Bryson_Biostat_HW4

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Biostatistics: Homework 4 Hannah Bryson Wed Feb 14 12:27:19 2024 EST Due Date: 2/19/2024 Instructions: For all hypothesis test problems, write H0 and H1 in words starting with ##. After the appropriate hypothesis test code, type the p-value and write decision and conclusion Question 1 Question 1a A cross between white and yellow summer squash gave progeny of the following colors. Color Observed White 155 Yellow 40 Green 11 For α = 0.10 , test if the data is consistent with some genetic model expectation of 12:3:1. #H0:Observed = Expected #H1: Observed =/= Expected O = c ( 155 , 40 , 11 ) E_p = c ( 12 / 16 , 3 / 16 , 1 / 16 ) chisq.test (O, p= E_p) ## ## Chi-squared test for given probabilities ## ## data: O ## X-squared = 0.32362, df = 2, p-value = 0.8506 ## p-value=0.8506 > alpha=0.10. We do not reject H0; the observed progeny phenotype are not significantly different from the expected. Question 1b The following table shows the population distribution and the number of participants in clinical data involving lung cancers. Please knit the file to see the table in better format.
Race/ethnic groups Number of patients in clinical trials Distribution of population White non-hispanic 3855 75.6% Hispanic 65 9.1% Black 311 10.8% Asian/pacific islander 54 3.8% American Indian/ Alaskan Native 12 0.7% Total 4297 100% Use α = 0.10 to test the claim that the description of the criminal trial participants fits well with the population distribution. #H0:Observed = Expected #H1: Observed =/= Expected O = c ( 3855 , 65 , 311 , 54 , 12 ) E_p = c (. 756 ,. 091 ,. 108 ,. 038 ,. 007 ) chisq.test (O, p= E_p) ## ## Chi-squared test for given probabilities ## ## data: O ## X-squared = 519.56, df = 4, p-value < 2.2e-16 ### p-value<2.2e-16 < alpha=0.10. We reject H0;the observed population distribution is significantly different from the expected. Question 1c According to Benford’s law, a variety of data sets include numbers with leading digits that follow the distribution shown in the following table: Leadin g digits 1 2 3 4 5 6 7 8 9 Benfor d’s distrib ution 28.1% 19.6% 10.5% 9.7% 9.9% 6.7% 5.8% 5.1% 4.6% When analyzing the leading digits of the magnitude of gradients from MRI of a patient, the following frequencies were found: 0, 15, 0, 76, 479, 182, 8, 23, and 0 correspond to the digits 1-9. Use $=$0.01 significance level to test the goodness of fit with Benford’s law whether \ the lending digits appear to fit the distribution.
#H0: Observed = Expected #H1: Observed =/= Expected O = c ( 0 , 15 , 0 , 76 , 479 , 182 , 8 , 23 , 0 ) E_p = c (. 281 ,. 196 ,. 105 ,. 097 ,. 099 ,. 067 ,. 058 ,. 051 ,. 046 ) chisq.test (O, p= E_p) ## ## Chi-squared test for given probabilities ## ## data: O ## X-squared = 2900.5, df = 8, p-value < 2.2e-16 ### p-value<2.2e-16 < alpha=0.01. We reject H0; the observed MRI frequency is significantly different from the expected. Question 2 Question 2a In investigating cold tolerance of adults of a species of tropical butterfly, 68 of the butterflies (32 female and 36 males) were subjected to a cold temperature. 20 of the female survived so are 14 males. Complete the contingency table. With α = 0.05 , test whether survival of females are different from males in the experimental cold temperature. ( H int : you need to complete following table and then you may have to write the values in matix form ) Female Male Total Alive Dead Total 68 #H0:female survived = males survived #H1: female survived =/= males survived A = matrix ( c ( 20 , 12 , 14 , 22 ), nrow= 2 ) A ## [,1] [,2] ## [1,] 20 14 ## [2,] 12 22 chisq.test (A) ## ## Pearson's Chi-squared test with Yates' continuity correction ## ## data: A ## X-squared = 2.8924, df = 1, p-value = 0.089 chisq.test (A) $ expected
