Two different formulations of an oxygenated motor fuel are being tested to study their road octane numbers. the variance of road octane number for formulation 1 is
σ1^2=1.5
, and for formulation, 2 it is
σ2^2=1.2
two random samples of size
n1=15andn2=20
are tested, and the mean road octane numbers observed are
ẋ1=89.6
and
ẋ2=92.50
. assume normality.
a. if formulation 2 produces a higher road octane number than formulation 1, the manufacturer would like to detect it. formulate and test an appropriate hypothesis using α = 0.05. what is the p-value?
b. explain how the question in part (a) could be answered with a 95% confidence interval on the difference in mean road octane number.
c. what sample size would be required in each population if you wanted to be 95% confident that the error in estimating the difference in mean road octane number is less than 1?

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .