8 question 5 8327 part 1 of 4 hw score 197 217 of 11 points...

Name: 8 question 5 8327 part 1 of 4 hw score 197 217 of 11 points points 0 of 1 a paired difference experiment yielded nd pairs of observations in each case what is the rejection region for testi 45862
Uploaded: 2022-02-22T02:51:52-08:00
Duration: 3 min 31 s
Channel: Jameson Kuper
Description: 8 question 5 8327 part 1 of 4 hw score 197 217 of 11 points points 0 of 1 a paired difference experiment yielded nd pairs of observations in each case what is the rejection region for testi 45862

8 Question 5, 8.3.27 Part 1 of 4 HW Score: 19.7%, 2.17 of 11 points Points: 0 of 1 A paired difference experiment yielded $n_d$ pairs of observations. In each case, what is the rejection region for testing $H_0: \mu_d = 2$ against $H_a: \mu_d > 2$? a. $n_d = 14, \alpha = 0.05$ t (Round to three decimal places as needed.)

Transcript

00:01 Here's a solution to 935.

00:02 We're given some hypotheses here, and it's a one -tailed test, and we're asked to find the rejection region for testing this particular hypothesis.

00:10 So we have different scenarios here, and these are right -tailed tests now.

00:14 So i'm going to draw a little picture here, and the degrees of freedom would be nine, and this alpha here, this represents this side over here, so that's 0 .05, right, that area there.

00:28 So we're going to use 1 minus since it's a right tilt test.

00:32 So 1 minus 0 .05 equals 0 .95, and that's the area.

00:37 Then we're going to use.

00:38 So you can actually use a table or a graph and calculator.

00:42 Since i don't have a table handy, i'm going to use a graph and calculator.

00:45 So if you go to inverse norm, i'm sorry, not inverse norm, inverse t, because we've got a t distribution, and put in that area of 0 .95.

00:53 You can put 0 .05, but it's going to give you the negative.

00:55 So we should have a critical value of positive.

00:57 And degrees of freedom is 9.

00:59 It's n minus 1, and then that gives us the critical region.

01:03 So 1 .833.

01:05 So t star equals 1 .833.

01:10 So the rejection region would be anything greater than 1 .833.

01:14 All right, very similar here.

01:16 So we have the same distribution.

01:18 This time the alpha equals 0 .10, which means the area is 1 minus that point 10, which is 0 .9.

01:26 And then the degrees of freedom is going to be 19, since we have 20 as our sample.

01:32 So either use a table or just go to inverse t on the calculator and put in your area of 0 .9 degrees of freedom is 19, and then that'll be 1 .3 to 8, we'll say...

Question

Please give Ace some feedback