A paired difference experiment produced the following...

A paired difference experiment produced the following results: nd =38 x1 =92 x2 =95.5 xd =-3.5 sd 2 =21 a. Determine the values of z for which the null hypothesis, m1 -m2 =0, would be rejected in favor of the alterna-tive hypothesis, m1 -m2 60. Use a=.10. b. Conduct the paired difference test described in part a. Draw the appropriate conclusions. c. What assumptions are necessary so that the paired dif-ference test will be valid? d. Find a 90% confidence interval for the mean difference md. e. Which of the two inferential procedures, the confidence interval of part d or the test of hypothesis of part b, pro-vides more information about the differences between the population means?

Transcript

00:02 Hello students welcome we have a question here and in this question this is given that z is a standard normal variable so we need to find the z for each situation to two decimal places and our post situation situation a is given as the area area to the left of c is zero 0 .972.

00:46 This is our first question.

00:48 So let's start solving this question.

00:50 To solve this part, first of all, what we need to do, we have to drop our graph, a curve, sorry for this lines.

01:03 So here, let me make a curve over here.

01:11 This is the excess line and here i have drawn a curve.

01:15 So the value of c will be like here, you have to write as equal to 0.

01:24 And at this point, it is 0 .972.

01:30 And here, this is zz is equal to z0.

01:37 And this is the lift of the curve.

01:43 And here the value will be 0 .0.

01:47 2 to 8.

01:49 So let's calculate this also from the probability of z is less than or equal to z 0 .9772.

02:05 So how do we get the value of 0 .0 to 0 .0 to 8? this will be as p c is less than equal to value of z -0 is 1 .9 .9.

02:24 997 that is equal to 0 .9772 so here we have our first equation and this is our second equation so now from equation one and two on comparing we can see z not is equal to 1 .999 777 that is equal to two, rounded off to two decimal places, round to two decimal place.

03:12 So therefore we can state our answer now in this way, z is equal to z0 will be equal to 2 .00.

03:24 This will be the answer for the part b.

03:29 And let's move on to the part b.

03:32 So in part b, we have a condition as the area between the area between 0 and z, 0 and z is 0 .4772.

04:00 And here z is 0 .4772...

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