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Single Factor ANQVAA product developer (s investigating the tensile strength ofa new synthetic fiber that will be used to make cloth for shirts. Strength is usually...

Question

Single Factor ANQVAA product developer (s investigating the tensile strength ofa new synthetic fiber that will be used to make cloth for shirts. Strength is usually affected by percentage of cotton used in the blend of materials for the fiber. The engineer conducts a completely random experiment with 5 levels of cotton content and replicates the experiment 5 times. The data is given below:Observations (Tensile strength).Cotton %15 20 25 30 3515 12 19 22 1111 18 18 19 1512 14 1917 19 25 1018 18 2

Single Factor ANQVA A product developer (s investigating the tensile strength ofa new synthetic fiber that will be used to make cloth for shirts. Strength is usually affected by percentage of cotton used in the blend of materials for the fiber. The engineer conducts a completely random experiment with 5 levels of cotton content and replicates the experiment 5 times. The data is given below: Observations (Tensile strength). Cotton % 15 20 25 30 35 15 12 19 22 11 11 18 18 19 15 12 14 19 17 19 25 10 18 18 23 11 Is the evidence to support the claim that cotton content affects the mean tensile strength with alpha = 0.05? Is the data normally distributed? Plot the residuals (with i run order; ii. Fitted values) Compare the treatment means using Tukey's method:



Answers

Tensile Strength Tensile strength is the amount of stress a material can withstand before it breaks. Researchers wanted to detcrmine the tcnsile strength of a resin cement used in bonding crowns to teeth. The researchers bonded crowns to 72 extracted teeth. Using a tensile resistance test, they found the mean tensile strength of the resin cement to be 242.2 newtons (N), with a standard deviation of $70.6 \mathrm{N}$. Based on this information, construct a $90 \%$ confidence interval for the mean tensile strength of the resin cement.
Source: Simonides Consani et al., "Effect of Cement Types on the Tensile Strength of Metallic Crowns Submitted to Thermocycling." Brazilian Dental Journal $14(3), 2003$

Following is a solution for number 16 and this is about tensile strength and we have 72 I guess. Pieces of tensile, I think it's something with like teeth. Um and Uh the sample of that was 242.2. Newtons of force on that residents cement and then the sample standard deviation 70.6. Now we only know the sample standard deviation were not given the population standard deviation. So since we don't know the population standard deviation, we have to use the tea interval instead of the Z interval. Now, had we been given the sigma, then we can go ahead and use the Z interval. But we weren't, we were given the sample standard deviation, We're gonna use the tea interval and we're asked to find the 90% confidence interval and then we're gonna interpret that now I'm gonna use technology but you can certainly use um the formula or any form of technology want, I'm gonna use the T. I. T. Four because it works pretty nicely. So if you go to stat and then air over two tests it's gonna be this eighth option down here, the T. Interval. So eight and make sure summary stats is highlighted there and I already have this, you know, worked out But 242.2 is your sample mean 70.6 is your sample standard deviation and was 72 and the sea levels .9 for 90%. So then we're going to calculate that in this top band here is the actual confidence interval. So 28-56. So let's go and write those numbers down. So to 28 0.33 All the way up to 256 0.7 Okay, so then the second part of this is actually interpreting it and these all kind of have the same style of interpretation the way we were interpreted confidence intervals, we say that We are 90 confident that the mean tensile strength of the resin cement is between To 28.33 and 2 56.07 Nunes.

I have been given a list off values and I have to find the standard deviations first in order to compute the F statistic. So this is how I do it. I take the value from the table off supplier A and substituted in this formula to find the standard deviation and the standard division that I get for Supplier A is this 4.43 Similarly, I do it for Supplier B as well. Over here I get three point 50179 Now, how do I find the F statistic? The statistic is, I'll just take two decimal places this square upon 3.503 point 50 square and the F statistic that comes out in this case is 1.60 This is equal to 1.60 This is my test statistic.

Here we are, given the following sample information. It's a size 16 sample averages 2160 and the sample standard deviation is 30. Now for part A were asked what hypotheses we would test to determine whether the population mean exceeds 2150. So, in this case, our no hypothesis is that the population mean is equal to 2150 and the alternative hypothesis is that the mean is greater than 2150. Next in Part B, we're told to assume that the population is approximately normally distributed. So what test statistic would we use to test the hypotheses from Part A. In this case, we would use the T test statistic, and that is equal to the sample average, minus the no hypothesis mean divided by the sample standard deviation over the square root of the sample size for Part C. We want to calculate the actual test statistic, and this is equal to 1.33 for Part D. We want the P value that's associated with this test statistic. The P value will be equal to the probability of the test statistic exceeding 1.33 and this comes out to about 0.102 and for Part E were asked at a significance level of 0.5 What conclusion would we draw regarding our hypotheses? So in this case, the P value is greater than Alpha and therefore we would fail to reject the null hypothesis, and the conclusion is that there is insufficient evidence based on the sample to suggest that the population mean is above 2000, 150.

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?


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