Statistical Exercises: Probability, Distribution & Confidence Intervals

Exercise 2.1A pnorm(42,35.29,4.49) 1-pnorm(42,35.29,4.49) 0.06753136 By hand: (45-35.29)/4.49= 1.49 prob= 0.9319 1-0.9319= 0.0681 Exercise 2.1B 1-pnorm(42,35.29,1.42) 1.148585e-06 By hand: 4.49/(sqrt(10))= 1.42 (standard error) (45-35.29)/1.42= 4.725 No probability because the value is too big Exercise 2.1C pnorm(41,35.29,4.49)-pnorm(33,35.39,4.49) 0.6010016 By hand: (41-35.29)/4.49=1.27 (33-35.29)/4.49= -0.51002 prob= 0.8980 prob= 0.3050 0.8980-0.3050= 0.593 Exercise 2.2A dbinom(0,20,0.53)+dbinom(1,20,0.53)+dbinom(2,20,0.53)+dbinom(3,20,0.53) 0.0005256292 Exercise 2.2B mean=(20*0.53) mean 10.6

sd=sqrt(20*0.53*0.47) sd 2.232039 normal_approximation=(3-mean)/sd pnorm(normal_approximation) 0.0003308714 The values are not that similar being 0.0005 and 0.0003. Although there is only a .0002 difference, there is a difference in r value and binomial distribution. Exercise 3.1A data=read.csv(file.choose()) data data=mfish mean(mfish) 23.59836 sd(mfish) 2.638705 hist(mfish,col="lightblue", xlim=c(0,30),ylim=c(0,250),breaks=15,xlab="Mosquito Fish Length (mm)",main="Histogram for Mosquito Fish Lengths") It is not normally distributed since the data can be seen as being skewed to the right.

Exercise 3.1B sample_means=replicate(500, mean(sample(fish$FishLength,10))) sample_means hist(sample_means, main="Histogram of Sample Means", xlab="Mean Length",breaks=20, col="lavender") sampling_distribution_mean=mean(sample_means) sampling_distribution_sd=sd(sample_means) sampling_distribution_mean sampling_distribution_sd Exercise 3.1C The data appears to be normally distributed

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Exercise 3.1D Sampling_distribution_mean 28.7258 Sampling_distribution_sd 2.180124 Exercise 3.1E standard_error=sd(mfish)/sqrt(10) standard_error 0.8344318 mu=mean(mfish) lower=mu-1.96*standard_error upper=mu+1.96*standard_error confidence_interval=c(lower,upper) Confidence_interval 21.96287 25.23385 count=0 for(i in 1:500) {if(lower<=sample_means[i]&&sample_means[i]<=upper) count=count+1} count/500*100 3.6 3.6% of the sample means calculated fall between the confidence interval which is 21.96287 and 25.23385 Exercise 3.1F 50 sample_means=replicate(500, mean(sample(fish$FishLength,50))) sample_means hist(sample_means, main="Histogram of Sample Means", xlab="Mean Length",breaks=20, col="lightpink") sampling_distribution_mean=mean(sample_means) sampling_distribution_sd=sd(sample_means) sampling_distribution_mean sampling_distribution_sd

The sampling distribution appears to be normal Sampling_distribution_mean 28.75564 Sampling_distribution_sd 0.9571983 Standard_error=sd(mfish)/sqrt(50) Standard_error .9669477 mu=mean(mfish) lower=mu-1.96*standard_error upper=mu+1.96*standard_error confidence_interval=c(lower,upper) Confidence_interval 26.86553 30.65597

count=0 for(i in 1:500) {if(lower<=sample_means[i]&&sample_means[i]<=upper) count=count+1} count/500*100 95.6 95% of the sample is between 26.86553 and 30.65597 100 sample_means=replicate(500, mean(sample(fish$FishLength,100))) sample_means hist(sample_means, main="Histogram of Sample Means", xlab="Mean Length",breaks=20, col="turquoise") sampling_distribution_mean=mean(sample_means) sampling_distribution_sd=sd(sample_means) sampling_distribution_mean sampling_distribution_sd

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The data is skewed to the right Sampling_distribution_mean 28.71972 Sampling_distribution_sd 0.6841676 standard_error=sd(mfish)/sqrt(100) Standard_error 0.2638705 mu=mean(mfish) lower=mu-1.96*standard_error upper=mu+1.96*standard_error confidence_interval=c(lower,upper) Confidence_interval 23.08117 24.11555 count=0 for(i in 1:500) {if(lower<=sample_means[i]&&sample_means[i]<=upper) count=count+1} count/500*100 0 95% of the values are between 23.08 and 24.12 Exercise 4.1A The data would be skewed to the right and would not be normally distributed library(psych) data=read.csv(file.choose()) data describe(bluegill$Length) vars n mean sd median trimmed mad min max range skew kurtosis se X1 1 888 120.03 41.95 135 119.98 53.37 46 220 174 -0.07 -1.37 1.41

hist(bluegill$length,col="hotpink",xlim=c(0,250),ylim=c(0,250),xlab="Bluegill Length", main="Bluegill Sunfish Lengths") This graph does not resemble a normal bell curve as can be seen in the graph above Exercise 4.1B qqnorm(bluegill$Length) qqline(bluegill$Length)

The data is not a straight line and deviates from the straight line Exercise 4.1C skew(bluegill$length) NA kurtosi(bluegill$length) NA There is no expected value for the skew or the kurtosi and thus the computed values are not close to the expected values Exercise 4.1D shapiro.test(bluegill$Length) Shapiro-Wilk normality test data: bluegill$Length W = 0.91704, p-value < 2.2e-16 Exercise 4.1E As can be seen in the above graphs, quartile plot, skew and kurtosi and the Shapiro-Wilk normality test we can conclude that the sample does not follow the normal distribution. Exercise 4.2A The data should be skewed towards the left library(psych) data=read.csv(file.choose()) data describe(data) hist(data,col="lightgreen",xlim=c(0,200),ylim=c(0,20),xlab="Income in thousands of dollars", main="Income in 1000's")

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This graph does not resemble a normal cell curve Exercise 4.2B qqnorm(income_data$Incomes) qqline(income_data$Incomes)

As can be seen at the beginning the data follows a straight line, however towards the end it skews from the line Exercise 4.2C skew(income_data$Incomes) 2.425454 kurtosi(income_data$Incomes) 6.828991 The computed skew value is 2.425454 and the computed kurtosi value is 6.828991, these values can be seen as being close to the expected values Exercise 4.2D shapiro.test(income_data$Incomes) Shapiro-Wilk normality test

data: income_data$Incomes W = 0.73049, p-value =3.199e-07 Exercise 4.2E As can be seen in the above graphs, quartile plot, skew and kurtosi and the Shapiro-Wilk normality test we can conclude that the sample does not follow the normal distribution.

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