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[2/4 Points]DETAILSPREVIOUS ANSWERSPODSTATS 11.E.024.MY NOTESASK YOUR TEACHERPRACTICE ANOTHERTo determine chocolate milk was as effective as other carponydrate rep ...

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[2/4 Points]DETAILSPREVIOUS ANSWERSPODSTATS 11.E.024.MY NOTESASK YOUR TEACHERPRACTICE ANOTHERTo determine chocolate milk was as effective as other carponydrate rep cement drinks, nine male cyclists performed an Intense workour followed by drink anc rest Period_ Ac the end of the period _ each cyclist performed an endurance trial where he exercised until exhausted and time exhaustion was measured: Each cyclst completed the entire regimen on two different days. On one day the drink provided was ch

[2/4 Points] DETAILS PREVIOUS ANSWERS PODSTATS 11.E.024. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER To determine chocolate milk was as effective as other carponydrate rep cement drinks, nine male cyclists performed an Intense workour followed by drink anc rest Period_ Ac the end of the period _ each cyclist performed an endurance trial where he exercised until exhausted and time exhaustion was measured: Each cyclst completed the entire regimen on two different days. On one day the drink provided was chocolate milk and on the other day the drink provided was carbohydrate eplecement drink. Data cons stent with summary quantities appear in the table below_ (Use statistical compute- package calculate the P-value Subtract the carbonydrate eplacement times from tne chocolate milk times, Round Your test statistic to two decima places, your df down the nearest who number; anc the P-value [nree decima places: Time Frhausion (minutes] Cyclist Chocolate 56.44/35.77 |26.3- 31.77 40.11 48.38 |49.03 Carbohydrate 42.68 47.89 |14.82 |25.98 6,41 Replacement 32.69 |23.02 35.28 39.56 00tt Is there >uffic ent evidence suggest that the mean time exhaustion greater after chocolate milk than after - Fponvdrate eplacement drink? Use significance level of 0.05 You may need use the approprlate table Appendix [0 answer rnis question_ Need Help? Read It Tuk lo @uutor



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Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. (b) Find the value of the chi-square statistic for the sample. What are the degrees of freedom? What assumptions are you making about the original distribution? (c) Find or estimate the $P$ -value of the sample test statistic. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence? (e) Interpret your conclusion in the context of the application. (f) Find the requested confidence interval for the population variance or population standard deviation. Interpret the results in the context of the application. Veterinary Science: Tranquilizer Jim Mead is a veterinarian who visits a Vermont farm to examine prize bulls. In order to examine a bull, Jim first gives the animal a tranquilizer shot. The effect of the shot is supposed to last an average of 65 minutes, and it usually does. However, Jim sometimes gets chased out of the pasture by a bull that recovers too soon, and other times he becomes worried about prize bulls that take too long to recover. By reading journals, Jim has found that the tranquilizer should have a mean duration time of 65 minutes, with a standard deviation of 15 minutes. A random sample of 10 of Jim's bulls had a mean tranquilized duration time of close to 65 minutes but a standard deviation of 24 minutes. At the $1 \%$ level of significance, is Jim justified in the claim that the variance is larger than that stated in his journal? Find a $95 \%$ confidence interval for the population standard deviation.

