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Consider the accompanying data on breaking load (kg/25 mm width) for various fabrics in both a unabraded condition and an abraded condition Use the paired test to t...

Question

Consider the accompanying data on breaking load (kg/25 mm width) for various fabrics in both a unabraded condition and an abraded condition Use the paired test to test Ho: LD versus Ha LlD 0 at significance level 0.01. (Use UD Mu 4A.)Fabric36.3 28.555.0 20.051.338.7 34.543.2 36.048.8 52.525.6 26.549.6 46.546.0Calculate the test statistic and determine the P-value (Round your test statistic to two decimal places and your P-value to three decima places.P-valueState the conclusion in the problem co

Consider the accompanying data on breaking load (kg/25 mm width) for various fabrics in both a unabraded condition and an abraded condition Use the paired test to test Ho: LD versus Ha LlD 0 at significance level 0.01. (Use UD Mu 4A.) Fabric 36.3 28.5 55.0 20.0 51.3 38.7 34.5 43.2 36.0 48.8 52.5 25.6 26.5 49.6 46.5 46.0 Calculate the test statistic and determine the P-value (Round your test statistic to two decimal places and your P-value to three decima places. P-value State the conclusion in the problem context. Reject Ho: The data suggests a significant mean difference in breaking load for the two fabric load conditions. Reject Ho: The data does not suggest a significant mean difference in breaking load for the two fabric load conditions. Fail to reject Ho: The data suggests a significant mean difference in breaking load for the two fabric load conditions_ Fail to reject Ho: The data does not suggest a significant mean difference in breaking load for the two fabric load conditions:



Answers

Testing the Difference Between Two Means (a) identify the claim and state $H_{0}$ and $H_{a}$, (b) find the critical value(s) and identify the rejection region(s), (c) find the standardized test statistic t, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Assume the samples are random and independent, and the populations are normally distributed. If convenient, use technology. Tensile Strength An cngineer wants to compare the tensile strengths of steel bars that are produced using a conventional method and an experimental method. (The tensile strength of a metal is a measure of its ability to resist tearing when pulled lengthwise. To do so, the engineer randomly selects steel bars that are manufactured using each method and records the tensile strengths (in newtons per square millimeter) listed below. Experimental Method: \begin{tabular}{|cccccccc} \hline 395 & 389 & 421 & 394 & 407 & 411 & 389 & 402 \end{tabular} 422 $\begin{array}{rrrrrrrr}416 & 402 & 408 & 400 & 386 & 411 & 405 & 389\end{array}$ Conventional Method: \begin{tabular}{rrrrrrr} 362 & 352 & 380 & 382 & 413 & 384 & 400 \\ \hline \end{tabular} $379 \quad 384 \quad 388$ $378 \quad 419$ $372 \quad 383$ At $\alpha=0.10,$ can the engineer support the claim that the experimental method produces steel with a greater mean tensile strength? Assume the population variances are not equal.

19, which in eight it's notice that me one is equal to Muto and each one is that me. One is not equal to Muto, so don't remind. The degree of freedom is equal to n one plus and two minus two is 10 plus 13. Minus two is 21 So the critical value corresponding. Tau alpha 4.1 with degree freedom 21 1 to date. So you think Temple five Critical Various possible negative 2.831 The second reason they contain all advantages. Smaller than negative 2.831 and larger than 2.83 More, uh, standard deviation is the square root off n minus one. So mine bye 22.3 or 12 square plus 12, which is into minus one times 14.5122 square over 10 plus 13 minus two to stem plus 13 minus two, which is 18.262 So the distance, which is X one minus six to so it 368.313 89.5385 Over square Rode off your father 18 0.262 is one over and one plus one over and to which approx negative 2.765 So if the value off the is in the rejection regions in the non deposit is reelected, so as the negative is bigger than negative 2.831 and smaller than 2.831 So we failed to reject, okay, they're not hypothesis, so there is no sufficient evidence to support, okay?

Okay, so here we are asked if there is a difference between this year's and last year's uh households. National Household Travel survey, if they travel more last year, before beer before that and were given that in the first year, it was 16.23 versus the second year, 17.69 with standard deviations of 4.6 and 4.42 respectively. Okay, so at first glance, it doesn't appear that they change significantly, but we will be evaluating this problem at the 5% significance level, which means that we are looking at a two tailed tests since we want to find out if they are equal or if they are not equal. So, I'm gonna do a two tailed test here. That means that we're looking at our normal distribution like this chop it off here, chop it off here. Yeah, I'd have to 5% here and the other half of the 5% there. If our statistic falls into one of these shaded regions, we will reject the null hypothesis. So what is our equation? It's going to be this blue equation right here, we'll go ahead and substitute those values of X one minus x to what we have is 16.23 minus 17.69 And the next thing we need to find out is what is S P. S. P. S are pooled. Uh Standard deviations, we can do this because our standard deviations are very close to each other and the sample sizes are large enough what our sample sizes. So I got to look up at the table. Okay, so we got 15 households and 14 households. All right, so first sample sizes 15 15 minus one is 14 and the standard deviation for that one is 4.6. So we square that and do the same thing with the other sample of 50. Okay. Of 14 minus ones. We have 13 times the standard deviation of 4.42 squared. And then we add those two. And what we get for RSP is 4.24 Great value. Okay, now we take one over each of the population or sample sizes, so that's gonna be 1/15 and won over 14. And then we and that gets us zero point negative 0.9 three. And then looking in the back of the book for our degrees of freedom, we have 27 degrees of freedom and a significance level is your point 0 to 5? Yeah, a critical value of 2.52 So obviously zero negative zero point and three is less than 2.52 And we're gonna take the absolute value since it's from both sides. Remember we're looking at that is a horrible normal graph, but are negative 0.93 lands right here and are shaded regions are right there and there. So what we do is we say that we fail to reject the null hypothesis.

