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A dynamic cone penetrometer (DCP) is used for measuring material resistance penetration (mrn /blow) as cone IS driven into pavement or subgrade. Suppose that for pa...

Question

A dynamic cone penetrometer (DCP) is used for measuring material resistance penetration (mrn /blow) as cone IS driven into pavement or subgrade. Suppose that for particular application it is required that the true mean DCP value for = cer- tain type of pavement be less than 30. The pavement will not be used unless there is conclusive evidence that the specification has been met. For random sample of Size n = 102_ the sample mean is 24.5 and the sample standard deviation is 21.2 A Q-Q plot of the

A dynamic cone penetrometer (DCP) is used for measuring material resistance penetration (mrn /blow) as cone IS driven into pavement or subgrade. Suppose that for particular application it is required that the true mean DCP value for = cer- tain type of pavement be less than 30. The pavement will not be used unless there is conclusive evidence that the specification has been met. For random sample of Size n = 102_ the sample mean is 24.5 and the sample standard deviation is 21.2 A Q-Q plot of the data shows that the distribution of the data cannot reasonably as- sumed to be normal (See Figure 1). Using these data, perform hypothesis test to assess whether this sample comes from population that satisfies the requirement Normal 0-Q Plot 8 J i 8 CCLL Theoretical Quantiles Figure 17: Q-Q Plot showing deviation from normality for data in problem 4. 0oO



Answers

A dynamic cone penetrometer (DCP) is used for measuring material resistance to penetration
$(\mathrm{mm} / \mathrm{blow})$ as a cone is driven into pavement or subgrade. Suppose that for a particular application it is required that the true average DCP value for a certain type of pavement be less than $30 .$ .
The pavement will not be used unless there is conclusive evidence that the specification has been
met. Test the appropriate hypotheses using the following data ("Probabilistic Model for the
Analysis of Dynamic Cone Penetrometer Test Values in Pavement Structure Evaluation," J. of Testing and Evaluation, $1999 : 7-14$ ):
$\begin{array}{lllll}{14.1} & {14.5} & {15.5} & {16.0} & {16.0} \\ {17.8} & {18.1} & {18.2} & {18.3} & {18.3} \\ {20.8} & {20.8} & {21.0} & {21.5} & {23.5} \\ {30.0} & {31.6} & {31.7} & {31.7} & {32.5} \\ {36.7} & {40.0} & {40.0} & {41.3} & {41.7} \\ {55.0} & {57.0}\end{array}$
$\begin{array}{llll}{16.0} & {16.7} & {16.9} & {17.1} \\ {18.3} & {19.0} & {19.2} & {19.4} \\ {23.5} & {27.5} & {27.5} & {28.0} \\ {32.5} & {33.5} & {33.9} & {35.0} \\ {41.7} & {47.5} & {50.0} & {51.0}\end{array}$
$\begin{array}{rr}{17.5} & {17.8} \\ {20.0} & {20.0} \\ {28.3} & {30.0} \\ {35.0} & {35.0} \\ {51.8} & {54.4}\end{array}$

In this question, we are basically told that we need the population mean to be less than 30. Before we can proceed, we cannot proceed unless there is conclusive evidence that the specification is met so inappropriate. No hypothesis to be tested here is that to assume that the mean is three, and we cannot proceed unless we have conclusive. We have conclusive evidence to reject this. No hypothesis. Yeah. So the alternative hypothesis, which would be the population mean, is less than 30 which we would need in order to proceed. So let's test these hypotheses at a significant level of 0.5 now, since the sample size given in the question is 52 that means we don't have to worry about the normality of the underlying population. With sample sizes of 52 we know that the means of such sized samples would be approximately normally distributed, and therefore the T test is an appropriate test to test these hypotheses. So let's do a T test in our. So the first step is to enter the sample information with this command, and then the second step is to run the T test. So the third argument in this command is new equals 30. That's specifying that the no hypothesis is that the population means 30. The second argument in this command pertains to the alternative hypothesis. So we are stating that the alternative hypothesis is that the population mean is less than 30. So if we enter this, we get our sample R t test results. I'll paste them here and we can see from our results that we have a P value of 0.23 which is pretty large and more specifically it's larger than our alpha level, are significant level mhm and therefore we would fail to reject the null hypothesis. Another way to look at this is the confidence interval, which goes from minus infinity to 31.6. So this 95% confidence interval actually includes the value of 30 which means that there is not conclusive evidence to suggest that the true mean is actually less than 30

