R Activity in BIOSTATISTICS

Autida, Trexia B.Sutliz, Larry J.

Torrejas, April Rose C.

BSE-BIOLOGY 3TTh 8:30 – 10:00 A.M.

> #ACTIVITY 1

> #1.1. Pick two different integers of your choice and assign x to be the smaller integer and y the bigger integer.> x<-3> y<-8

> #1.2. Find the results of these R commands:> #a> x+y[1] 11> x<-5> y<-9> x+y[1] 14> #b> sqrt(x)[1] 2.236068> #c> x^2[1] 25> #d> y-5[1] 4> #e> y-x[1] 4> #f> x/y[1] 0.5555556> #g> x*y[1] 45> #h> y*7[1] 63> #i> log(y)[1] 2.197225> #j> factorial(x)[1] 120

> #1.3. Find the value of the following expressions.> #a.> sqrt(x^2+y^2)[1] 10.29563> #b> sqrt((y-x)/(x*y))[1] 0.2981424> #c> ((x*y)/y)^2[1] 25

> #d> factorial(y)/(2*factorial(x))[1] 1512

> #1.II.DATA ENTRY> #1. Suppose you list your commute times for two weeks (10 days) and you listed the following times in minutes,

17 16 20 24 22 15 21 15 17 22a) Enter these numbers into R.b) Find the longest commute time and the minimum commute time.c) Arrange the data in increasing order.d) List the number categories

> #a> H<-c(17,16,20,24,22,15,21,15,17,22)> H [1] 17 16 20 24 22 15 21 15 17 22> #b> max(H)[1] 24> #c> min(H)[1] 15> #d> sort(H) [1] 15 15 16 17 17 20 21 22 22 24> #f> table(H)> H15 16 17 20 21 22 24 2 1 2 1 1 2 1

> #2. Your cell phone bills vary from month to month. Suppose your year has the following monthly amounts

460 330 390 370 460 300 480 320 490 350 300 480

a) Enter this date into a variable called phone bill.b) How much have you spent this year on cell phone bill?c) What is the smallest amount you spent in a month?d) What is the largest?e) Give the amounts greater than 400.

> #a> phone.bill<-c(460,330,390,370,460,300,480,320,490,350,300,480)> phone.bill [1] 460 330 390 370 460 300 480 320 490 350 300 480> #b

> sum(phone.bill)[1] 4730> #c> min(phone.bill)[1] 300> #d> max(phone.bill)[1] 490> #e> phone.bill[phone.bill>400][1] 460 460 480 490 480> #3 Suppose 4 people are asked three questions; their wight (lbs), height (cm), and gender. The data are as follows:

Weight Height Gender150 65 female135 61 female210 70 male140 65 female166 61 male

a) Enter the data in R.b) Extract a data frame which holds only the weight and the height column..c) Extract the information of the tallest person.d) Make a table with assigned names (of your own choice) on the 4 people.

> #3.a> Weight<-c(150 ,135,210,140,166)> Height<-c(65,61,70,65,61)> Gender<c("female","female","male","female","male")Error: object 'Gender' not found> Gender<-c("female","female","male","female","male")> df<-data.frame(Weight,Height,Gender)> df Weight Height Gender1 150 65 female2 135 61 female3 210 70 male4 140 65 female5 166 61 male

> #3.b> df[,1:2] Weight Height1 150 652 135 613 210 704 140 655 166 61> #3.c> df['3',] Weight Height Gender3 210 70 male> #3.d> row.names(df)<-c("Trexia","April","Larry","April","Stefan")> df Weight Height GenderTrexia 150 65 femaleApril 135 61 femaleLarry 210 70 maleApril 140 65 femaleStefan 166 61 male

> #ACTIVITY 2> #2.1.> hours<-c(5,20,8,11,15,10,6,8,5,14,5,8,6,10,9,13,9,9,10,20,6,7,20,7,20,6,8,9,5,6)> hist(hours,col="green")> #2.2> X<-c(2,3,4,6,8,9,10,10,11,12)> Y<-c(4,5,6,8,8,10,10,11,12,12)> plot(X,Y,type="o",xlab="X",ylab="Y",ylim=c(1,15),main="X and Y",col="green")

#2.1. The number of hours by selected 30 high school students on a computer games per week are recorded below.

