Watch corporation of switzerland claims that its watches on average will neither gain nor lose time during a week. a sample of 18 watches provided the following gains (+) or losses (−) in seconds per week. -.20

.20

-.40

-.32

.20

-.23

.00

.25

-.10

-.37

-.61

.10

-.30

-.64

.40

-.10

-.68

-.20

(a-1)
is it reasonable to conclude that the mean gain or loss in time for the watches is 0? use the .10 significance level. at a level of .10 significance, we reject h0: ? = 0 if t < or t > . (negative amounts should be indicated by a minus sign. round your answers to 3 decimal places.)

(a-2) the value of the test statistic is . (negative amount should be indicated by a minus sign. round your answer to 2 decimal places.)
(a-3) (click to select)cannot rejectreject h0: ? = 0. the p-value is (click to select)between 0.02 and 0.05between 0.05 and 0.1greater than 0.2between 0.1 and 0.2

the entire expression should be subtracted from 30. to subtract the quantity, distribute the negative sign, which changes the signs on the terms. so, the –2 changes to positive and gets added to 30 instead of subtracted from it.

hope this ! ^w^

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