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1. Consider the following observations on shear strength of a joint bonded in a particular manner:30.0 4.4 33.1 73.7 36.6 109.966.781.5 22.2 40.416.4aDetermine the ...

Question

1. Consider the following observations on shear strength of a joint bonded in a particular manner:30.0 4.4 33.1 73.7 36.6 109.966.781.5 22.2 40.416.4aDetermine the value of the sample mean:b Determine the value of the sample median: Why is it so different from the mean?

1. Consider the following observations on shear strength of a joint bonded in a particular manner: 30.0 4.4 33.1 73.7 36.6 109.9 66.7 81.5 22.2 40.4 16.4 a Determine the value of the sample mean: b Determine the value of the sample median: Why is it so different from the mean?



Answers

The compressive strength of samples of cement can be modeled by a normal distribution with a mean of 6000 kilograms per square centimeter and a standard deviation of 100 kilograms per square centimeter. (a) What is the probability that a sample's strength is less than $6250 \mathrm{Kg} / \mathrm{cm}^{2} ?$ (b) What is the probability that a sample's strength is between 5800 and $5900 \mathrm{Kg} / \mathrm{cm}^{2} ?$ (c) What strength is exceeded by $95 \%$ of the samples?

All right. In this problem we are told that the random variable is normally distributed and has means 6000 and standard deviation sigma equals 100. Basically information. Hello, we want to answer a three seat. This question is testing our understanding of random variables. Specifically normal variables or normally distributed variables. Facade, let's first review relevant information to the normal distribution which will allow us to answer agency. So we cannot these scores on the probabilities using a Z table. Yeah, because the tables are used for standard distributions. We have to make the conversion from our random variable X here to a standard normal distribution. We do that using conversion Z equals x minus mu over sigma. Specifically, you can utilize the symmetry of the normal curve as well as total area in the normal curve being one to answer questions eight through see below as well. So first to find the probability that x is less than 60 to 50 we're looking for the probably that Z is less than a certain value. Probably that 60 we can convert 62 52. The Z score of 62 59 6100 equals 2.5. That's the probably the less than 2.5 is not 0.62 but rather one minus 10.62 by symmetry. So if we can fix this answer now we write this as one minus 10.62 equals 0.9938. Next proceeding to part B one of the probably between 585901st as I've done here, we convert to the relevancy scores as I did in part, a probability falls between these scores. 1.1359 from is the table finally, and see what strength is exceeded by 95% of samples. We want to find the sample where 95% is to the right. That is probably the greatest seen articles. 950.95 from a normal distribution table. This is not equals negative 1.64. To convert to X. We reverse this equation to obtain X equals 100 times the Not for 6000 equals 58 36.

So for this question, the population is 45 and nine 45 common I and the sample size is too, because you're picking so random, so s o n requested to. And then the first question I want you to figure out, um, the first question wants to figure out the value of the population Media. Andi, mind you, the media is always the media number. When I arrange, the data are not sending order. And from four, five and nine, you could definitely tell that night is meeting number. So definitely, um, the population media, the population media is you go to five. And that's the answer. The second question, You do remember that the two values a randomly selected replacements from 457 on. Um, we're going to write the samples off the possible, um, the possible send the samples of the possible I'm I'm gonna write the samples of the possible outcomes in order to describe the sampling distribution of the sampling. So the samples, the samples would be oh, four comma four, four comma, five for common night, $5 for five form of life and five common night also. No, I come up before night. Come up five and I come tonight so that their nine deny probabilities the problem. Total probabilities equal tonight, right? And then to construct the table for sampling distribution of media who need to find the sample meet media in order to my sugar table. And we know that it Number of probability, easy for tonight. And in order to find the sample media, we're gonna right the sample over a year. And then we're gonna have the samples. The nice samples, which is four come before. Welcome of five. Welcome a night 54 five from of five. Fire from my life. Yes, also night coming for like on the fly in nine Kalmunai on. And then we're gonna find the sample media for each of them in order to finish off the table. So the sample media and we know that the media is the, um the some off the number of divided by the actual number, the actual the total numbers we have. And since we have two figures because we have to, um, we're picking soon numbers the random so is too. And then we have, like, two samples for every samples. So four plus four developed by two is gonna be four. So that's gonna be the something something media four plus five they don't have the right to is going to be 4.5. That's going to be a sample media. The 4.94 point four plus nine when it bites, who is going to be six points? All right, that's sample media. Why, of course, for the bye bye. Service 4.5 on this one is fight. This one is seven seven. Just 16.5. This one several. And this one is night. And then the probability is gonna be the last thing we're gonna do. We know that they're night there, nine on samples. So each release is gonna be one of the night, and that's gonna be answered for the rest of the table. So I'm just gonna click the check. Yeah, and then the question also wants us to combine the values of the media, and they are the same. So combining the values and ideas are the same. We're gonna have a sample media sample media. Yeah, yeah, yeah. We're gonna have sample media. I mean, the other side, we're gonna have to probably t. Yeah. So obviously is gonna be, um, the sample medians from the top. You can see it for four. Um, 4.5, talking five, um, 5.0, 6.5, 7.0. And then I concert. Mhm. Yeah. Anybody with your fault is because one time one of the night. 4.5 focus twice. So overnight five or close ones one of the night. 6.5 occurs twice overnight. The seven because slice until the night. 90 Clockers ones, One of the nine. That's it. The next question wants us to find the mean off the sampling distribution of the sample media and the mean off the sample distribution of the sample begin is going to be is going to be this song. The some off the total, some of the total off the night sample media divided by the number of samples where by a number of samples. And this is gonna be equal. So you adding all of this together the 4456 months, five. The way. I don't have that together. It gives you 54 and the number of samples of nine. And that's Why my night? And that's six. That's the mean off the sampling distribution of the sample median. And the last question, Um, sorry about this. The last question say's, um, based on the preceding results is a sample media unbiased estimator for the population media. The answer is no. The sample media is not unbiased estimator of the population medium because the published So the answer is no, because the population media is no e. Quote the meat off the sub media. Remember, the population medium is the one we solved first, which is on five, and it's not a call. So the median actually the mean of the sample media, okay?

