics Z= sample mean sample standard edviation Under the null hypothesis One-Sample Z Test of mu = 28.2 vs > 28.2 The assumed standard deviation = 5.9 N 33 Mean 31.70 SE Mean 1.03 95% Lower Bound 30.01 Z 3.41 P 0.000 Hence Z=3.41 and corresponding p value is 0.00 which is less than 0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that the BMI is significantly higher than 28.2. 3 Peak expiratory flow (PEF) is a measure of a patient's ability to expel air from the lungs. Patients with asthma or other respiratory conditions often have restricted PEF. The mean PEF for children free of asthma is 306. An investigator wants to test whether children with chronic bronchitis have restricted PEF. A sample of 40 children with chronic bronchitis is studied and their mean PEF is 279 with a standard deviation of 71. Is there statistical evidence of a lower mean PEF in children with chronic bronchitis? Run the appropriate test at a=0.05. Solution: Null hypothesis H0: 306 against alternative hypothesis H : 1 < 306 Since the sample size n=40>30 therefore test will be Z test and test statistics is sample standard edviation / sample sample mean Sample mean=279 and sample standard deviation =71. Under the null hypothesis the test result is: One-Sample Z Test of mu = 306 vs < 306 Z= The assumed standard deviation = 71 N 40 Mean 279.0 SE Mean 11.2 95% Upper Bound 297.5 Z -2.41 P 0.008 From the result we see Z=-2.41 and corresponding p value is 0.008< 0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that < 306 which means there is a statistical evidence of a lower mean PEF in children with chronic bronchitis. Considering again the study in Problem 3, a different investigator conducts a second study to investigate whether X there is a difference in mean PEF in children with chronic 4 bronchitis as compared to those without. Data on PEF are collected and summarized below. Based on the data, is there statistical evidence of a lower mean PEF in children with chronic bronchitis as compared to those without? Run the appropriate test at a=0.05. St M d Nu ea D mbe Gro n e r of up P v Chil E P dren F E F Chr onic 28 6 Bro 25 1 8 nchi tis No Chr onic 31 7 25 Bro 9 4 nchi tis Solution: Null hypothesis H0: 1 2 against the alternative H1: 1<2 where 1, 2 are the population mean of children with chronic bronchitis and without bronchitis. since sample size n =25+25=50>30 therefore the test will be two sample independent Z test and test x 1 x 2( 1 2) 2 2 s1 s2 statistics is Z= where s1 and s2 are the standard deviation of the two sample and ( + ) n1 n2 n1 and n2 be the sample size of the two sample. Under the null hypothesis the test results as follows: Two-Sample Z-Test Sample 1 2 N 25 25 Mean 281.0 319.0 SE Mean 14 15 StDev 68.0 74.0 Difference = mu (1) - mu (2) Estimate for difference: -38.0 95% CI for difference: (-78.4, 2.4) Z-Test of difference = 0 (vs not =): Z-Value = -1.89 P-Value = 0.02938 From the result Z=-1.89 and the corresponding p value =0.02938<0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that 1<2 which means there is a statistical evidence of a lower mean PEF in children with chronic bronchitis as compared to those without. 5 A clinical trial is run to investigate the effectiveness of an experimental drug in reducing preterm delivery to a drug considered standard care and to placebo. . Pregnant women are enrolled and randomly assigned to receive either the experimental drug, the standard drug or placebo. Women are followed through delivery and classified as delivering preterm (< 37 weeks) or not. The data are shown below. Is there a statistically significant difference in the proportions of women delivering preterm among the three treatment groups? Run the test at a 5% level of significance. Preterm Experimental Standard Drug Placebo Delivery Drug Yes 17 23 35 No 83 77 65 Solution: Solution: Chi-Square Test Preterm Delivery Observed Frequencies Column variable Experimental Standard Placebo Total Drug Yes No