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(a) [15pts] An experiment was done to test method for reducing faults on telephone lines (Welch 1987). Ten matched pairs of areas were used: The following table sho...

Question

(a) [15pts] An experiment was done to test method for reducing faults on telephone lines (Welch 1987). Ten matched pairs of areas were used: The following table shows the fault rates for the test areas (X) and for the control areas (Y):Test (X) 676 206 230 256 280433 337 466 497 512 Control (Y): 88 570 605 617 653 2913 924 286 1098 982(i) Compute the mean difference construct 98 % confidence interval for the true mean difference of fault rates in the test; and control areas_(ii) Using (i) to co

(a) [15pts] An experiment was done to test method for reducing faults on telephone lines (Welch 1987). Ten matched pairs of areas were used: The following table shows the fault rates for the test areas (X) and for the control areas (Y): Test (X) 676 206 230 256 280433 337 466 497 512 Control (Y): 88 570 605 617 653 2913 924 286 1098 982 (i) Compute the mean difference construct 98 % confidence interval for the true mean difference of fault rates in the test; and control areas_ (ii) Using (i) to conclude at a = 0.02 whether the mean differences are different from zero (b) [10 pts] Biological effects of magnetic fields are matter of current concern and research: Inan early study of the effects of strong magnetic field on the development of mice in 5 cages, each containing three 30-day-old albino female mice were subjected for period of 12 days to field with an average strength of 80 Oe/cm. Thirty other mice in similar cages were NOT placed in a magnetic field and served as controls The following table shows the weight gains in grams, for each cage Magnetic Field present (X): 22.8 10.2 20.8 21.3 12.1 Magnetic Field Not present (Y): 23.5 19.5 21.0 18.5 Use the Mann-Whitney test to compare the two weight gains testing the following hypothesis: Ho: X and Y have identical distributions H;:There distributions are dif ferent Conclude at ( 0.10



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?

It's a kind of an interesting question if we we would be assuming that the anxiety level if we are going from easy to difficult is equal to the anxiety level. If the questions were difficult to easy and alternately, we just want to know if there's a difference and easy to difficult and not equal to difficult to easy. And so we're going to assume that that difference is actually zero so that the, I'll just write E G minus D. E. That that difference is zero. And when I put this into my calculator, um we have the lists have different sample sizes. We have that first group of the E. T. D. I have, there are 25 numbers versus the D two E. There are 16 numbers and we could use the degrees of freedom of 15, but I'm going to use the degrees of freedom from the formula. And that degrees of freedom ends up being uh 30 almost 39. So 38 point well, I'm just going to call it 39 it's approximately 39 degrees of freedom. And that test statistic that we're going to get, we need to take the mean, which was 27.1152 minus the mean of the other group, which was 31.7 to 8125 And then divided by the square root of. And the first standard deviation all around that a bit is 6.857 one square, divided by the sample size, which was 25. And then the second standard deviation was 4.26 square divided by the sample size of 16. And when I got that test statistic, the test statistic came out to be, I'll just read it up here negative 2.6 566 And so it came out down here. And since we're doing a two tailed test, we also use this one. And so what's the likelihood of getting a test statistic in this distribution that is less than or equal to that negative 2.6566 That gives us this tale and then we want the other tales to double it. And that p value comes out to be a 0.114 So at a 5% significant level, this is smaller than 5%. So we would have sufficient evidence to reject now. Yeah. And say that the mean anxieties are different so that they're different. The means are different. However, at a 1% significance level we would fail to reject the novel. Mhm. And we'd have to say here that they're actually not. They appeared to be the same. So it does depend on our significance level. How did pick you want to be. But it is an interesting idea in all my years of teaching, I never thought about the arrangement of having them all go easy, too difficult or difficult to easy. So what interesting concept.

Okay. In this question, we're going to be comparing the duck nesting boxes and seeing if one situation appears to be better than the other. So for questioning the sample proportion of ducks that hatch was 2 70 out of 4 74 4.5696 Almost 57% hatched. Now when we run our confidence interval based on that sample proportion we've got Rpf plus or minus are critical. Z score times the square root of the P hat times one minus P. A. Over the sample size and the 1.96 times the square root part ends up being point oh 446 And when we add and subtract that from the 0.5696 we get an interval from 5 to 5 2.61 for two. So we're 95% confident the proportion of all ducks hatching in Group one is between 52 a half percent and 61.4% for the second set of nesting boxes are sample proportion is to 70 out of 805 or 8050.3354 And then our confidence interval is going to be 0.3354 plus or minus our margin of error, which is point oh 3 to 6, which is going to give us an interval from 0.3028 up to 368 So we see right there that the proportion of ducks hatching from the set to nesting boxes is much lower than nest than the next one boxes. So, and let her see, we're going to see how big this difference actually is. Is it a big deal that there is this difference between the two nesting boxes? So we're using this great big formula here for our confidence interval for the difference of two proportions I've plugged in our all of our numbers are proportions are sample sizes and our Z score 1.96 or the star there. And the difference comes out to be, the difference of the means is 0.2342 and then plus or minus the margin of error. All of this should work out to be point oh +55 So that when we find the average difference for the the confidence interval for the average difference, it's going to be from 0.17922 point 2892 So we are 95% confident that the difference in the proportions of the the hatching of the docks is somewhere between 18% and 29%. This is a big difference. So we would say that yeah, all three confidence intervals 30.2 group one nesting boxes where they are well separated and hidden will yield a higher proportion of hatching ducks.

Hi. This question would be laughed toe to find Chi square and justice is six for two different samples. So for the 1st 1 I used many tab on. I found that guys square is equal to seven point 16 degree. Freedom is nine because the real freedom is equal and minus one, and evil is 1758 on when we change the sample to 20 we found that. So the statistics is the team board 96 So and increases okay on the P value, it becomes now going forth to eight. So it decrease and that's actually take care of part C and the So what will have been for that? This is a six, as is gonna increase because, as you know, from the formula, it is and minus one here, this where with us in the square. So I'm assuming this part this fixed constant in some sense. So when you're in keys in Chi Square or the just statistics is gonna increase, and that's gonna have a consequence, which is we're gonna have ah better P values, which is, um, less big value here


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