A random sample of six salespersons that attended a...

A random sample of six salespersons that attended a motivational course on sales techniques was monitored three months before and three months after the course. The table shows the values of sales (in thousands of dollars) generated by these six salespersons in the two periods. Salesperson Before the Course After the Course 1 207 234 2 287 287 3 209 192 4 337 349 5 155 182 6 199 183 Assume that the population distributions are normal. Find a 90% confidence interval for the difference between the two population means. What is the confidence interval estimate of the mean difference (μ after - μ before)? ___ ≤ μD ≤ ____ (fill in the blanks, round to two decimal places as needed and show your work)

Transcript

00:01 So there are two cases is given in this question.

00:05 And the first salesperson before they are selling 207, 207 products.

00:13 And after this education, they are selling 200.

00:17 I mean, he is selling 234.

00:20 So what we have to do, we have to just find the difference between them, the difference them, which is 234 minus 207, which is 27.

00:31 And this is 0 and this one is negative this is 17 and this one is which is 12 and this one is 27 great and this one is negative 16 great so this is the difference of them so what we have to do first of all let's find the mean difference let's say this is the bar that is equal to the sum of the values here which are 27 plus 0 minus 17 plus 12 plus 27 and minus 16 and divided by there are six numbers here just divide by 16 6 so we got 33 negative this is 33 i think this is negative 33 negative 6 and 6 33 divided by 6 which is equal to 5 .5 great so we got the difference here let's find this column here this is the i minus 12 .5 and take the square of this one here i mean sorry this is 5 .5 because the mean difference is 5 .5 so 27 minus this is 5 .5 which is this one and take the square of this item which is 462 .25 great and the next one uh zero minus so this is 5 .5 and take the square of this number that is 30 .25 and this is negative 17 so this is negative 17 and minus 5 .5 that would be and take the square of this number which is 506 .25 and 12 minus 5 .5 and take the square of this number 2 which is 42 .25 and this is 27 minus 5 .5 and take the square of this number that is 462 .25 and this is negative 16 minus 5 .5 and take the square of this number that would be 462 .25.

03:02 Great.

03:03 So the next thing, so we need to just find the standard deviation of this difference.

03:08 So the standard deviation of the difference.

03:17 So the standard deviation of the difference, which is equal to the sum of all values here.

03:23 This is from i to 6, and this is the i minus the bar squared.

03:30 And divided by the number of item here, let's say this is n.

03:34 So we just got the square of all these numbers.

03:37 When i just add these numbers together, let me just add them up.

03:40 This is 462.

03:42 Point 25 plus 30 .25 plus 50 .25 plus 506 .25 plus this is 42 .25 plus and 462 .25.

04:02 And 4162 .25.

04:06 That is equal to 1 ,965 .5 and divided by the number of term here.

04:14 So let's divide this number by six and take the square of this number here.

04:19 Let's take the square of this number here.

04:22 And we got the value which is 18 points.

04:26 Let's say this is 010.

04:29 Great.

04:30 So this is 18 .10.

04:32 By the way, this is the standard deviation here.

04:38 So what we have to do, the next step, we need to find the confidence interval.

04:43 So because the question is being asked, find the confidence interval for this one.

04:49 So let me just recall the formula confidence interval, which is equal to the d bar plus...

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