STAT 503 – Statistical Methods for Biology 2026/2027 Homework 5 Key (8pages)

STAT 503 – Statistical Methods for Biology Homework 5 Key 30 Points. Due at 11:59 PM on Friday, October 28, 2026 Please use complete sentences unless the question is marked with an asterisk (*). Round answers to 3 decimal places Please show the key process of calculation. [I have included > 3 decima places in some answers. Give full credit if students' answers are off due to a rounding error.] 1. [Total: 9 points] One of the great discoveries of biology is that organisms have a class of genes called “regulatory genes,” whose only job is to regulate the activity of other genes. How many genes does the typical regulatory gene regulate? A study of interaction networks in yeast (S. cerevesiae) came up with the following data for 109 regulatory genes (Guelzim et al. 2002). Using R to do this problem: First imported the data #17 “Regulator_gene_data” in Brightspace, under the “Data for R” module, then answer the following questions: Number of genes regulated Frequency 1 20 2 10 3 7 4 7 5 8 6 8 7 5 8 2 9 4 10 4 11 3 12 4 13 5 14 1 15 2 16 1 17 3 18 2 19 2 20 3 22 3 This study source was downloaded by 100000857965443 from CourseHero.com on 07-14-2026 15:21:32 GMT -05:00 https://www.coursehero.com/file/183764411/stat503-Fall2026-hw5-keypdf/ STAT 503 – Statistical Methods for Biologists Modified: 2012-10-30 2 25 1 26 1 28 1 29 1 37 1 a. What type of graph should be used to display these data? [1 point] A histogram or cumulative frequency distribution. b. What is the estimated mean number of genes regulated by a regulatory gene in the yeast genome? (Hint: Using the mutate function in tutorial 5. [2 points,1 point for calculation, i.e. students show which formula they used, and 1 point for R coding] E(X)=∑????????????(????????)) =8.3 genes. Sample R codes: [Students’ code might be different from mine, but they should get the same results if it is correct] rm(list=ls()) library(dplyr) genepath<-file.choose() gene<-read.csv(genepath) reg_gene<-mutate(gene,p_x=gene$count/109) reg_gene Ex<-sum(reg_gene$Number.of.genes.regulated*reg_gene$p_x) Ex c. What is the standard error of the mean if the variance of the regulatory gene in the yeast genome is 56.508? [1 point] SE(????̅)=√????????????(????) √???? [0.5 pts] = √56.508 √109 ≈0.720 genes[0.25 for unit,0.25 correct value] Or R codes: sqrt(56.508)/sqrt(109) d. Explain what this standard error measures. [1 point] The standard error is 0.072 genes. It explains the spread of the sampling distribution of the mean number of genes regulated. e. What assumption are you making in part (c)? [1 point] That we have a simple random sample of the total population of regulatory genes. Here we do not require the underlying population distribution is normal since the sample size is sufficient large with n=109, so CLT applied here. This study source was downloaded by 100000857965443 from CourseHero.com on 07-14-2026 15:21:32 GMT -05:00 https://www.coursehero.com/file/183764411/stat503-Fall2026-hw5-keypdf/ STAT 503 – Statistical Methods for Biologists Modified: 2012-10-30 3 f. Find a 95% confidence interval for the population mean. [2 points] C=0.95, ???? = 1 − ???? = 0.05, ???? 2 = 0.025[0.25pts] Therefore, by Z table or R, ????????⁄2 = 1.96[0.5pts] ????̅± ????????⁄2 ???? √???? [0.5pts] = 8.3 ±1.96*0.72=8.3±1.418≈ (6.889,9.711) genes [0.5pts for two bounds, 0.25 pts for unit] Or R codes: > C <- 0.95 > sigma<-sqrt(56.508) > sigma [1] 7.51718 > xbar <- 8.3 > n <- 109 > z<- qnorm((1-C)/2,lower.tail = FALSE) > z [1] 1.959964 > c(xbar-z*sigma/sqrt(n),xbar+z*sigma/sqrt(n)) [1] 6.888796 9.711204