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Program stimulate weight love diet-modification paricipated showin In the following Iist. Ghen thiot Ten individuals have In the program polnts) Participaatioor 0f ...

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Program stimulate weight love diet-modification paricipated showin In the following Iist. Ghen thiot Ten individuals have In the program polnts) Participaatioor 0f thedifference / 41: Their weight both before and atter differences 17 and standard deviation MCan Guhicct FIO 646 EJ pol Kcluta 190 prorram particular diet-modification the claim that this (7 points) there evidence t wepgbt r eduction? Use 0,05 mean welght particular diet modification program will effectivc producing evidence t0 suppo

program stimulate weight love diet-modification paricipated showin In the following Iist. Ghen thiot Ten individuals have In the program polnts) Participaatioor 0f thedifference / 41: Their weight both before and atter differences 17 and standard deviation MCan Guhicct FIO 646 EJ pol Kcluta 190 prorram particular diet-modification the claim that this (7 points) there evidence t wepgbt r eduction? Use 0,05 mean welght particular diet modification program will effectivc producing evidence t0 support the claim that this points) there least 10 pounds? Use 0,05. result in = mean welght loss of at



Answers

When 40 people used the Weight Watchers diet for one year, their mean weight loss was 3.0 lb and the standard deviation was 4.9 lb (based on data from “Comparison of the Atkins, Ornish, Weight Watchers, and Zone Diets for Weight Loss and Heart Disease Reduction,” by Dan singer, et al., Journal of the American Medical Association, Vol. 293, No. 1). Use a 0.01 significance level to test the claim that the mean weight loss is greater than 0. Based on these results, does the diet appear to be effective? Does the diet appear to have practical significance?

Looking at the before and after weight of a weight loss program were given is that we have a correlation coefficient R which is equal to 0.876. Now, would this be enough for us to say that there's linear correlation between the before and after weights? Well, for that we can say yes, and that's because our correlation coefficient of 0.876 is very close to one. So we know that this is going to arrange from negative 12 positive one. So we've got a fairly positive linear correlation going on right here. So we could say that yes, there, we could justify linear correlation. However, if we wanted to know if this value of our gives explains any amount of effectiveness in reducing weight based on this program? Well, we're only given this value of art which is a correlation coefficient. So we know that there's positive linear correlation between these before and after weights. However, we can't say that it's necessarily effect because in order to say that it's effective, we would be implying some form of causation and we know that correlation does not imply causation correlation just explains our linear relationship. We would have to rule out any other possible variables that could have caused these people to lose weight. So we can't say based solely on this correlation coefficient that the weight loss program is effective. We would need to know a little bit more than just our correlation coefficient

Hi, if you read the doctor here, God has given us the delivery supplementing that purple have lose weight Jumping up and down. A member of mine, 50 adult males. We have so and its identity going to. That's only for disagreement and find a mean way Lol. After eight weeks to be 0.9 lb Saturation weight loss of 1.2. We have to start out here like spotted over nine at 10.2. Select the population. Mean the friends. We take that zero to music that you know, nothing works requires little point more. So help people lose weight. So therefore we have money for this is that is coming actually munis the Puerto zero and already because this is gonna be of unions were than by rather than zero because it helps lose people wait so more and more weight to lose that. You noticed. And for his amusement where that deal and you know, where the media, news, radio already policies. And I warned the B part here That type of alpha, 0.1 level of significance. So alpha is no 0.1 Is the mean weight loss of the .9 lbs significant. Well, that you go for Peeta statistic, there is even a 3.9 negative zero Over us. 10 point to overrule man would be to calculate that. And from that you find out the p value to the technology. So we get to see there's a matter of 3.853. It will three points 853 Technology. We get the p value. Do you want people didn't want Rwandan peace but it has another fuck right. Therefore Bill ejecta. My hypothesis reject now people didn't, that means there will be a weight loss. There would be a weight loss to reject Malibu. This is so we can say In the b. part it says that the radius of 0.9 is significant. Yes be part of it is significant right now, european fire. Yes, significant with more of the C. Part. Now you think that I mean weight loss of Lebanon is worth the expense and commitment of our daily diet supplement. Yes it is worth the expense and commitment of a daily like the supplement. By the year we got a weight loss supplement and all this all the things that product is helping in weight loss so we can serve eight law has particular significance. Yes it has practical significant more of an RV park for the part. Now the same status parameters and the stigmata or B. We'll be back and the student exported select calculate the value and be released or table. You will be as ah Jonah 2.9 Negatives legal over 1.2 7.2 over no reported connects to that. So we get a statistic for an equal to 40.791 Your .791 between technology will be value accordingly. That's coming up to BP will lose your point to 1000 your point who want them and that is great America President Falco. We paid to reject. You can't reject not a positive right? That means weight loss is not significant. Weight loss is now north significant writing audience. People really get not. So weight loss is not practically significant in this case. So where's the infective? If we have end is ingredients If we have and is including. So in that case can empty to uh fail to reject at north. It decreases right? And we keep on going towards uh we just uh uh well, I mean if anything ingredient see behavior and is increased here in the B bar. So anything priest and we go towards significant weight loss becomes significant and we go on towards detection of knowledge policies, right? We're going towards directional militant. Thank you.

