Two researchers run identical experiments except researcher a collects twice as many points as researcher b. for a specific value x, researcher a estimates a y value of and researcher b estimates a y value of . we would expect that researcher a's 95% prediction interval around to be, in general, a) wider than researcher b's 95% prediction interval around y'b. b) narrower that researcher b's 95% prediction interval around y'b. c) the same as researcher b's 95% prediction interval around y'b. d) wider than researcher b's 95% prediction interval around y'b or narrower than researcher b's 95% prediction interval around y'b. with equal probability

B

Step-by-step explanation:

Consider an event A happening. If we do not have enough data to estimate its actual probability, we may choose a range 0.6 to 0.9 as a first case which indicates we are quite sure it will most likely occur. If however, we have enough data, we may estimate a range of 0.7 to 0.8 as a second case that is more certain on its actual likelihood of occurrence.

Say the actual probability of the event is given as 0.75, in the first case, we can infer the probability interval as 0.75 ± 0.15 (as 0.75-0.15=0.6 and 0.75+0.15=0.9 for the lower and upper bounds respectively). In the second case, we can infer the probability interval as 0.75±0.05 (as 0.75-0.05=0.7 and 0.75+0.05=0.8 for the lower and upper bounds respectively).

Thus, we can see that with more certainty of the event happening (with more data in this case), the probability or prediction intervals are lower.

Hence, in the experiment, we will observe a narrower prediction interval for researcher A who has more (twice as many points) data than researcher B who has fewer points.

hope this !

step-by-step explanation:

d = distance between moon and sun, which in this problem is the hypotenuse of a right triangle.  

the distance y is the side opposite to angle x in the right triangle.   use the definition of the sine function from basic trigonometry.

sin(x) = y/d ⇒

d = y/sin(x)