5

Tensile sirength tests were carried out on tWo different grades of wire rod, resulting Ihe accompanying dataSample Mean Sample (kg/mm?)Sample SizeGradeAISI 1064129X...

Question

Tensile sirength tests were carried out on tWo different grades of wire rod, resulting Ihe accompanying dataSample Mean Sample (kg/mm?)Sample SizeGradeAISI 1064129X= 103.7 S1 - 1.1AISI 1078n=129Y=123.8 52 - 2.2Does the data provide compelling evidence for concluding that true average strength for Ihe 1078 grade exceeds Ihat for the 1064 grade by more than kglmm?? Test the appropriate hypotheses using signillcance level of 0.01_ State the relevant hypothases:Ho; H1cs4 - K1oz8 Ha: [1054 41078 - -1

Tensile sirength tests were carried out on tWo different grades of wire rod, resulting Ihe accompanying data Sample Mean Sample (kg/mm?) Sample Size Grade AISI 1064 129 X= 103.7 S1 - 1.1 AISI 1078 n=129 Y=123.8 52 - 2.2 Does the data provide compelling evidence for concluding that true average strength for Ihe 1078 grade exceeds Ihat for the 1064 grade by more than kglmm?? Test the appropriate hypotheses using signillcance level of 0.01_ State the relevant hypothases: Ho; H1cs4 - K1oz8 Ha: [1054 41078 - -10 Ho' /10e4 - 41o78 -10 Hi' [1054 41078 Ho: [1ce4 41o78 -10 Hi' /1054 01078 -10 Ho: L1ce4 ` 41078 Ha' /1084 41078 Ho: H1c84 - 01078 = H: [1054 41078 Calculale Ihe los statislic and P-value. (Round your test statistic two decimal places and your P-value our decimal places:} Pvalue Slale Ihe concluslon Ine problam conlext;



Answers

One method for straightening wire before colling it to make a spring is called "roller straight-
ening." The article "The Effect of Roller and Spinner Wire Straightening on Coiling Perfor-
mance and Wire Properties" (Springs, 1987: $27-28$ ) reports on the tensile properties of wire.
Suppose a sample of 16 wires is selected and each is tested to determine tensile strength
$\left(\mathrm{N} / \mathrm{mm}^{2}\right) .$ The resulting sample mean and standard deviation are 2160 and 30 , respectively.
(a) The mean tensile strength for springs made using spinner spinner straightening is 2150 N/mm?
(b) Assuming that the tensile strength distribution is approximately normal, what test statistic
would you use to test the hypotheses in part (a)?
(c) What is the value of the test statictic for this data?
(d) What is the $P$ -value for the value of the test statistic computed in part $(c) ?$
(e) For a level. 05 test, what conclusion would you reach?