Total Preterm Delivery Yes No Total Drug 17 23 45 83 77 66 100 100 111 Expected Frequencies Column variable Experimental Standard Drug Drug Placebo 27.33119 27.33119 30.33762 72.66881 72.66881 80.66238 100 100 111 Data Level of Significance Number of Rows Number of Columns Degrees of Freedom 85 226 311 Total 85 226 311 0.05 2 3 2 Results Critical Value Chi-Square Test Statistic p-Value 5.991465 16.07015 0.000324 Reject the null hypothesis Since P-value is less than 0.05, so we can conclude that there is a statistically significant difference in the proportions of women delivering preterm among the three treatment groups Solution: Solution: Z Test of Hypothesis for the Proportion Data Null Hypothesis = 0.32 Level of Significance 0.05 Number of Successes 45 Sample Size 111 Intermediate Calculations 0.40540540 Sample Proportion 5 0.04427595 Standard Error 7 Z Test Statistic 1.92893415 2 Upper-Tail Test Upper Critical Value p-Value Reject the null hypothesis 1.64485362 7 0.02686951 9 Since P-value is less than 0.05, so we can conclude that the proportion of women delivering prematurely is significantly higher in the placebo group. Solution: 1 BMI is measured in a sample of patients, with a mean of 31.7 and a standard deviation of 5.9. How many subjects would be needed to ensure that a 95% confidence interval estimate of BMI had a margin of error not exceeding 2 units? Solution: Given that margin of error E 2 sample 2 sample standard edviation Critical Z value at 5% level of significance * sample 2 Z0.05/2* 5.9 Z table at 5 level of significance Z sample 2 1.96* 5.9 sample 0.05/2 =Z0.025=1.96) 1.965.9 =5.782 2 2 Sample size 5.782 =33.43 33 Sample size 33 So 33 subjects would be needed at least to ensure that a 95% confidence interval estimate of BMI had a margin of error not exceeding 2 units. 2 The mean BMI in patients free of diabetes was elsewhere reported as 28.2. The researcher producing the data in question 1 wonders if the BMI in his patients free of diabetes is higher than this reported number. Based on the data in Problem 1 is there evidence that the BMI is significantly higher that 28.2? Use a 5% level of significance. Solution: Null hypothesis H0: 28.2 against the alternative H : 28.2 1 The test statistics will be Z test because n=33>30 and the test statistics Z= sample mean sample standard edviation Under the null hypothesis One-Sample Z Test of mu = 28.2 vs > 28.2 The assumed standard deviation = 5.9 N 33 Mean 31.70 SE Mean 1.03 95% Lower Bound 30.01 Z 3.41 P 0.000 Hence Z=3.41 and corresponding p value is 0.00 which is less than 0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that the BMI is significantly higher than 28.2. 3 Peak expiratory flow (PEF) is a measure of a patient's ability to expel air from the lungs. Patients with asthma or other respiratory conditions often have restricted PEF. The mean PEF for children free of asthma is 306. An investigator wants to test whether children with chronic bronchitis have restricted PEF. A sample of 40 children with chronic bronchitis is studied and their mean PEF is 279 with a standard deviation of 71. Is there statistical evidence of a lower mean PEF in children with chronic bronchitis? Run the appropriate test at a=0.05. Solution: Null hypothesis H0: 306 against alternative hypothesis H : 1 < 306 Since the sample size n=40>30 therefore test will be Z test and test statistics is sample standard edviation / sample sample mean Sample mean=279 and sample standard deviation =71. Under the null hypothesis the test result is: One-Sample Z Test of mu = 306 vs < 306 Z= The assumed standard deviation = 71 N 40 Mean 279.0 SE Mean 11.2 95% Upper Bound 297.5 Z -2.41 P 0.008 From the result we see Z=-2.41 and corresponding p value is 0.008< 0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that < 306 which means there is a statistical evidence of a lower mean PEF in children with chronic bronchitis. Considering again the study in Problem 3, a different