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7 CarvatVt15V >1SQuestion 545ptsTne [exia dcterminc %hetherwomen takes more averaze persona tanc thanmnen Based on its P-value; conclusion using 0 Gatc P-valuc 0...

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For Exercises $9-14,$ use the Wilcoxon signed-rank test to test each hypothesis. Compulsive Gamblers A group of compulsive gamblers was selected. The amounts (in dollars) they spent on lottery tickets for one week are shown. Then they were required to complete a workshop showing that the chances of winning were not in their favor. After they complete the workshop, test the claim that, at $\alpha=0.05,$ the workshop was effective in reducing the weekly amount spent on lottery tickets. $$ \begin{array}{l|ccccccc}{\text { Subject }} & {\mathrm{A}} & {\mathrm{B}} & {\mathrm{C}} & {\mathrm{D}} & {\mathrm{E}} & {\mathrm{F}} & {\mathrm{G}} & {\mathrm{H}} \\ \hline \text { Before } & {86} & {150} & {161} & {197} & {98} & {56} & {122} & {76} \\ \hline \text { After } & {72} & {143} & {123} & {186} & {102} & {53} & {125} & {72}\end{array} $$

So we're giving a problem that we have to find at the 5% significance level. The sample size is told to be of 10. So and is equal to 10. Which implies that are Significant value or critical value is at five and 10 and we're losing one till test. So we have our critical value of W. C. is equal to 44. Now when we take a look at this test that we're supposed to do it's one tailed. However which tail is it? We are asked to say that one is more effective than the other. So mm everything is mostly positive here and we find what are our values? Mhm. Use the following data set which gives the additional sleep in hours obtained by 10 patients who used love level high school. So I mean high bromide. So that's gonna be so the control group minus the other one or not. The control group. So what we'll just do here is calculate the both ends since I'm not sure if it's a To a left tail or right tail. So we'll go ahead and find out that right tail is at 44 and then the other tail is going to be 10 times 11 divided by two. So that's gonna be 55 minutes 44 Which is gonna be 11. Very handy. Okay. So if it's less than 11 won't reject and if it's greater than 44 will also reject. Sound good? So um we got 10 students at the positive values so 1.9 plus 0.8 plus 1.1 plus 0.1 was 4.4 plus 5.5 plus 1.6 plus 4.6 plus 3.4. And we get 23.4 for our w nut. Okay. And or critical value that just leads us straight in the middle. So at alpha equals .05 We failed to reject the null hypothesis and this was just a right tail test that are evaluation is insignificant. So now if we tried it again at the 1% significance level, mhm at point a one that changes our values to be mhm uh 50. So we got 50 right here and then we do 10 tons 11 divided by two. So 55 -50, which is going to be five. And once again 23.4 still lands us right in there. So we failed to reject the my hypothesis once more.

Hi. We're gonna start with stating our null hypothesis, which is at the mail. Old mule is equal to, uh, $5400 on our claim, which is the alternative. Is that that the mean is Liston $5400 on the critical value. We can find it from the table. Ah, As, um minus two point or 52 on. That's why. Because the degree of freedom is 27 on. We are looking at a point or 25 and as you know, it is one Tim. And it just looked it and that this value is Ah, this one. And we can commuted by this for my expose. What is that? Yeah, the average that we find and minus a mule divided by s, which is assembled. Send a deviation, divide by square, root off, and you find out that it is minus 1.262 on. Here's a graphical AIDS toe to visualize the idea. So here's the critical value and anything be lowered. It's gonna be, um, and the rejection area. But we happens to be greater than that. The the critical value. So we don't reject the now on and for summarizing the results, You can say there is no enough evidence to support the claim that the average course is the Stan 54 $5400.

Okay, well, cooks in ST test. So we're in space. Complete the table. No difference. Which in your one in here, too? Snake Youssif Dating president. A negative too naked 73 Make it to 10. Negative did. It's Who is it? AIDS? And it's a 27 pay, much less full. The difference is dirty. A to 73. It's in. Get it, too. Hey, Jimmy Serving Enduring from the lowest Belarus. One positions when created. Same. So it's super and fire two point by for by six 78 Send Bring snake its seeds. Was the two so quantify? I think it's good one because of age. It gets full against seven and it's a 2.5. It is by against is the help of disease. But it is age notes. The local cost for school districts. Uh, seem ah, Laker course for school. Just drinks. I didn't say h one the Legal Coast from score District. A different Sorry most. Oh, scored this cream uh, friend and a chauvinistic name. It didn't when you're so Kate Friendly critical elevator, okay and is a course of age identifies the crown still 0.5 for one sailed test. A critical Vader is a Fortune six. So we present toe. Finally test really must Wendy. Some offerings. They impose it, too. The ring so is in quotes. Some is the quality, so point fire and the negative bring some isn't quite so. It's negative. Six Less because if one less no kids who eight less they issue seven sir sneakers it full does biggest seven plus thinking to to clarify, Bless. There is a fire, and this is important, too. Naked suit. They didn't play claims by day, much less full. The neck is here. Bring some. I think it's too is a pencil? I was three. Did three point by and since the positive bring some is the lowest, it is a lot hottest. Big, but in convey untested Valerie Decrease part of it. Since our test available, Chase 2.5 is less than the Crisco very, which is six. We reject and concluded days enough evidence was a politic name that is a difference in the local post for school district. Ah is enough. Everything's Teoh separate 18. It's the is a different in the leg. Our coast squared district

This question covers constructing normal probability plots. So we're given data about horse races, winning times of one mile horse races and were asked to construct the probability plots. Mhm. Um determine whether there are outliers and if the plot is normal, some normality assessing normality. So to construct a probability plot, we must first arranged this and or her and referring to table three. There are two things we must do. Yeah. So you want to order order just data um in increasing order and refer to table three for the value. This is where the normal probability plot comes from. Yeah. Mhm. So if we refer to table, it's 1.7 values, negative 1.7 values all the way to 1.7. And if you have a calculator you can just input this and sort list to get this in order. Um I believe it begins with 93.37 and it ends with one a 1.9 or so. Mhm. But once you have that Eugene just plot these points to determine if it's a line or not. When you have plotted the points, it will look something like this. And now we want to determine if there's outliers. Okay, so outliers don't fold the trend. If there was a dot out here, we would consider it an outlier because it's not following this general trend. But in this case there is no such dot. So there's no outliers. And finally, we want to see if this is normal, normally distributed. And the condition for normal distributions is that our trend must be linear, and in this case it is linear. So we know that it is normally distributed. Mm hmm. Home


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