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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

3.53.74.23.93.34.11.84.82.93.1

Another random sample of n2 = 12 regions inthe Rocky Mountain states gave the following violent crime rates(per million population).

x2: Rocky Mountain CrimeRate

3.54.14.55.33.34.83.52.43.13.55.22.8

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 μ1 − μ2. Donot use rounded values. Round your answer to three decimalplaces.)

Answer & Explanation Solved by verified expert
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(3.5   ,3.7,   4.2,   3.9,   3.3,   4.1,   1.8,   4.8,   2.9,   3.1)
x1
length(x1)
x2 <- c(3.5   ,4.1,   4.5,   5.3,   3.3,   4.8,   3.5,   2.4   ,3.1,   3.5   ,5.2,   2.8)
length(x2)
mean(x1)
sd(x1)
mean(x2)
sd(x2)

Output:

length(x1)
[1] 10
> x2 <- c(3.5   ,4.1,   4.5,   5.3,   3.3,   4.8,   3.5,   2.4   ,3.1,   3.5   ,5.2,   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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