investigator conducts a second study to investigate whether X there is a difference in mean PEF in children with chronic 4 bronchitis as compared to those without. Data on PEF are collected and summarized below. Based on the data, is there statistical evidence of a lower mean PEF in children with chronic bronchitis as compared to those without? Run the appropriate test at a=0.05. St M d Nu ea D mbe Gro n e r of up P v Chil E P dren F E F Chr onic 28 6 Bro 25 1 8 nchi tis No Chr onic 31 7 25 Bro 9 4 nchi tis Solution: Null hypothesis H0: 1 2 against the alternative H1: 1<2 where 1, 2 are the population mean of children with chronic bronchitis and without bronchitis. since sample size n =25+25=50>30 therefore the test will be two sample independent Z test and test x 1 x 2( 1 2) 2 2 s1 s2 statistics is Z= where s1 and s2 are the standard deviation of the two sample and ( + ) n1 n2 n1 and n2 be the sample size of the two sample. Under the null hypothesis the test results as follows: Two-Sample Z-Test Sample 1 2 N 25 25 Mean 281.0 319.0 SE Mean 14 15 StDev 68.0 74.0 Difference = mu (1) - mu (2) Estimate for difference: -38.0 95% CI for difference: (-78.4, 2.4) Z-Test of difference = 0 (vs not =): Z-Value = -1.89 P-Value = 0.02938 From the result Z=-1.89 and the corresponding p value =0.02938<0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that 1<2 which means there is a statistical evidence of a lower mean PEF in children with chronic bronchitis as compared to those without. 5 A clinical trial is run to investigate the effectiveness of an experimental drug in reducing preterm delivery to a drug considered standard care and to placebo. . Pregnant women are enrolled and randomly assigned to receive either the experimental drug, the standard drug or placebo. Women are followed through delivery and classified as delivering preterm (< 37 weeks) or not. The data are shown below. Is there a statistically significant difference in the proportions of women delivering preterm among the three treatment groups? Run the test at a 5% level of significance. Preterm Experimental Standard Drug Placebo Delivery Drug Yes 17 23 35 No 83 77 65 Solution: Solution: Chi-Square Test Preterm Delivery Observed Frequencies Column variable Experimental Standard Placebo Total Drug Yes No Total Preterm Delivery Yes No Total Drug 17 23 45 83 77 66 100 100 111 Expected Frequencies Column variable Experimental Standard Drug Drug Placebo 27.33119 27.33119 30.33762 72.66881 72.66881 80.66238 100 100 111 Data Level of Significance Number of Rows Number of Columns Degrees of Freedom 85 226 311 Total 85 226 311 0.05 2 3 2 Results Critical Value Chi-Square Test Statistic p-Value 5.991465 16.07015 0.000324 Reject the null hypothesis Since P-value is less than 0.05, so we can conclude that there is a statistically significant difference in the proportions of women delivering preterm among the three treatment groups Solution: Solution: Z Test of Hypothesis for the Proportion Data Null Hypothesis = 0.32 Level of Significance 0.05 Number of Successes 45 Sample Size 111 Intermediate Calculations 0.40540540 Sample Proportion 5 0.04427595 Standard Error 7 Z Test Statistic 1.92893415 2 Upper-Tail Test Upper Critical Value p-Value Reject the null hypothesis 1.64485362 7 0.02686951 9 Since P-value is less than 0.05, so we can conclude that the proportion of women delivering prematurely is significantly higher in the placebo group. Solution: Solution: Null hypothesis H0: 1 2=3 against the alternative H1: any of them are not equal. Since the sample size n=25+25=50>30 therefore the test will be two sample independent Z test and test x 1 x 2( 1 2) s 12 s 2 2 statistics is Z= where s1 and s2 are the standard deviation of the two sample and ( + ) n1 n2 n1 and n2 be the sample size of the two sample. Under the null hypothesis the test results as follows: Solution: a) Y=54.6-2.22*7 = 39.06 b) Y=54.6-2.22*0 = 54.6 Solution: Solution: We assume that censoring