Mhm. In this video, let's look at constructing a 90% confidence interval for the population mean tensile strength. If we randomly selected 72 items and we found the sample mean tensile strength To be 242.2 newtons With a sample standard deviation of 70.6 newtons. So when we're looking at trying to make an estimate of the population mean with our constructing our confidence intervals, the first thing we do is find our point estimate for the population mean. And the point estimate for the population mean is the sample mean? Which here is given to us as 242.2. Now, once we have that, we look and see that we have a sample size that is 72, which is a large sample size. So we don't need to have any information about the distribution of the population when our sample size is large because by the central limit theorem that the sampling distribution of our sample means will be approximately normal with large sample sizes. So we can use our formula for finding our lower bound of our confidence interval by X bar minus T sub alpha over to times S over the square root of n. Um and then also finding our upper bound of our confidence interval by X bar plus teeth of alpha over two times S over the squared event. And x bar we have is the to 42.2 s. We have is the 70.6 And we have is the 72. So what we need to calculate before we can do the full amount with the confidence interval, is this test evolve over to now some courses are set up so the students or the person working with the information would just be allowed to use a scientific calculator and tables and charts to be able to find their critical values. Other courses are set up where the students are allowed to use a graphing calculator and you can utilize the information and the um programs within the graphing calculator to do these statistical inferential statistical methods here, I'm going to show you both ways. So if we're looking at finding this teeth of alpha over to value the first thing I want to do is find alpha, alpha is one minus the confidence level in decimal form. So for 90% confidence interval my confidence level is 90% and in decimal form that's a 900.90. And when i subtract that 1 -10 is 0.10. Now alpha over to you just actually take that alpha and specifically divided by two. So 0.10 divided by two is 0.05. Now, next with this, we also need our degrees of freedom degrees of freedom for this application is and -1. So for this particular question, my n recall is 70 two. So it's 72 -1 or my degrees of freedom is 71. Now when you look at AT chart it will skip over that um 71. So you need to find your critical value by looking across the column heads and with the student's T distribution page, the legend of it for many of them. Is that the area to the right of your critical value is what's the column heading? And then you look under the .05 for that and across from 71 if it's on the table. If not, you get to your closer value and find your teeth the buffalo over to If you have your graphing calculator that you can use, you would go second and the bars key to get two distributions, you would curse her down to inverse T. And the area in your calculator is area left of the critical value. What alpha over two is our area right of it. So area left of it is one minus that. 10.5, which would be 0.95. So it's already in there. From a previous problem 0.95 is the area that we will have. Now notice you don't put in the confidence level, You have to find your alpha divided by two and go 1- that number for the entry for your area. Now, degrees of freedom I have is 71. And then when I curse her and paste it And then push enter one more time to have it calculate it. It gives me that my degrees that my critical value to stable for over two is equal to 1.67. Now we are going to go ahead and put our values in. So the sample mean Is the 2 42.2. So we have 242.2 minus the 1.67 critical value that we found. And that's lower down here. I'm sorry, it's right here. And then times s Which in this case is 70.6. Yeah, divided by The Square Root of an N. is 72. And then close up parentheses. That's a lower bound. And then the upper bound is 242.2 then plus The 1.67 times The 70.6 divided by the square root of 72. So this amount that we're subtracting off of the point estimate and adding to the point um point estimate, that's your margin of error. So after uh ever just asked you for the margin of error you would calculate that part of it. But now going through the calculations here, you're going to get a lower bound of to 28.33 And an upper bound of to 56.07. Newton's. Now if you are allowed to use your graphing calculator, you can also find this interval without having to go through the formulas. So if you go stat button that's right underneath your delete cursor right to tests. Now we're doing confidence intervals. So we curse her down to the intervals and we have the tea is our critical value. So we do t interval push enter. We have stats not data because we've gotten the pre calculated mean and standard deviation data would be if we had the individual numbers. So we push enter on the stats Then for our sample mean that's the 242 Point to our sample. Standard deviation is 70.6, Our N is 72. Our confidence level is .90 remember you get that from what it told you to do the construct the confidence interval for cursor, down push enter to calculate and when it reports it it gives it to you as an open interval. So the lower bound is our to 28.33 and then comma the upper bound is to 56.07 like we had. So if you were to write this in a sentence you would say a 90 confidence interval four. The population mean tensile strength Yeah, Perfect. Mhm. Is between yes, 228.33 and 256.07. Newtons. Another way they might ask you to interpret it is that you're 90% confident that the population mean is somewhere between 288-28.33 and 256.07

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?

Uh, h note in that new one is mother than or equal to me too. And each one is that new one is bigger than mutant. So the degree of freedom, which is the minimum off anyone Where? This one in two minus one. So is 16 and 13. So the minimum 13 and the critical value in the row Corresponding toe degree. Freedom is off. Ableto appoint one on table five. So the critical value is 1.35 So the distances is Is that x one bar minus X tow bar over. Squared off S one squared over N one plus is two squared over into. She is equal to 3.429 So at 3.429 is it's a bigger than 135 So we reject this offer is won t This value is from table five

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.