5 10 5 13 6 620 6 8 9 7 88 8 6 9 20 9

11 5 10 10 7 515 14 9 20 20 6

a) Enter the data into R.b) Make a histogram indicating the title and labels.

>#2.2. Enter the following data in R and create a line graph.Make your graph appears more attractive.

x 2 3 4 6 8 9 10 10 11 12y 4 5 6 8 8 10 10 11 12 12

X<-c(2,3,4,6,8,9,10,10,11,12)Y<-c(4,5,6,8,8,10,10,11,12,12)plot(X,Y,type="o",xlab="X",ylab="Y",ylim=c(1,15),main="X and Y",col="blue")

> 2.3. Load the built-in data set cars by typing<cars

in the command line. Make a line graph which indicates the points on thegraph and make appropriate labels and title.

> cars speed dist1 4 22 4 10

3 7 44 7 225 8 166 9 107 10 188 10 269 10 3410 11 1711 11 2812 12 1413 12 2014 12 2415 12 2816 13 2617 13 3418 13 3419 13 4620 14 2621 14 3622 14 6023 14 8024 15 2025 15 2626 15 5427 16 3228 16 4029 17 3230 17 4031 17 5032 18 4233 18 5634 18 7635 18 8436 19 3637 19 4638 19 6839 20 3240 20 4841 20 5242 20 5643 20 6444 22 6645 23 5446 24 7047 24 9248 24 9349 24 12050 25 85

> plot(cars$speed,cars$dist,type="o",xlab="Speed",ylab="Distance",ylim=c(0,120),main="Speed and Distance",col="black")

>#ACTIVITY 3>#A.

>#3.1. You want to by reconditioned cellphone and find that over three months of watching at Gaisano mall, you see the following prices (suppose the cellphones are all similar)

9000 9500 9400 9400 10000 9500 10300 10200Use R commands to find

a) the mean valueb) the medianc) What is the variance?d) The standard deviation?

> prices<-c(9000,9500,9400,9400,10000,9500,10300,10200)> mean(prices)[1] 9662.5> median(prices)[1] 9500> var(prices)[1] 205535.7> sd(prices)[1] 453.3605

>#3.2. Fifteen randomly selected statistics students were asked for the number of hours they spent in studying at night. The resulting data are as follows:

2 2 3 1 4 5 2 3 2 4 3 2 1 1 1.5

> study<-c(2,2,3,1,4,5,2,3,2,4,3,2,1,1,1.5)

> #a. Arrange the data in increasing order

> sort(study) [1] 1.0 1.0 1.0 1.5 2.0 2.0 2.0 2.0 2.0 3.0 3.0 3.0 4.0 4.0 5.0

>#b. Make a table of the possible categories. > table(study)study 1 1.5 2 3 4 5 3 1 5 3 2 1

>#c. What is P8 ? D4 ? Q1?

> quantile(study,0.08)8% 1 > quantile(study,0.40)40% 2 > quantile(study,0.25) 25% 1.75

>#B.

>#3.3.Load the built-in data set trees by typing > trees

> trees Girth Height Volume1 8.3 70 10.32 8.6 65 10.33 8.8 63 10.24 10.5 72 16.45 10.7 81 18.86 10.8 83 19.77 11.0 66 15.68 11.0 75 18.29 11.1 80 22.610 11.2 75 19.911 11.3 79 24.212 11.4 76 21.013 11.4 76 21.414 11.7 69 21.315 12.0 75 19.116 12.9 74 22.217 12.9 85 33.818 13.3 86 27.419 13.7 71 25.720 13.8 64 24.921 14.0 78 34.522 14.2 80 31.723 14.5 74 36.324 16.0 72 38.325 16.3 77 42.626 17.3 81 55.427 17.5 82 55.728 17.9 80 58.329 18.0 80 51.5

30 18.0 80 51.031 20.6 87 77.0

>#a. Use summary command on the data. What does it give?