of data is done after the death occurs. Terminology: tj = Time of jth death dj = No. of death at 'tj' nj = No. of people alive just before jth death. j = Probability of death at tj. MEN j tj dj nj j = dj / nj 1 - j S(t) 1 5 1 15 0.0667 0.9333 0.9333 2 12 1 13 0.0769 0.9231 0.8615 3 14 1 12 0.0833 0.9167 0.7897 4 23 1 10 0.1000 0.9000 0.7108 5 29 1 5 0.2000 0.8000 0.5686 1 0 29 WOMEN j tj dj nj j = d j / nj 1 - j S(t) 1 19 1 10 0.1000 0.9000 0.9000 2 20 1 8 0.1250 0.8750 0.7875 3 21 1 7 0.1429 0.8571 0.6750 4 24 1 4 0.2500 0.7500 0.5063 1 0 < t < 19 0.9 19 < t < 20 0.7875 20 < t < 21 Survival Function = S(t) 0.675 21 < t < 24 0.5063 t > 24 1 BMI is measured in a sample of patients, with a mean of 31.7 and a standard deviation of 5.9. How many subjects would be needed to ensure that a 95% confidence interval estimate of BMI had a margin of error not exceeding 2 units? Solution: Given that margin of error E 2 sample 2 sample standard edviation Critical Z value at 5% level of significance * sample 2 Z0.05/2* 5.9 Z table at 5 level of significance Z sample 2 1.96* 5.9 sample 0.05/2 =Z0.025=1.96) 1.965.9 =5.782 2 2 Sample size 5.782 =33.43 33 Sample size 33 So 33 subjects would be needed at least to ensure that a 95% confidence interval estimate of BMI had a margin of error not exceeding 2 units. 2 The mean BMI in patients free of diabetes was elsewhere reported as 28.2. The researcher producing the data in question 1 wonders if the BMI in his patients free of diabetes is higher than this reported number. Based on the data in Problem 1 is there evidence that the BMI is significantly higher that 28.2? Use a 5% level of significance. Solution: Null hypothesis H0: 28.2 against the alternative H : 28.2 1 The test statistics will be Z test because n=33>30 and the test statistics Z= sample mean sample standard edviation Under the null hypothesis One-Sample Z Test of mu = 28.2 vs > 28.2 The assumed standard deviation = 5.9 N 33 Mean 31.70 SE Mean 1.03 95% Lower Bound 30.01 Z 3.41 P 0.000 Hence Z=3.41 and corresponding p value is 0.00 which is less than 0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that the BMI is significantly higher than 28.2. 3 Peak expiratory flow (PEF) is a measure of a patient's ability to expel air from the lungs. Patients with asthma or other respiratory conditions often have restricted PEF. The mean PEF for children free of asthma is 306. An investigator wants to test whether children with chronic bronchitis have restricted PEF. A sample of 40 children with chronic bronchitis is studied and their mean PEF is 279 with a standard deviation of 71. Is there statistical evidence of a lower mean PEF in children with chronic bronchitis? Run the appropriate test at a=0.05. Solution: Null hypothesis H0: 306 against alternative hypothesis H : 1 < 306 Since the sample size n=40>30 therefore test will be Z test and test statistics is sample standard edviation / sample sample mean Sample mean=279 and sample standard deviation =71. Under the null hypothesis the test result is: One-Sample Z Test of mu = 306 vs < 306 Z= The assumed standard deviation = 71 N 40 Mean 279.0 SE Mean 11.2 95% Upper Bound 297.5 Z -2.41 P 0.008 From the result we see Z=-2.41 and corresponding p value is 0.008< 0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that < 306 which means there is a statistical evidence of a lower mean PEF in children with chronic bronchitis. Considering again the study in Problem 3, a different investigator conducts a second study to investigate whether X there is a difference in mean PEF in children with chronic 4 bronchitis as compared to those without. Data on PEF are collected and summarized below. Based on the data, is there statistical evidence of a lower mean PEF in children with chronic bronchitis as compared to those without? Run the appropriate test at a=0.05. St M d Nu ea D mbe Gro n e r of up P v Chil E P dren F E F Chr onic 28 6 Bro 25 1 8 nchi tis No Chr onic 31 7 25 Bro 