Similar Solved Questions

5 answers
The graph of f' (1) .the derivatire of f(r) is shown below: Find the critical number(s) of f(r}-20-40I = {-3,0,3}OI =0n} 6z
The graph of f' (1) .the derivatire of f(r) is shown below: Find the critical number(s) of f(r} -20 -40 I = {-3,0,3} OI =0 n} 6z...
1 answers
Exercise 5.2.7 In each case show that the statement is true or give an example showing that it is false.If {x, y} is independent; then {x, Y, x+y} is in- dependent:b. If {x, %, 2} is independent; then {y, 2} is indepen- dent.If {y, 2} is dependent; then {x, %, 2} is dependent for any xd. If all of I. X2, Xk are nonzero; then {x1, X2, X} is independent:
Exercise 5.2.7 In each case show that the statement is true or give an example showing that it is false. If {x, y} is independent; then {x, Y, x+y} is in- dependent: b. If {x, %, 2} is independent; then {y, 2} is indepen- dent. If {y, 2} is dependent; then {x, %, 2} is dependent for any x d. If all ...
5 answers
0 > u W cull? TION 1 can INvo = ol othor cells and 3 nolnuvo a1 nuciou'satomgrns canau
0 > u W cull? TION 1 can INvo = ol othor cells and 3 nolnuvo a 1 nuciou' satomgrns canau...
5 answers
Consider the function f(r) = Ta(z) = &x4 Sz2 _ 1, the fourth Chehyshev polynomial. Approximate the two positive roots using the bisection method with tolerance 10-10 . The roots of Ta(z) are Tj cos '2i+l7) 0 <j <3 Show that your approximations are indeed within the tolerance of the true values. You may find it useful to plot the function:.
Consider the function f(r) = Ta(z) = &x4 Sz2 _ 1, the fourth Chehyshev polynomial. Approximate the two positive roots using the bisection method with tolerance 10-10 . The roots of Ta(z) are Tj cos '2i+l7) 0 <j <3 Show that your approximations are indeed within the tolerance of the tr...
5 answers
~/1 pointsAUFCAT7 6.4.054Write the given equation in the form Y = k sin(x a) , where the measure of a is in degreessin * -cos
~/1 points AUFCAT7 6.4.054 Write the given equation in the form Y = k sin(x a) , where the measure of a is in degrees sin * - cos...
5 answers
For all functions f, if a<b, f(a)<o,f(b )> 0, then there must be a number €, with a<c<b and f(c)-0.Select one: TrueFalse
For all functions f, if a<b, f(a)<o,f(b )> 0, then there must be a number €, with a<c<b and f(c)-0. Select one: True False...
5 answers
Locate any inflection points of g. Select the correct choice below and if necessary; fill in the answer box t0 complete your choice_0A An inflection point occurs at t = (Use comma t0 separate answers as needed.) 0 B There are no inflection points for g:.
Locate any inflection points of g. Select the correct choice below and if necessary; fill in the answer box t0 complete your choice_ 0A An inflection point occurs at t = (Use comma t0 separate answers as needed.) 0 B There are no inflection points for g:....
5 answers
Question 4 (1 point) The Krebs is a pathway in which there is cycling of glucose due to glycolysis in muscle and gluconeogenesis in liver:TrueFalseQuestion 5 (1 point) The anaerobic conversion of 1 mol of glucose to 2 mol of lactate by fermentation is accompanied by a net gain of:1 mol of NADH1 mol of ATP2 mol of ATP2 mol of NADHNone of the above
Question 4 (1 point) The Krebs is a pathway in which there is cycling of glucose due to glycolysis in muscle and gluconeogenesis in liver: True False Question 5 (1 point) The anaerobic conversion of 1 mol of glucose to 2 mol of lactate by fermentation is accompanied by a net gain of: 1 mol of NADH 1...
5 answers
(10 points) How are elcctrospray and atmospherc pressute ionization techniques diflerent? Explain E using echcmatics
(10 points) How are elcctrospray and atmospherc pressute ionization techniques diflerent? Explain E using echcmatics...
1 answers