In exercise nine. Yeah. Considering a case where a college athletic program, which is to establish the average increase in the total weight of an athlete um A little bit an athlete can can lived in three different lifts are following a training program for four for about six weeks. We're told that 24 randomly selected athletes, meaning that the sample size here, N is 24. So those 24 athletes were placed on a program and they exhibited a mean gain of £473. So that means the sample mean there is £47.3. We also told That the standard deviation was £6.4. Mhm Yeah, We are going to use this information to construct a 90% confidence interval for the main increase in lifting capacity all athletes would experience if placed on that training program. And for this we shall assume that increases among all athletes are normally distributed. Now the first step is to determine the formula that will be used to get the confidence interval. This here, in this case being the 90% confidence interval for the population means since the population standard deviation is unknown, we shall use the following formula X. Bar. Class or minus the margin of error. And given by the critical value of T. For the given level of significance, multiplied by s divide the square root of n. No. We want to determine the level of significance here, Alpha, which is equal to 1 -0.9, which equals 0.1. And we have that, it's going to be 0.05. And for the T distribution, we need to determine the degrees of freedom Using N -1, which in this case is 24 -1. And that equal 20 uh 24 minutes one. Oh I'm sorry. When We read that question, we noticed that the sample size there is 25 and not 24. So change that and this as well. So So the sample size is 25 and therefore the degrees of freedom here would be 25 -1, which equals 24. So If we check the critical value of T, that corresponds to the given level of significance at 24° of freedom, it will be 1.711. Now we're ready to substitute our values into the formula Expo 47.3 plus or minus the Critical Valley 1.711 multiplied by S, which is 6.4 divided by the square root of 25 which is the sample science. And the margin of error calculated simplifies to 2.19 and so this is in pounds. The confidence in the 90% confidence interval for the mean um increase in lifting capacity, all athletes would experience if they were placed in this training

Number two, A dietician compares the effects of to diet pills using 25 participants taking each pill. She calculates a statistic W by computing a mean difference in weight loss, expressed and pooled standard deviation units. The dietician calculates W to be equal to negative 0.7 in her experiment, 1000 simulations, assuming no real difference between the two pills gives the plot below for W. So we have this plot in which we have all these simulated values of the statistic W. What is the estimated P value for her observed value of W equals negative 0.7. For the null hypothesis of no difference in effect versus the alternative hypothesis that there is a difference in the effect. So the first thing I want you to realize is the no hypothesis is no difference. The alternative is different, not indicating left side or right side. We have a W of negative 0.7 and if there is no difference, W should be at zero. So we want to find this P value. So looking on the plot, let's find negative 0.7 negative 0.7 um has 12 above it. And we want to take everything more extreme than that, which is everything to the left. So everything to the left of negative 0.7 would be 12 seven and one. But because we want to do purely difference, we want an extreme value negative 0.7 or less, or positive 0.7 or more. So to the right of 0.7 on the right side of the simulation, we have five. And we have to this gives us a total of 27 out of the total number of simulations which there are 1000 simulations. That gives us a p value of 0.0 to seven. And looking at our answer choices 0.27 is option E.


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