9 4 nchi tis Solution: Null hypothesis H0: 1 2 against the alternative H1: 1<2 where 1, 2 are the population mean of children with chronic bronchitis and without bronchitis. since sample size n =25+25=50>30 therefore the test will be two sample independent Z test and test x 1 x 2( 1 2) 2 2 s1 s2 statistics is Z= where s1 and s2 are the standard deviation of the two sample and ( + ) n1 n2 n1 and n2 be the sample size of the two sample. Under the null hypothesis the test results as follows: Two-Sample Z-Test Sample 1 2 N 25 25 Mean 281.0 319.0 SE Mean 14 15 StDev 68.0 74.0 Difference = mu (1) - mu (2) Estimate for difference: -38.0 95% CI for difference: (-78.4, 2.4) Z-Test of difference = 0 (vs not =): Z-Value = -1.89 P-Value = 0.02938 From the result Z=-1.89 and the corresponding p value =0.02938<0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that 1<2 which means there is a statistical evidence of a lower mean PEF in children with chronic bronchitis as compared to those without. 5 A clinical trial is run to investigate the effectiveness of an experimental drug in reducing preterm delivery to a drug considered standard care and to placebo. . Pregnant women are enrolled and randomly assigned to receive either the experimental drug, the standard drug or placebo. Women are followed through delivery and classified as delivering preterm (< 37 weeks) or not. The data are shown below. Is there a statistically significant difference in the proportions of women delivering preterm among the three treatment groups? Run the test at a 5% level of significance. Preterm Experimental Standard Drug Placebo Delivery Drug Yes 17 23 35 No 83 77 65 Solution: Solution: Chi-Square Test Preterm Delivery Observed Frequencies Column variable Experimental Standard Placebo Total Drug Yes No Total Preterm Delivery Yes No Total Drug 17 23 45 83 77 66 100 100 111 Expected Frequencies Column variable Experimental Standard Drug Drug Placebo 27.33119 27.33119 30.33762 72.66881 72.66881 80.66238 100 100 111 Data Level of Significance Number of Rows Number of Columns Degrees of Freedom 85 226 311 Total 85 226 311 0.05 2 3 2 Results Critical Value Chi-Square Test Statistic p-Value 5.991465 16.07015 0.000324 Reject the null hypothesis Since P-value is less than 0.05, so we can conclude that there is a statistically significant difference in the proportions of women delivering preterm among the three treatment groups Solution: Solution: Z Test of Hypothesis for the Proportion Data Null Hypothesis = 0.32 Level of Significance 0.05 Number of Successes 45 Sample Size 111 Intermediate Calculations 0.40540540 Sample Proportion 5 0.04427595 Standard Error 7 Z Test Statistic 1.92893415 2 Upper-Tail Test Upper Critical Value p-Value Reject the null hypothesis 1.64485362 7 0.02686951 9 Since P-value is less than 0.05, so we can conclude that the proportion of women delivering prematurely is significantly higher in the placebo group. Solution: Solution: Null hypothesis H0: 1 2=3 against the alternative H1: any of them are not equal. Since the sample size n=25+25=50>30 therefore the test will be two sample independent Z test and test x 1 x 2( 1 2) s 12 s 2 2 statistics is Z= where s1 and s2 are the standard deviation of the two sample and ( + ) n1 n2 n1 and n2 be the sample size of the two sample. Under the null hypothesis the test results as follows: Solution: a) Y=54.6-2.22*7 = 39.06 b) Y=54.6-2.22*0 = 54.6 Solution: Solution: Chi-square critical value with (row-1)(column -1) = (3-1)93-1) = 4 degrees of freedom = 9.49 Calculate chi-square: Find all expected frequencies: (use the method shown above) Treatment 1.....symptoms worsened=15, no effect=13.67, symptoms improved=21.33; total 50 Treatment 2.....symptoms worsened=15, no effect=13.67, symptoms improved=21.33 ; total 50 Treatment 3.....symptoms worsened=15, no effect=13.67, symptoms improved=21.33 ; total 50 --------Total ....... ........ ......... ........ . .45 ...... .... .41 .......... ....... ........ ...--------.64 .........150 Chi-squared: (O(i)-E(i) )^2 / E(i) = (22-15)^2/15+(14-13.67)^2/13.67+(14-21.3... (9-15)^2/15+(12-13.67)^2/13.67+(29-21.33... =11.356 11.356 > 9.49 Reject H0 , we have significant evidence, =0.05, to show that treatment and response are not independent. Solution: Solution: Apparently it's absolute weight loss (after - before), given your reported s.d. You haven't said whether you're testing for weight loss or weight gain, so I'll assume that you're interested in any weight change and use a two-tailed z test. At the 0.05 level of significance, z = +/- 1.96. The 95% confidence interval about zero (no change) is 0 +/- 1.96 (3.27 / sqrt(6) ) = +/- 2.61 The mean change = +0.67, which is well within the bounds of the CI, so this is not a significant result at the 0.05 level of significance. Alternatively, p = 1 - { CDF(0.67 / (3.27 / sqrt(6)) ) - CDF(-0.67 / (3.27 / sqrt(6)) ) } = 1 - (0.6915 - 0.3085) = 0.617 so that the result is only significant at the 0.05 level of significance. Solution: Step 1. Set up hypotheses and determine level of significance. H0: The populations of self-reported improvement in pain who took new drug and those who took standard drug are equal H1: The populations of self-reported improvement in pain who took new drug and those who took standard drug are not equal. =0.05 Step 2. Select the appropriate test statistic. The test statistic for the Mann Whitney U Test is U. Step 3. Set up decision rule. The critical value for this two-sided test with n 1=n2=6 and =0.05 is found in Table and is 5. The decision rule is Reject H0 if U < 5. Step 4. Compute the test statistic. Total Sample (Ordered Smallest to Largest) New Standard Ranks New Drug Standard Drug New Standard 4 2 1 1 5 3 2 3 3 4 2 3 3 1 2 3 4 2 3 6.5 2 3 3 6.5 3 6.5 3 6.5 4 10 4 10 4 5 10 12 R1=39 U1 n1n 2 n1 (n 1 1) 6(7) R 1 6(6) 39 18 2 2 U 2 n1n 2 R2=30 n 2 (n 2 1) 6(7) R 2 6(6) 30 27 2 2 The test statistic is U = 27. Step 5. Conclusion. Do not reject H0 because 27 > 5. We do not have statistically significant evidence at =0.05 to show that populations of self-reported improvement in pain who took new drug and those who took standard drug are not equal. Solution: A: False B: False Solution: Hypotheses of interest : Ho: d (mean difference) = 1 - 2 = 0 against Ha : d 0. where 1, 2 are mean age of children assigned to the new drug and placebo respectively. Test: Paired T test. Test Statistics : t = [ (x1-bar - x2-bar ) - ( 1 - 2 ) ] / { [s1^2 + s2^2 ] / n} Here t = [ (12.4 - 13 ) - 0 ] / [[3.62 + 42] / 125] = -1.25 Here our test statistics is student's t statistic. The degrees of freedom is the number of independent estimates of variance = 124 Tabulated value of t for 124 degrees of freedom at 5% level of significance for two-tailed test is +/- 1.97. Decision: Since the modulus of calculated value of t ( |-1.25| = 1.25 ) is less than the modulus of tabulated value of t (|-1.97| = 1.97 ), so it is inside the acceptance region of the test and thus is not significant. Hence we are failed to reject the null hypothesis at 5% level of significance. Conclusion: At 5% level of significance we can conclude that there is not a statistically significant difference in the ages of children assigned to the treatments. Solution: Solution: We assume that censoring of data is done after the death occurs. Terminology: tj = Time of jth death dj = No. of death at 'tj' nj = No. of people alive just before jth death. j = Probability of death at tj. MEN j tj dj nj j = dj / nj 1 - j S(t) 1 5 1 15 0.0667 0.9333 0.9333 2 12 1 13 0.0769 0.9231 0.8615 3 14 1 12 0.0833 0.9167 0.7897 4 23 1 10 0.1000 0.9000 0.7108 5 29 1 5 0.2000 0.8000 0.5686 1 0 29 WOMEN j tj dj nj j = d j / nj 1 - j S(t) 1 19 1 10 0.1000 0.9000 0.9000 2 20 1 8 0.1250 0.8750 0.7875 3 21 1 7 0.1429 0.8571 0.6750 4 24 1 4 0.2500 0.7500 0.5063 1 0 < t < 19 0.9 19 < t < 20 0.7875 20 < t < 21 Survival Function = S(t) 0.675 21 < t < 24 0.5063 t > 24