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## [,1] [,2] ## [1,] 16 18 ## [2,] 16 18 ### p-value=0.089 > alpha=0.05. We do not reject H0; the female survival rate is not significantly different from the male survival rate. Question 2b In a study of patients infected with Cl o st r id ium di f f i ci l e , 8 out of 37 patients treated with Vancomycin were cured from the infection while 13 out of 26 treated with fecal transplant (from healthy person to restore gut bacteria) were cured. Construct a contingency table. With α = 0.05 , test whether one treatment method has better outcome than the other. First make a table and enter the values. cured not cured Total Vancomycin 8 29 37 fecal transplant 13 13 26 Total 21 42 63 #H0: vancomycin = fecal transplant #H1: vancomycin =\= fecal transplant B = matrix ( c ( 8 , 13 , 29 , 13 ), nrow= 2 ) B ## [,1] [,2] ## [1,] 8 29 ## [2,] 13 13 chisq.test (B) ## ## Pearson's Chi-squared test with Yates' continuity correction ## ## data: B ## X-squared = 4.3304, df = 1, p-value = 0.03744 chisq.test (B) $ expected ## [,1] [,2] ## [1,] 12.333333 24.66667 ## [2,] 8.666667 17.33333 ### p-value=0.03744 < alpha=0.05. We reject H0; the methods of treatment are significantly different.
Question 3 Question 3a The national Cesarean delivery rate for a recent year was 32.2% (number of live births performed by Cesarean section). A random sample of 100 birth records from three large hospitals showed the following results for type of birth. Test for homogeneity of proportions using α = 0.10. Cesarean Non cesarean Hospital A 44 56 Hospital B 28 72 Hospital C 39 61 #H0: p1=p2 #H1: p1=/=p2 C = matrix ( c ( 44 , 56 , 28 , 72 , 39 , 61 ), nrow= 3 , byrow= TRUE ) C ## [,1] [,2] ## [1,] 44 56 ## [2,] 28 72 ## [3,] 39 61 chisq.test (C) ## ## Pearson's Chi-squared test ## ## data: C ## X-squared = 5.7486, df = 2, p-value = 0.05646 chisq.test (C) $ expected ## [,1] [,2] ## [1,] 37 63 ## [2,] 37 63 ## [3,] 37 63 ### p-value=0.05646 < alpha=0.10. We reject H0; the proportion of the people who have had Cesareans vs. people who have not had a Cesareans are different Question 3b On average, 79% of American fathers are in the delivery room when their children are born. A physician‘s assistant surveyed 300 randomly selected first-time fathers to determine if they had been in the delivery room when their children were born. The results are shown here. Present Not Present
Present Not Present Hospital B 60 15 Hospital C 57 18 Hospital D 56 19 At α = ¿ 0.05, is there enough evidence to reject the claim that the proportions of those who were in the delivery room at the time of birth are the same? #H0: p1=p2 #H1: p1=/=p2 D = matrix ( c ( 66 , 9 , 60 , 15 , 57 , 18 , 56 , 19 ), nrow= 4 , byrow= TRUE ) D ## [,1] [,2] ## [1,] 66 9 ## [2,] 60 15 ## [3,] 57 18 ## [4,] 56 19 chisq.test (D) ## ## Pearson's Chi-squared test ## ## data: D ## X-squared = 5.0003, df = 3, p-value = 0.1718 chisq.test (D) $ expected ## [,1] [,2] ## [1,] 59.75 15.25 ## [2,] 59.75 15.25 ## [3,] 59.75 15.25 ## [4,] 59.75 15.25 ### p-value=0.1718 > alpha=0.05. We do not reject H0; the proportion of fathers present during delivery are significantly different from the fathers not present
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