Um were given a model for the distribution of the random variable. Why? Which is the time to pavement city? But told that x one time to failure between running. That's two times of failure you to trans verse scripting. We're told these two random variables are assumed to be independent. Then we're told us running variable. Why is the minimum of these two random variables That's when the next to so the probability of fears would be to either one of these. Those is assumed the an increasing function of Time team and this would seem to make sense right? The longer the pavements out, more spokesman the weather and you think the more likely it's going to fail? Sure, now mhm to making assumptions mhm form of the CDF. Each node was detained to be five of a plus B t over the square to C plus B T plus E. T squared the fires, the standard normal, chaotic distribution function now it's cold were given the values of the parameters. A, B, C D V cracking and also different days or running were asked, defying the probability of statement failure within five years and also within 10 years. Okay, so first I'll consider the case of five years. Well, let f one and F two the accumulate distribution functions of X one and X two, respectively. Probability of pavement failure within two years. Mhm probability. That's why his last century, quarter T this is equal to the same as a probability at the minimum of X one, and x two is less than or equal to t. And this, of course, is the same as one minus probability that the minimum of X one next to is greater than teeth. Now notice that the minimum of two numbers could be greater than a number if and only if both of those numbers are greater than that number. This is the same as one minus ability. The X one is greater than tea, and that's two is great beauty. And since we've assumed that X one next to our independent follows that this is equal to one minus the probability that X one is greater than tea times, the probability that X two is great in T, which is the same as one minus one, minus the probability that X one is less than or equal to t times one minus the probability that X two is less than a quarter T which we know. It used to be one minus one minus form of tea times one minus F two of teeth. You're using the king of the distribution functions that we were given all that. This is equal to one minus one minus. Bye. Um, and this is for X one. We have negative. 25.49 plus 1.15 TV over square root. Um, 4.45 minus 1.78 t plus 0.17 won t squared, Mrs. Simply coming from the values that we were given for the constants times one minus cap If I, uh Mrs for the random variable x two negative 21.27 plus 0.3 to 5 Team over the square root of 972 minus 0.2 18 +100022 T square. Now if we are to evaluate a T equals five, see the probability that why less than or equal to five, then we have the argument inside. The first five is imagining it. The argument inside first five correspondent. F one is imaginary. However, we haven't defined dysfunction for imaginary numbers. Now, if we evaluate AT T equals 10 then we have the probability that why is less than or equal to 10. This reduces toe one minus one, minus five. Approximately negative. 7.22 times one minus phi of approximately negative 21.14 Mhm incredibly small number. Both of these really are. So this is really approximately equal to one minus one times one, which is zero. But there's a zero chance that we cracked within 10 years, which, okay there within 10 years, which is clearly nonsense. Obviously, there should be a chance that will fail within 10 years.

All right. So what do we have this time? The types off raw materials used to construct stone tools they were found at an archaeological site called Casa de Rito are shown to us. They're given to us. We have a table and a random sample off 1486 stone tools was obtained from an ongoing excavation site. Now, what we have to do is we have to use a 1% level of significance. Okay, 1%. Level of significance means what are ALPA is 0.1 okay, and one person level of significance to test the claim that the original distribution off raw materials fits the distributions at the current excavation site. So what is going to be a little hypothesis in this case, Arnel hypothesis is that the regional distribution of raw materials that the regional distribution off raw materials fits the distribution of the current excavation site? It's the distribution. It's the distribution off the current excavation site off the current excavation site. This is what we're going to write in an AL hypothesis. And what do we write for our alternative hypothesis? It will be that the regional distribution off raw materials off raw materials does not fit. Or I can say the reason of distribution of raw materials and the distribution off. The current excavation sake are not the same. Our e can say that they are not similar, that they're not similar. Okay, All right. Now what do we have now? We have a table that is given to us. So let's just look at the table closely. We have raw material. Okay? We have raw material. One of the raw materials that we have The first one is Basil. The next one is obsidian. Then we have welded tuff welded tuff. Then we have paternal short. Or let me just write. This is PC. Okay. And then the last one is the other category. The other category. Okay, then we have reasonable percent off stone tools. All right, then. We have the regional percent off stone tools. Okay, on the stone tools. All right. This is 61.3%. 61.3%. Then we have 10.6% 10.6%. Then we have 11 point 4%. Then we have 13 points, 1% and then we have 3.6%. All right, Now, what is the observed number of tools at current excavation site? So let me just write this as the observed values, the observed values. Okay, these are 906. 906. Then these are 1 62 then these are 1 68 then 1 97 and 53. Now, what is the total? I need this total because this is going to be my sample size, my total sample size. And this student is already given to me as 1486 Right. Okay, so this is 1486 Okay, now, the first step in calculating the guys question district is goingto be to find the expected value e. Now, how exactly do I find the expected value? How do I find the expected values? These are going to be for each category. Expected values are given by the sample size in the sample size that is n multiplied by the probability off its category or the proportion that we have. Right. So the probability the probability off each category off each category. Okay. All right. So what is my sample size? My sample sizes. This oil 1 46. Okay, these are going to be the expected values. Okay, these will be the expected values. All right. So what is going to be the expected value for the first category? If I take my calculator, this is going to be in. That is one for 86 multiplied by now. I want 61.3% off 1486 So this is going to be 61% 61.3%. 0.631 This is 9. 37.666 Or let me describe this as nine. 37.67 Okay, the next one stand 10.6% off for 1/4. 861486 multiplied by 10.6. Divided by 100. This is 1 57.516 1, 57.516 Okay, then we have 11.4% off. 1486 multiplied by 0.114 This is 1 69.4041 16, 9 0.404 Then we have 13.1%. 1486 multiplied by 0.131 Right. That in 0.1% this is 1 94.666 Recognized this as 1 94.67 Yeah, and then I have 3.6% or 0.0 36 multiplied by 1486 This is 53.496 This is 53.496 but I could access 53 point Faith. Okay, These are my expected values. Now I want to find the guys question distinct. This column is going to be for individual chi square values, right? For individual chi square values. Okay, Now, what is the formula for Chi Square statistic? The overall case question The stick is going to be first one. Find the individual high school values of the formula for that is observed minus the expected, or we consider the difference of the observed and expected square upon the expected value. We do this for all the categories and in the end of the some of them, all up, and this will give us the highest question to stick for our problem. Okay, so let us do that. Now. The difference is observed and expected. This is 9 37.67 minus 906 I found this. Now I squared the study 1.6 71.67 and I divide this by 9. 37.67 9. 37 point 67 Okay, so this is 1.696 or I could write. This is 1.71 point 07 Okay, moving on to the next category. This is the difference between 1. 62 1. 57.516 We square this 4.484 and divide this by 1. 57 point 516 which is one point. Sorry. This is 0.12764 I can write this as 0.13 Okay, then we have 1 68 and 1 69.404 We square this, okay? And then we divide this by 1. 69.0 40.1 16 So this is 0.11 Okay, let me just write this much. Many of the difference between 1 97 and 1 94 minus 1 94.67 We square this 2.33 and divide this by 1. 90 4.67 which is 0.278 or regulate this as 0.3 Then we have the difference between 53.5 and 53. We square this and we divide this by 53.5. So this is 0.0 467 reconnected says 0.0 05 All right, now I need to some all of these individuals high school values, right? We saw, were you? The submission is going to give us the final chi square statistic. So this is 1.7 Yes. 0.13 less 0.11 plus 0.3 plus 0.5 Now, this is 1.246 on my guys. Question stick is 1.246 All right, so now we have this s one point 1.246 All right, now, what is do we need now? We need the degrees of freedom that is DF now. This is given by number off categories. Number off categories minus one Know how many categories do we have? If you look away? We have basal obsidian welded tuff than PC or spc Paternal short on the other. So these are the five categories, so number of categories is five five minus one is four. So this becomes four. So my diesel fuel it was for okay. Now you can either use a chi square table and with the help of the table, you will get a P value range off the values. You will not be able to find the exactly value in order to find exactly value. You need the guys for statistic, the degrees of freedom. And along with this, you need a statistical software like Excel or SPS or our python or whatever statistical tools that you're using. Like over here. I'm using an online calculator, so this will give me an exactly value. So my thigh square statistic is 1.26 If I'm not mistaking 1.246 1.2 for six and my degrees of freedom is five minus one, which is four. Okay, My significant level. There's al 50.1 Andi, I calculate this and write P value is coming out to be 0.87 my p value. My peopling is 0.87 My Alfa was 0.1 right? Yeah, I also was 0.1 All right, So what I can say over here is that since my P value is much greater than Alfa right, I can say that I will fail to reject that. I feel to reject my null hypothesis. H not right. So I can say that I do not have I do not half enough statistical evidence enough to this critical evidence to suggest to suggest that to suggest that the regional distribution off raw materials that the regional this line is very important. This is the answer the regional distribution, the reasonable distribution off raw materials. We'll draw materials and the distribution and the distribution from the current excavation site from the current excavation site and the current excavation site are different. Okay, so this is my answer. I fail to reject my inner life. But this is aged not And I say that I do not have enough statistical evidence to suggest that the regional distribution off raw materials and the distribution from the Context Commission site are different. So this is an ally prosthesis. We're saying that it fits, and the alternative is that it does not fit right. So we do not have enough evidence to suggest that the distributions don't fit. This is going to be an answer.

Okay, So we know that the horizontal and vertical areas from aiming at a point target are independent of each other and each have a normal distribution para meters. I mean off zero under standard deviation off one. Also, given that the density function F V is V over seating squared times E to the power monetary screwed over to signal skirt for one V is great in zero on, otherwise chicken to do so Basically for views. Liston culture. So no. Do we know that the standard norm alleviation distribution is? Trust me. Sure. She was one way consult be. I calculated the proportion of darts that land between 25 millimeters of the target. Okay, that's just given by the entry of 0 to 25. No FV TV. Did you go thio actual 0 25 of the over 20 squared. I am e to the minus fee squared over two times 20 squared TV. So if you succeed for V in terms of tea, get his tea. She was a negative V squared over two terms. 20 screamed, and we know the D key is therefore negative to v just differentiating Curator over two times 20 squared. This will then reduce this integral. You too. And to go who is you to 0.78 1 to 5 e to the power of negative Chief teach you. So if you're just playing the boundaries way, know that Did you bring this or just have the minus coming front? Something negative. Same boundaries. Oh, sure, which will give negative zero from 4578 plus one, which is equal to 0.542 which is the proportion of darts that land within 25 millimeters.


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