> summary(trees) Girth Height Volume Min. : 8.30 Min. :63 Min. :10.20 1st Qu.:11.05 1st Qu.:72 1st Qu.:19.40 Median :12.90 Median :76 Median :24.20 Mean :13.25 Mean :76 Mean :30.17 3rd Qu.:15.25 3rd Qu.:80 3rd Qu.:37.30 Max. :20.60 Max. :87 Max. :77.00

>#b. What is the total of all Girths?

> sum(trees$Girth)[1] 410.7

>#c. Give the table of frequencies of the heights.

> table(trees$Height)

63 64 65 66 69 70 71 72 74 75 76 77 78 79 80 81 82 83 85 86 87 1 1 1 1 1 1 1 2 2 3 2 1 1 1 5 2 1 1 1 1 1

>#d. Give the volume of the trees greater than 40.

> trees$Volume[trees$Volume>40][1] 42.6 55.4 55.7 58.3 51.5 51.0 77.0

>#3.4. Below is a cost and return analysis in marketing tomato using different packaging material.

>#a. Enter the data in R which gives a similar table.

>Packaging.material<-c("Control","Cellophane","Wooden.box")> Product.quantity<-c(20,20,20)> Gross.Income<-c(400,540,700)> Cost.Pesos<-c(275,258,821)> Net.return<-c(124,281,378)> df<-data.frame(Packaging.material,Product.quantity,Gross.Income,Cost.Pesos,Net.return)> df Product.quantity Gross.Income Cost.Pesos Net.returnControl 20 400 275 124Cellophane 20 540 258 281Wooden Box 20 700 821 378

>#b.What is the mean cost?

> mean(Cost.Pesos)[1] 451.3333

>#c) Which material has the highest net return?

> df[3,] Product.quantity Gross.Income Cost.Pesos Net.returnWooden Box 20 700 821 378

>#d) What is the standard deviation of the gross income?

> sd(Gross.Income)[1] 150.1111

>#ACTIVITY 4

>#4.1. Find the value of:

>#a. 7P7 > factorial(7)/factorial(7-7)[1] 5040

>#b) 99 C 66 > choose(99,66)[1] 1.974439e+26

>#c. 9C6 > choose(9,6)[1] 84

>#d. 10P5 > factorial(10)/factorial(10-5)[1] 30240

>#e. 50C50> choose(50,50)[1] 1

>#4.2. Let z be a standard normal random variable. Find the following probabilitiesa. P ( z < 1.65 )b. P (-0.25 < z < 1.64)c. P( z > 1.91)

#a> pnorm(1.65)[1] 0.9505285

>#b> pnorm(1.64)-pnorm(-0.25)

[1] 0.5482037

>#c> 1-pnorm(1.91)[1] 0.02806661

>#B

>#4.3. A set of scores in a Statistics examination is approximately normally distributed with a mean of 74 and a standard deviation of 7.9. find the probability that a student received a score between 75 and 80.

> pnorm(80,74,7.9)-pnorm(75,74,7.9)[1] 0.2258569

>#4.4. A multiple choice quiz has 10 questions, each with four possible answers of which only one is the correct answer. What is the probability that sheer guesswork would yield at most 1 correct answer?

> y<-c(0,1)> sum(dbinom(y,10,.25))[1] 0.2440252

>#4.5. A family has 6 children. Find the probability P that there are

>#a.3 boys and 3 girls

> dbinom(3,6,0.5)[1] 0.3125

>#b.fewer boys than girls.

> #let x=number of boys where boys are fewer than girls> #let n=number of trials=6> #let p=probability of getting a boy=0.5> x<-c(2,1,0)

>sum(dbinom(x,6,0.5))[1] 0.34375

>#ACTIVITY 5

>#5.1. Students use many kinds of criteria when selecting course. “Teacher who is a very easy grader “ is often one criterion. Three teachers are scheduled to teach statistics.