In Exercises $1-10$, match the function with its graph (a-j). A.(GRAPH CAN'T COPY). B.(GRAPH CAN'T COPY). C.(GRAPH CAN'T COPY). D.(GRAPH CAN'T COPY). E.(GRAPH CAN'T COPY). F.(GRAPH CAN'T COPY). G.(GRAPH CAN'T COPY). H.(GRAPH CAN'T COPY). I.(GRAPH CAN'T COPY). J.(GRAPH CAN'T COPY). $$y=-2 \sin \left(\frac{1}{2} x\right)$$
In Exercises $1-10$, match the function with its graph (a-j). A.(GRAPH CAN'T COPY). B.(GRAPH CAN'T COPY). C.(GRAPH CAN'T COPY). D.(GRAPH CAN'T COPY). E.(GRAPH CAN'T COPY). F.(GRAPH CAN'T COPY). G.(GRAPH CAN'T COPY). H.(GRAPH CAN'T COPY). I.(GRAPH CAN'T CO...
1 answers
For Exercises 59–70, use a calculator to approximate the logarithms. Round to four decimal places. $$\log 6$$
For Exercises 59–70, use a calculator to approximate the logarithms. Round to four decimal places. $$\log 6$$...
5 answers
Problem 1 (20 points)Two students start simultaneously at opposite ends of a straight 170m track and run toward each other: Runner A, runs at a steady 1Omls and runner B runs at constant 6 mls. When and where do they meet: Diagram:
Problem 1 (20 points) Two students start simultaneously at opposite ends of a straight 170m track and run toward each other: Runner A, runs at a steady 1Omls and runner B runs at constant 6 mls. When and where do they meet: Diagram:...
5 answers
The phone company had to change their model to reduce themistakes. For the next generation of the phone they want to have areplacement rate of maximum 5.05%. After some testing they foundout that they reduced the mean, so that this worked out.What is the value of the mean in this new situation?
The phone company had to change their model to reduce the mistakes. For the next generation of the phone they want to have a replacement rate of maximum 5.05%. After some testing they found out that they reduced the mean, so that this worked out. What is the value of the mean in this new situation?...
5 answers
A cylinder of mass 8.7 kg, and radius 31 cm is rotating about anaxis along the length of the cylinder touching its side. How muchtorque is required to increase its angular momentum by 50% in 3seconds if the the initial angular velocity was 10.8 rad/s?HINT: Use parallel axis theorem to find the moment of inertia ofthe cylinder about the required axis.
A cylinder of mass 8.7 kg, and radius 31 cm is rotating about an axis along the length of the cylinder touching its side. How much torque is required to increase its angular momentum by 50% in 3 seconds if the the initial angular velocity was 10.8 rad/s? HINT: Use parallel axis theorem to find the m...
5 answers
The least-squares regression line given above said to be line "hich best fics" the sample data The term best fits" is used because the line has an equation that minimizes the which for these data total E squJICs regression sum of squares crror Sum Sovarcs 64.9) the value of the residual (Round vour answer to
The least-squares regression line given above said to be line "hich best fics" the sample data The term best fits" is used because the line has an equation that minimizes the which for these data total E squJICs regression sum of squares crror Sum Sovarcs 64.9) the value of the residu...
5 answers
Complete the Algorithm below that approximates the integral ] = Ji f(r)drIrput:vcctors and equal longth ropresonting tho data sct Componcnts of oqually spacod with X(1)-a and X(N)-b Y_if(x_:) for all approxicacion tho incogral based Cozp _ Trap Rule I ] CompositoTrapozoid(XOutput functionTest this Algorithm on the integral and the set of Data of Exercise 2, with Greatest value in the Table and f=&(x)(1 log(x+2)) and write YOur answer format long
Complete the Algorithm below that approximates the integral ] = Ji f(r)dr Irput: vcctors and equal longth ropresonting tho data sct Componcnts of oqually spacod with X(1)-a and X(N)-b Y_if(x_:) for all approxicacion tho incogral based Cozp _ Trap Rule I ] CompositoTrapozoid(X Output function Test th...

-- 0.018360--