An economics instructor would like to test the theory that spraying lavender essential oil in a room while studying will improve test scores. One group of participants studies in a room that releases 2 ounces of lavender every 3 to 4 hours. The second group studies in a room that does not release any essential oils. Both groups study the same amount of time leading up to all course exams. The instructor records the number of points each group gets on a total of 4 exams during a 5 week summer course. Which best describes the scale of measurement for the dependent variable in this experiment.

-1/2 +2

step-by-step explanation:

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.