In the following video, we're going to use chefs test uh to determine which of these means is different for very good, good and fair for toothpaste cost per per ounce or something. So they give you the formula of the book and Uh they involve x bar so the average, so here's the data set. Very good. Good and fair and I found the average here for the very good. It's .455 for the good, it's .606. And then for the fair it's .46. Now I have the technology to do this for me and that's nice. But if you are running out of time or something you can kind of look at these and what I would do is you're you're looking at which one is the most different which which means the most different. And it appears that this good is pretty far off the other ones. So I'm just going to make a wild guess. I mean I've already done it but I'm going to make a wild guess that Very good and good. They seem to be pretty far apart, so I think that's going to be the big difference. All right, So the next thing I did was I found the sample variance, because that's what you also need with that formula. And so that was S T D E V dot s. That's the sample variance of the first of the very good. Of the good and then of the fair, so it's the standard deviation and then squared. So I needed to find the variance squared and then I did these in minus ones. So in is the sample size for very good? Well, there were six minus one, is five, so that's where I get 54 and three. And then I found the S. S. W. The way you find ssw you just simply take the n minus one times the sample variance. That's why I found the sample variance. So it's five times 50.287 Plus. And it's the summation of all those plus four times .01 098 plus three Times .011467. And that gave me a uh within as 0.9267 And then I found the summation of n minus one. Which was that five plus four plus plus three. Okay, so that's basically everything that I need um to use the formula and then it also says for a critical value. Use the critical value found in an example five and that was 3.885 I won't show you how to do that again, but it's 0.33 point 885 And then times k minus one where K. Is the number of categories. So. Very good. Good. Fair. There are three categories. So I multiply that by two and I got 7.77 So if the um chefs statistic is larger than that critical value then that's the difference. That's the that's a significant difference in means if it's smaller than than it's not significant. So here we have these numbers and it is very good and good. That are the most different. So that's kind of what I thought it was going to be because that's larger than 7.77 Now how did I get that? Well I applied the formula so it's the two means subtracted and then squared. So it's the difference of the two means that's where I get the 8 10 minus B. 10 quantity squared divided by It's the SSW. Which is this a 19 Divided by the c. 19. So divided by that summation of N. I -1 and then times one over N plus one over N. So that's one over the number of very good which was six plus one over. Um the number of goods which was five. And that gave me this 8.055 and that's how I did this one and this one just different numbers. Um but you know, I won't go through that because it's essentially the same thing. But this is the statistic that is larger than that 7.77. So very good is um there is a significant difference between very good and good for the toothpaste problem.

What's up? Sat cats in this video, we are given some raw data and we are asked to perform a one way Unova. And if the another test is significant, we are asked to them perform either a chef or a turkey test. So we are given this raw data and this is, ah, weekly er visits between different hospitals. So in order to perform an Innova, I'm gonna be using the data analysis package on Excel, which is also what the book uses. So we're gonna do a nova single factor, and our input range is gonna be all of our data. And we do have labels in the first row, so make sure that's checked. We are doing this at a 0.5 alfa levels to make sure that coincides with what kind of tests your doing. And then our output ranges just where it's gonna summarize those results for us. So says the results of our one way and over test and what we really want to look at is our critical value and our test statistic. So we will have a significant result if our test value is larger than our critical value. So is 3.76 larger than 3.68 Yes, it ISS, so we can reject our null hypothesis. So to summarize our results, we would say there is enough evidence to conclude that there is at least one mean that is different from another mean. So because we have a significant unova TAAS, we can go ahead and perform a post talk test. And if you look, um, our sample sizes are equal, so we're going to be doing a two key test. And the test statistic for this test is found by taking the mean differences and dividing that by the square root of our mean square within groups multiplied. Fine. I'm sorry. Did. And then we're gonna divide that by end, and then our critical value is found with the end tables. Let's go back to our Excel spreadsheets so we can do some calculations. So let's do Let's do our pairs first. So we're gonna pair ex and why? Why? And Z and the and X So let's take the mean difference. So we're gonna take mean difference of X and why, why? And see and Z and facts, and then we're going to take the square root of our means square with end groups to buy by our sample size, which is all six. All right. And then we can calculate our test statistic by taking our numerator divided by our denominator. Okay, so these are our test statistics. I noticed that one of them's negative, but, um, an F value is always gonna be positive. So we can just ignore that and just consider it a positive value. So to find the critical value, we are going to go to our and table. So Kay is the number of means we have, which is three. So we have three categories. We have hospital X, Y, and Z and V is our total population. So the sample size all of our sample sizes together. So six times three. And we're going to subtract K from that. So we're gonna take 18 minus three, so it's gonna be 15. So que is three and V is 15. So 3.67 is our critical value. All right, let's go to our white board so we can write this down. All right, so, in orderto figure out if we have a significant para not We're going to see if our test statistic is larger than our cook about you. So it was 1.77 larger than 3.67 No. Is 2.10? No. Is 3.87? Yes. So we have one significant pair. It's between hospitals, uh, X and Z. So there is sufficient evidence to claim, uh, a mean difference exists between hair three and one. All right, that's it for this video. You guys, I hope you learned a lot. And also in its time.


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