During a test period, an experimental group of 10 vehicles using an 85 percent ethanol-gasoline mixture showed mean CO2 emissions of 667 pounds per 1000 miles, with a standard deviation of 20 pounds. A control group of 14 vehicles using regular gasoline showed mean CO2 emissions of 679 pounds per 1000 miles with a standard deviation of 15 pounds. At α = 0.05, in a left-tailed test (assuming equal variances) the test statistic is:. A. 1.321.
B. -2.508.
C. -2.074.
D. -1.717.

-1.683

Step-by-step explanation:

Given :

Group 1 :

x1 = 667 ; n1 = 10 ; s1 = 20

Group 2 :

x2 = 679 ; n2 = 14 ; s2 = 15

The test statistic assuming equal variance :

x1 - x2 / √[Sp² * (1/n1 + 1/n2)]

sp² = [(n1 - 1)*s1² + (n2 - 1)*s2²] ÷ (n1 + n2 - 2)

Sp² = [(10 - 1)*20² + (14 - 1)*15²] = 296.59

Test statistic =

(667 - 679)/ √[296.59 * (1/10 + 1/14)]

-12 / 7.1304978

Test statistic = - 1.682

5.6%

step-by-step explanation:

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v