The following is solution video to number 26 which is two sample T tests comparing meantime loss in the workforce with stressors and intimidators. And you just got to be really careful here. The only kind of weird thing is They kind of switch up the order on you so X one represents the intimidators. In the next two represents the stressors. And we want to see if the stressors, so I'm actually gonna write that down. So this is intimidators and this is stressors. Okay, So it says is there enough evidence to suggest that the stressor is greater than the intimidator? So keeping that in mind, we're going to write less than some you won Is less than you two, Even though it says greater than but the orders just switched, if that makes sense? So the greater than in this case means less than for our alternative, since the order is opposite. Okay, so the first part of this is just verifying that these means. And standard deviations are four and 2.38 for intimidators, and then 5.5 and 2.784 stressors. So I went ahead and it says to use a calculator, so I'm using a T 84 if you go to stat and then edit here are the list, so L one is the intimidator and then L two is the stressor. So if you go back to stat over to couch and then it's one of our stats, we can change that to L one and this will be the four and the 238 that was over here. So four and 238 And then go back to stat couch. One bar stats, change that to L two calculate, and that's where we get to 55 and the 27855 to 78 So we verified that that is the way to do it. So then the second part of it is to kind of sort of not formally conduct a hypothesis test but essentially come up with the conclusion here and it is gonna be a two sample T test. We're gonna use the calculator since we already have the data input anyway. So it's two sample T test for the alpha is point oh five. And the alternative we already talked about, the alternative is less than not greater than but less than so. Um if you go back to stat and then we're gonna air over two tests and it's the fourth one down is the T. Test. Two sample T test because we don't know the standard deviation for the population, We don't know sigma, we only know as um so sincerity of the data, I'm just going to keep it as data instead of doing the summary stats. L one is the first list. L two is the second list and you can keep those frequencies as one and then we're gonna change this alternative to less than you two. And then pulled is usually no unless they tell you otherwise and then we're gonna go and calculate and you can have that test statistic there if you want but you really don't need to because the p value is all you need. So it's 0.14. So let's write that down. So the p values 0.14 and what we do is we explicitly compare the P value with the alpha. In this case 0.14 is greater than point oh five. So that means whenever the p values greater than alpha, we fail to reject H not so we failed to reject h. Not whenever the p values greater than alpha at the p values less than alpha, then we will reject the null hypothesis. So keep in mind that the null hypothesis is generally always saying that these two means are equal. We can we're failing to reject that, so we're accepting that to be the truth. So let's go ahead and complete this hypothesis test. And we're gonna say there is not sufficient evidence to suggest that the mean time lost due to stressors is greater than the meantime lost due to intimidators. That's spelled right? I have no idea. Looks like it's not so intimidators. Okay, that's why math teachers aren't spelling enthusiast. So intimidators um now you could also say there is not sufficient evidence to suggest that the meantime lost due to intimidators is less than the meantime loss due to stressors. If you want to keep to that less than thing. I just wrote it like this because that's the way the book had it. So there is not enough evidence to say that these two means are different or that one means more than the other. So we're accepting the truth that these two means are probably about the same.

There's this problems. Meteo first convert a table of numbers in two seconds. So we have a table of minutes. So we have 1993 1995 in 1997. So it's a year, and we have the time. It's the first time is 26. They're all 26 minutes. Um so if we get 26 minutes time, 60 seconds, we can get the number of seconds And those just a minute part which is gonna be at 15. 60 and then adding the 58 38 at 15. 16. Um, but we're gonna get our 1993 value, which is going to be a 16 18 0.38 in 1995. I'm going to be adding 43.53 So me, 16 3.53 in the 1997 were adding 27.85 It's 15. 87.8, right? Okay, So, using a chart like the one they have in the book, we have it. We have a 1993. We have 1995. I'm scared of something like what's in the book? 1997. So it's gonna start kind of higher in 16. 18 a little lower 1995 and a little lower in 1997. Okay, so it's gonna be somewhere between the 16. 20 in 16. 16. Can they have? The rest of the dash is going down. Okay, that's that's about what it would look like approximately. Okay, so, um, gets the part b once, but he is the midpoint formula on 1993 and 1995 to find 1994. It's time the 1994. We'll look at the halfway point of this two times so that we're into 16. 18. You create the 16 03 three e. I don't know about Teoh. May 16.955 That's how many seconds it would be predicted for an approximation for 1994. Okay, um, so the loss making Peter Percentage era. So the actual time from 1994 was 26 minutes 52 23 sorts. Gonna add 52. 23 to ar 15 16 be 16 12.23 So I'm going to 16. 12 playing to three Prince attract 16 10.955 And divide that by 16. 12.23 Cancer in 16. 12.23 But now I'm using my calculator. 16. 10.9 Popeye, then all divided by 16. 12.2 green multiplied by 100 and our percent is gonna be 0.0 79 percent. It's a very small percent air on that one. Okay. Said my estimate was just slightly too low. Okay, um, case of part C wants me do the same thing for 95 to 97 to figure out 1996 is It's during 1996 in our halfway point between those two. So I'm gonna have 16/3 0.53 plus 15 87 28 5 What about two? So the estimate is gonna be 15 95 0.69 okay. And then, Ah, the percentage air we're gonna calculate Based off of the time given in part C, which is 26 minutes and 38 seconds. A 38.8. That's how I will be 15 98 point OK. 30. 15. 98 0.8 minus 15 95 0.69 Do. About about 15. 98 point away. So actual minus approximation, divided by actual. It's been in the calculator now and so 15. 98 point away my 15. 95.69 and then divided by 15 98. Going away looks by that about 100 and our percent ERA B 0.1 1496%. Okay, and so our estimate is just a little bit too low. Uh, actually, I was too low both times. It was just off by just off by a lot less than 1%. Okay, and then it says it is in the percent office is a little bit higher. Okay, it's in 1998. The time was an extra 22.75 seconds. So let's figure that out in seconds. I'm going to be 15 82.7 times. Okay, so it's our 1st 1 Was 16 12 approximately for 1994 when they gave us 15 98 for 96. So after the first year went down six seconds the next year went down nine seconds. The next year went down five seconds and then 11 seconds and then five seconds. So because where we haven't really seen a lot of leveling off yet. Um, uh, it mean, like, the very last time, it only went down five seconds, but because it went down five seconds before and then jumped again, it's It's hard to say just on the amount of data that they have. Um, but I would say that there's gotta be some maximum speed, so it should level off to the fastest, like, humanly possible to run it like no errors made. No missteps. There should be like a absolute fastest. Like if you designed a robot to run it, it wouldn't just keep leveling off to zero. Okay, well, thank you very much.