The first thing we have to compute is the differences between each element from each population. So 11 minus eight is equal to three seven minus eight is equal to negative. 19 minus six is equal to three. 12 minus seven is equal to five. 13 minus 10 is equal to 3 15 minus 15 is equal to 0 15 minus 14 is equal to one. And now we have to find this is the answer to part A. Now we're asked to find the bar which is just the mean of the differences so that is equal to each individual mean the sum of each individual need over the number. Sorry. The sum of each individual difference it over the number of differences which is equal to In this case, the sum of each of these individual data points over the number elements which is seven, which is equal to two. And now we have to find a sample standard deviation which is equal to the square root of the some of the difference between each individual difference and our mean difference squared over the number of differences minus one soldier to a new page. For this equal to the square root of the Somme. Each individual difference minus mean difference over a number of differences minus one which is equal to approximately 2.817 This is the answer to part C. And now we're asked to find a point estimate for the difference of population means. And this is so the difference of population means is also our deep are which is equal 22 And now we have to come up with a a confidence interval. So we're asked to find a 95% confidence interval. So to do this, we will use, um, the following formula. Our confidence interval equals D bar plus or minus a T to t statistic. Because we're not given a population ah, standard deviation. We compute the sample standard deviation a T statistic for Alfa over to where Alfa equals one minus the confidence level. So that is equal to one minus 10.95 equal 2.5 So de bar plus or minus our T statistic for half of our Alfa times wth east andr deviation, the difference is over our sample size. So in our situation, D bar is equal to two plus or minus. I'm just going to write t of 0.0 to 5 for now because Alfa over to is 0.5 over to which is your 0.0 to 5 times 2.817 over the square root of seven. In order to find our T statistic, we have to compute a degrees of freedom and our degrees of freedom is equal to end minus one, which is equal to seven minus one, which is equal to six. Now, using a tea table, we can find out where a significant Slobo of Sierra 0.25 associated with a degrees of freedom of six lies. And we get that we have a confidence interval of two plus or minus 2.447 times 2.817 over the square root of seven. And that means our final answer is our were 95% confident that the true average difference for our two populations lies between our lower end of our confidence. Interval 0.747 and our upper end of our conference interval 3.9253

So the question here basically gives us the data set, and it wants us to do some various statistical analysis in order to, um, solve this particular problem. So for part A here, it wants us to calculate the mean. So what is the mean? So the mean here is essentially going to be the average of all values here, and this is gonna be calculated by taking the total sum over. And we're and is gonna be the sample size. So for the first data set here, we're going to get a value of nine here on the second data set here. We're going to get a value of seven here. So for part B here, it wants us to compute to standard deviation. So in order to do so, we just need to plug it into the typical standard deviation, um, formula. And this is going to give us the values of for one here, um, 2.284 And for data set to we're gonna get 1.7889 here. So see here wants us to calculate the point of difference between the two population means So in order to do that we essentially need to find the difference between, um one and two. So we're gonna take the absolute value of one minus two here. And this is gonna be too for part, do you? Here, which is the last part of this question here. It wants us to, um basically asked us more or less what the 90% confidence interval is. So firstly, in 95 90% confidence interval here is going to be a sort of range by which there is a 99 90% probability, rather that this particular, um range is going to contain the true value. So in this case, it's going to contain the true mean. So what we get when we calculated is we firstly, just take the bottom 10% and the top to the top 10% here. And what we get is going to be, um, negative 0.1689 to 4.1689 here. And this is gonna be the answer to our question here


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