Astatistics instructor who teaches a lecture section of 160 students wants to determine whether students have more difficulty with one-tailed hypothesis tests or with two-tailed hypothesis tests. on the next exam, 80 of the students, chosen at random, get a version of the exam with a 10-point question that requires a one-tailed test. the other 80 students get a question that is identical except that it requires a two-tailed test. the one-tailed students average 7.81 points, and their standard deviation is 1.06 points. the two-tailed students average 7.64 points, and their standard deviation is 1.33 points.

There is no evidence that there is no significant difference between the sample means

Step-by-step explanation:

given that a  statistics instructor who teaches a lecture section of 160 students wants to determine whether students have more difficulty with one-tailed hypothesis tests or with two-tailed hypothesis tests. On the next exam, 80 of the students, chosen at random, get a version of the exam with a 10-point question that requires a one-tailed test. The other 80 students get a question that is identical except that it requires a two-tailed test. The one-tailed students average 7.81 points, and their standard deviation is 1.06 points

The two-tailed students average 7.64 points, and their standard deviation is 1.33 points.

Group   One tailed X     Two tailed Y  

Mean7.81007.6400

SD1.06001.3300

SEM0.11850.1487

N80      80  

(Two tailed test)

The mean of One tailed X minus Two tailed Y equals 0.1700

t = 0.8940

 df = 158

p value =0.3727

 p is greater than alpha 0.05

There is no evidence that there is no significant difference between the sample means

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.

c

step-by-step explanation: