Question

A random sample of n₁ = 10 regions in New England gave the following violent crime rates (per million population).

x₁: New England Crime Rate

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 in the Rocky Mountain states gave the following violent crime rates (per million population).

x₂: Rocky Mountain Crime Rate

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 normal in both regions. Use a calculator to calculate x₁, s₁, x₂, and s₂. (Round your answers to two decimal places.)

1. Find S1

2. What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate. (Test the difference μ₁ − μ₂. Do not use rounded values. Round your answer to three decimal places.)

Answer 1

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