> Professor.1<-c(12,16,35)> Professor.2<-c(11,29,30)> Professor.3<-c(27,25,15)> grades<-data.frame(Professor.1,Professor.2,Professor.3)> grades

Professor.1 Professor.2 Professor.31 12 11 272 16 29 253 35 30 15> row.names(grades)<-c("A","B","C")> grades Professor.1 Professor.2 Professor.3A 12 11 27B 16 29 25C 35 30 15> chisq.test(grades)

Pearson's Chi-squared test

data: gradesX-squared = 21.318, df = 4, p-value = 0.0002739

>#5.2. Test the hypothesis that the average running time of films produced by company exceeds the running time of films produced by company 1 by 10 minutes against the one sided alternative that the difference is more than 10 minutes. Use a 0.10 level of significance and assume the distributions of times to be approximately normal.

> company1<-c(102,86,98,109,92)> company2<-c(81,165,97,134,92,87,114)> t.test(company1,company2,mu=10,alt="greater",conf.level=0.90)

Welch Two Sample t-test

data: company1 and company2t = -1.8689, df = 7.376, p-value = 0.9491alternative hypothesis: true difference in means is greater than 1090 percent confidence interval: -29.62057 Infsample estimates:mean of x mean of y 97.4 110.0

before<-c(9,12,6,15,3,18,10,13,7)after<-c(9,17,9,20,2,21,15,22,6)t.test(before,after,mu=0,alt="greater",paired=T,conf.level=0.90)

>#3. As an aid for improving student’s habits, nine students were randomly selected to attend a seminar on the importance of education in life. The table shows the number of hours each student studied per week before and after the seminar. At ?=0.10, did attending the seminar increase the number of hours the students studied per week?

Before 9 12 6 15 3 18 10 13 7After 9 17 9 20 2 21 15 22 6

> before<-c(9,12,6,15,3,18,10,13,7)> after<-c(9,17,9,20,2,21,15,22,6)> t.test(before,after,mu=0,alt="greater",paired=T,conf.level=0.90)

Paired t-test

data: before and aftert = -2.8, df = 8, p-value = 0.9884alternative hypothesis: true difference in means is greater than 090 percent confidence interval: -4.663128 Infsample estimates:mean of the differences -3.111111

#ACTIVITY 6> #6.1. An educator wants to see how the number of absences a student in her class has affects the students' grade.> #a) Enter the data in R.> #b) Make a scatter plot on the data.> #c) Plot the regression line on the scatter plot.> #d) Use lm command for the regression analysis.> #e) What is the equation of the regression line?> #f) What is the estimated grade when the student has 7 absences?(Note: use the equation of the line formula and input it in R)> no.of.absence<-c(10,12,2,0,8,5)> final.grade<-c(70,65,96,94,75,82)> plot(final.grade ~ no.of.absence)> lm(final.grade ~ no.of.absence)

Call:lm(formula = final.grade ~ no.of.absence)

Coefficients: (Intercept) no.of.absence 96.784 -2.668

> a<-c(96.784)> b<-c(-2.668)> x=no.of.absence> y=(a+(b*x))> no.of.absence<-c(7)> y=a+(b*no.of.absence)> y[1] 78.108

>#6.2. Consider the following data of weight (g) and length (cm) of milkfish:

Weight 150 139 100 145 121 128 143 155 138 153Length 28 25 20 27 23 20 28 28 26 29

> #1. Generate a scatter plot on the data (length vs.weight).> #2. Fit the regression line on the scatter plot.> #3. Carefully examine the regression plot. What does this indicate?> #4. Find the correlation coefficient r and interpret.> weight<-c(150,139,100,145,121,128,143,155,138,153)> length<-c(28,25,20,27,23,20,28,28,26,29)> plot(length~weight)> plot(length~weight)> lm(length~ weight)

Call:lm(formula = length ~ weight)Coefficients:(Intercept) weight 1.1075 0.1771 > a<-c(1.1075)> b<-c(0.1771)> y=a+(b*weight)> y

