A random sample of n1 = 10 regions in New England gave the following violent crime...

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Statistics

A random sample of n₁ = 10 regions in NewEngland gave the following violent crime rates (per millionpopulation).
x₁: New England CrimeRate
3.5 3.7 4.2 3.9 3.3 4.1 1.8 4.8 2.9 3.1
Another random sample of n₂ = 12 regions inthe Rocky Mountain states gave the following violent crime rates(per million population).
x₂: Rocky Mountain CrimeRate
3.5 4.1 4.5 5.3 3.3 4.8 3.5 2.4 3.1 3.5 5.2 2.8
Assume that the crime rate distribution is approximately normalin both regions. Use a calculator to calculatex₁, s₁,x₂, and s₂. (Round youranswers to two decimal places.)
1. Find S1
2. What is the value of the sample test statistic?Compute the corresponding z or t value asappropriate. (Test the difference Î¼₁ âˆ’ Î¼₂. Donot use rounded values. Round your answer to three decimalplaces.)

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4.5 Ratings (871 Votes)

Solution-1

use mean function in R studio to get mean and sd function in R to get the standard deviation

x1 <- c(3.5Â Â ,3.7,Â Â 4.2,Â Â 3.9,Â Â 3.3,Â Â 4.1,Â Â 1.8,Â Â 4.8,Â Â 2.9,Â Â 3.1)
x1
length(x1)
x2 <- c(3.5Â Â ,4.1,Â Â 4.5,Â Â 5.3,Â Â 3.3,Â Â 4.8,Â Â 3.5,Â Â 2.4Â Â ,3.1,Â Â 3.5Â Â ,5.2,Â Â 2.8)
length(x2)
mean(x1)
sd(x1)
mean(x2)
sd(x2)

Output:

length(x1)
[1] 10
> x2 <- c(3.5Â Â ,4.1,Â Â 4.5,Â Â 5.3,Â Â 3.3,Â Â 4.8,Â Â 3.5,Â Â 2.4Â Â ,3.1,Â Â 3.5Â Â ,5.2,Â Â 2.8)
> length(x2)
[1] 12
> mean(x1)
[1] 3.53
> sd(x1)
[1] 0.8287206
> mean(x2)
[1] 3.833333
> sd(x2)
[1] 0.9413079

mean of x1=3.53

standard deviation of x1=s1=0.83

mean of x2=3.83

standard deviationof x2=s2=0.94

Solution-2:

test statistic

t=x1-x2/sqrt(s1^2/n1+s2^2/n2)

t=(3.53-3.833333)/sqrt(0.8287206^2/10+0.9413079^2/12)

t=-0.8035032

t=-0.804

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