We want to connect the chi square test variants and construct confidence intervals as follows. We have a population acts with variance, sigma squared equals nine. A random samples taken from the population has size and equals 23 sample standard deviation or variance s squared equals 1.9. We want to in one test the claim sigma squared is less than nine at 5% significance level. We proceed to step 83 days all first and a. We stayed alpha and our hypotheses, alpha is a significant level. Hypotheses are are given variants in our claim. Thus we have alpha 830.5 sigma squared equals nine. H. A single square is less than nine. Next we calculate our chi square value, our degree of freedom in our state. The assumptions for our normal distribution chi squared equals m minus one squared over. Signal squared equals 8.8 to the degree freedom is anyone in 2022. And we're assuming here that X is normally distributed. And see, we estimate the p value for chi square and degree of freedom given from a chi square table. We see that peace between 0.5 point 01 Thus we conclude in D and E to reject a channel because it is difficult to alpha, which means we have evidence to support H. I. Finally into we construct a 90% confidence interval for the standard deviation sigma. This confidence interval is given by the formula left. We have an N. S square but we need to X squared and X. L. These are the chi squared values for degree of freedom 22. Let's put the distribution of chi square into two tails on left Area. on the right track. You squared .05 as well. These excuse square next elsewhere values given on the right, plugging them in gives interval 1.53 is less than sigma, is less than 2.54.

In this problem we want to conduct a high square tested variance and construct confidence interval as follows. We have a population X with variants. Sigma squared equals to 25. A random sample of sand size and equals 10 has sample variance X squared equals 5 76. We want to test the claim. Sigma squared is greater than 2 25 at 1% significance. We proceed these types a three below conductive chi square test variants. So in a way we set our alpha level which is our confidence of our significance and we set our hypotheses. Thus we have all 4.189 sigma squared equals to 25. H A sigma squared is greater than 2 25 our claim and be we calculate the chi square value the degree of freedom and state the assumptions we're making a better distribution. We have a chi square value of n minus one squared over sigma squared equals 23.3 or four degree of freedom is n minus one equals nine. And we assume x is normally distributed in order to conduct this test and see we estimate the p value. This is the probability of observing the given chi square value given the degree of freedom From a Chi Square table. This is not the less than .005. Thus we conclude in D any to reject H not because P is great less than or equal to alpha, which means that we have some evidence to support the null hypothesis. AJ Yeah. Finally, in part two, we construct a 95% confidence interval for the population standard deviation sigma. The interval formula is given on the left. We have an S square but we need x squared and x squared away from a chi square table for the given degree of freedom. Nine excuse square satisfies probability high greater than chi squared you is 90.25 XL square square is the left most tail with 0.25 area. Thus we have X squared you an expert. L as given on the right. Thus, we conclude our Interval is 16.5 less than six miles less than 43.8.


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