[1] 27.6725 25.7244 18.8175 26.7870 22.5366 23.7763 26.4328 28.5580 25.5473[10] 28.2038> cor(length,weight,method="pearson")[1] 0.8939989> # the r value 0.893999 is closer to 1 which indicate that there is a positively high linear relationship between weight and length.> #6.3.> #a. Estimate the equation of the regression line.> #b. Predict the moisture content of the raw material if the relative humidity is 50.> #c. Compute the sample coefficient of determination and interpret.> x<-c(46,53,37,42,34,29,60,44,41,48,33,40)> y<-c(12,14,11,13,10,8,17,12,10,21,9,13)> lm(y~ x)

Call:lm(formula = y ~ x)

Coefficients:(Intercept) x -0.7367 0.3133

> a<-c(-0.7367)> b<-c(0.3133)> y=a+(b*x)> # b)> x<-c(50)> y=a+(b*x)> y[1] 14.9283> # c)> x<-c(46,53,37,42,34,29,60,44,41,48,33,40)> y<-c(12,14,11,13,10,8,17,12,10,21,9,13)> cor(y,x,method="pearson")[1] 0.7612409> # the r value 0.7612409 is closer to 1 which indicate that there is a positively high linear relationship between x and y.

>#ACTIVITY 7 The following data represent the scores in the final examination obtained by 4 students in mathematics, English, and biology:

Student

SubjectsMathematic

s English Biology1 68 57 612 83 94 863 72 81 594 55 73 66

Use a 0.05 level of significance t test the hypothesis that

>#a.the course are of equal difficulty;>#b.the students have equal ability.

> Math<-c(68,83,72,55)> Eng<-c(57,94,81,73)> Bio<-c(61,86,59,66)> student<-c("1","2","3","4")> #a> df<-data.frame(student,Math,Eng,Bio)> df student Math Eng Bio1 1 68 57 612 2 83 94 863 3 72 81 594 4 55 73 66> subj<-stack(df)Warning message:In stack.data.frame(df) : non-vector columns will be ignored> subj values ind1 68 Math2 83 Math3 72 Math4 55 Math5 57 Eng6 94 Eng7 81 Eng8 73 Eng9 61 Bio10 86 Bio11 59 Bio12 66 Bio> > anova(lm(values~ind, data=subj))Analysis of Variance Table

Response: values Df Sum Sq Mean Sq F value Pr(>F)ind 2 154.5 77.25 0.4407 0.6568Residuals 9 1577.8 175.31 > #b> stud<-c(1,2,3,4,1,2,3,4,1,2,3,4)> New<-data.frame(subj,stud)> New values ind stud1 68 Math 12 83 Math 23 72 Math 34 55 Math 45 57 Eng 16 94 Eng 27 81 Eng 38 73 Eng 49 61 Bio 110 86 Bio 211 59 Bio 3

12 66 Bio 4> anova(lm(values~ind+stud, data=subj))Analysis of Variance Table

Response: values Df Sum Sq Mean Sq F value Pr(>F)ind 2 154.50 77.25 0.3947 0.6863stud 1 12.15 12.15 0.0621 0.8095Residuals 8 1565.60 195.70 >#7.2. The strains of rats were studied under 2 environmental conditions for their performances in a maze test>#Use a 0.05 level of significance to test the hypothesis that>#a.there is no difference in error scores for different environments;>#b.there is no difference in error scores for different strains; the environments and strains of rats

> Environment<-c("free","free","free","free","restricted","restricted","restricted","restricted")> Bright<-c(28,12,22,23,72,32,48,93)> Mixed<-c(33,83,36,14,60,89,35,126)> Dull<-c(101,94,33,83,136,120,38,153)> Performances<-data.frame(Environment,Bright,Mixed,Dull)> > Performances Environment Bright Mixed Dull1 free 28 33 1012 free 12 83 943 free 22 36 334 free 23 14 835 restricted 72 60 1366 restricted 32 89 1207 restricted 48 35 388 restricted 93 126 153> Rats<-stack(Performances)Warning message:In stack.data.frame(Performances) : non-vector columns will be ignored> Rats values ind1 28 Bright2 12 Bright3 22 Bright4 23 Bright5 72 Bright6 32 Bright7 48 Bright8 93 Bright9 33 Mixed10 83 Mixed11 36 Mixed12 14 Mixed13 60 Mixed14 89 Mixed15 35 Mixed16 126 Mixed

17 101 Dull18 94 Dull19 33 Dull20 83 Dull21 136 Dull22 120 Dull23 38 Dull24 153 Dull> New.rats<-data.frame(Rats,Environment)> New.rats values ind Environment1 28 Bright free2 12 Bright free3 22 Bright free4 23 Bright free5 72 Bright restricted6 32 Bright restricted7 48 Bright restricted8 93 Bright restricted9 33 Mixed free10 83 Mixed free11 36 Mixed free12 14 Mixed free13 60 Mixed restricted14 89 Mixed restricted15 35 Mixed restricted16 126 Mixed restricted17 101 Dull free18 94 Dull free19 33 Dull free20 83 Dull free21 136 Dull restricted22 120 Dull restricted23 38 Dull restricted24 153 Dull restricted> #a> anova(lm(values~Environment, data=New.rats))Analysis of Variance Table

Response: values Df Sum Sq Mean Sq F value Pr(>F) Environment 1 8067 8066.7 5.5509 0.02779 *Residuals 22 31971 1453.2 ---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1> #b> anova(lm(values~ind, data=New.rats))Analysis of Variance Table

Response: values Df Sum Sq Mean Sq F value Pr(>F) ind 2 11834 5917.2 4.4059 0.02525 *Residuals 21 28203 1343.0 ---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

> anova(lm(values~ind+Environment, data=New.rats))Analysis of Variance Table

Response: values Df Sum Sq Mean Sq F value Pr(>F) ind 2 11834.3 5917.2 5.8771 0.009824 **Environment 1 8066.7 8066.7 8.0121 0.010334 * Residuals 20 20136.3 1006.8 ---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

>#7.3. The data of the initial weight of quail (Coturnix Japonica) subjected to different light bulb are given below.

Treatment

Observation1 2 3 4

A 75 76 78 80B 79 76 75 76C 79 78 75 78D 76 78 76 78

Analyze the data using alpha 0.01. Compare it when alpha is 0.05. Interpret your results.

> Obs1=c(75,79,79,76)> Obs2=c(76,76,78,78)> Obs3=c(78,75,75,76)> Obs4=c(80,76,78,78)> Treatment=c("A","B","C","D")> Trmt=data.frame(Treatment,Obs1,Obs2,Obs3,Obs4)> Trmt Treatment Obs1 Obs2 Obs3 Obs41 A 75 76 78 802 B 79 76 75 763 C 79 78 75 784 D 76 78 76 78> LAREXIL=stack(Trmt)Warning message:In stack.data.frame(Trmt) : non-vector columns will be ignored> Treat=data.frame(LAREXIL,Treatment)> Treat values ind Treatment1 75 Obs1 A2 79 Obs1 B3 79 Obs1 C4 76 Obs1 D5 76 Obs2 A6 76 Obs2 B7 78 Obs2 C8 78 Obs2 D9 78 Obs3 A10 75 Obs3 B11 75 Obs3 C12 76 Obs3 D

13 80 Obs4 A14 76 Obs4 B15 78 Obs4 C16 78 Obs4 D

> anova(lm(values~ind, data=Treat, conf.level=0.99))Analysis of Variance Table

Response: values Df Sum Sq Mean Sq F value Pr(>F)ind 3 8.1875 2.7292 1.065 0.4002Residuals 12 30.7500 2.5625 Warning message:In lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) : extra argument ‘conf.level’ is disregarded.

> anova(lm(values~ind, data=Treat, conf.level=0.95))Analysis of Variance TableResponse: values Df Sum Sq Mean Sq F value Pr(>F)ind 3 8.1875 2.7292 1.065 0.4002Residuals 12 30.7500 2.5625 Warning message:In lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) : extra argument ‘conf.level’ is disregarded.