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Test the data in the file labeled 'Ch10_prob4.txt' in the Filesarea of Canvas to see if there is a difference between the means ornot, then answer the fol...

Question

Test the data in the file labeled 'Ch10_prob4.txt' in the Filesarea of Canvas to see if there is a difference between the means ornot, then answer the following. Use alpha = 0.05.The four lists come from populations that are not normallydistributed..Group of answer choicesTest Statistic = [ Choose] 7.72699E-09 126.94547085170467 14.35889406 2.461906e-27 2.65067651 Reject Ho 1.76708E-08 2.65

Test the data in the file labeled 'Ch10_prob4.txt' in the Files area of Canvas to see if there is a difference between the means or not, then answer the following. Use alpha = 0.05. The four lists come from populations that are not normally distributed.. Group of answer choices Test Statistic = [ Choose ] 7.72699E-09 126.94547085170467 14.35889406 2.461906e-27 2.65067651 Reject Ho 1.76708E-08 2.65164033 15.01927392 FTR Ho p-value = [ Choose ] 7.72699E-09 126.94547085170467 14.35889406 2.461906e-27 2.65067651 Reject Ho 1.76708E-08 2.65164033 15.01927392 FTR Ho Decision = [ Choose ] 7.72699E-09 126.94547085170467 14.35889406 2.461906e-27 2.65067651 Reject Ho 1.76708E-08 2.65164033 15.01927392 FTR Ho ZOOM 17.1 11.72613404 24.8 18.59.5 16.78656803 25 8.915.3 9.847127597 20.4 13.918 9.611185568 23.5 18.713.7 8.885791055 15.2 11.812.2 9.152294522 19.9 11.49.2 14.83447204 16.5 16.311 8.889538652 20.7 16.618.5 100.9090661 20.1 1213.8 9.621001462 21.6 1815.7 8.882244647 22.2 12.39.6 13.95676766 17.1 14.613.6 8.824437892 17.9 1412.6 8.517012256 19.9 14.713.5 9.449773605 27.3 1611.6 9.502676149 20.4 17.912.4 10.36470451 17.8 15.717.7 12.96391686 28.7 18.118.1 11.61136677 22.6 16.312.6 10.8876194 20.9 19.98.6 16.63983801 19 13.210.5 12.48845987 18.1 16.314.8 9.239120102 22.1 17.410.6 8.359493017 20 14.621.2 11.15651349 23.7 18.414 11.24952984 23.4 17.514.5 10.46379325 19.2 16.811.8 9.193895335 21.3 15.911.9 7.366538532 23.1 12.67.8 8.945479198 20.5 16.69.1 9.0529341 18.1 15.618.4 13.34066344 21.5 14.814.2 10.06355674 20.5 15.715.5 9.002574489 22.7 15.412.3 13.29540389 20 17.114.2 9.116575784 22.3 16.916.2 10.19419644 25.1 16.614 8.770311406 18.5 15.912.4 11.01216229 23.7 18.912.2 10.86981161 19.8 18.216.4 10.44922735 22.6 16.213.7 11.84471124 19.8 17.216.5 10.32527921 26.8 16.916.8 8.605075425 20 11.410.8 8.082408277 11.5 9.511.1 10.82241215 19.3 17.610.3 15.97270874 20.6 13.912.1 10.25814048 17.6 13.617.4 11.04419226 26.6 2211.6 9.241626936 20.8 16.59.8 11.54309241 21.3 14.7



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.

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.

The following is a nova test based on the mean salaries for different metropolitan areas. So the alternative or the null hypothesis is that all the means are the same. So there are six metropolitan areas, I think it goes Chicago, Dallas Miami, Denver san Diego and Seattle. Uh So the null hypothesis is that all the means are the same. And then the alternative is that at least one of them is different. The second step is to find the critical value and you can do that using either software or a table, But they're essentially three things you need. The first thing is your alpha value, your significance level and that's usually given to you the problem and that's .05. Then you need the degrees of freedom for the numerator and the degrees of freedom for the denominator. And the way you find that Is the degrees of freedom for the numerator is the number of categories -1. So there were six cities that we looked at our metropolitan areas, so 6 -1° of freedom would be five for the numerator. And then for the denominators, the total number of data values minus the number of categories. So there were 36 data values minus the six metropolitan areas. So 30 is your degrees of freedom for the denominator. So that should be enough to use a table. But I use a calculator and I wrote a program in here called inverse. F. I'm not going to show you how to how to write the program. You can youtube it if you wish. Um But this is what I do. So um I put in my area which is my alpha value, my degrees of freedom is five and then my degrees of freedom for the denominator is 30 and That gives me my critical value. About 2.534 2534 is my critical value. I call f. star. So 2.534. Okay so anything greater than 2.534. We reject the annual hypothesis that all the means are the same And anything less than 2.534. We failed to reject meaning the h not is true. Okay so the second step is to find the F statistic and there's a formula but it's a bit of a mess. I always use software you know technology is a great thing. So if you go to stat and you can type in your data values. So these are the mean salaries um So again L1 I think was Chicago and then this is the mean salary for Dallas Miami Denver San Diego and Seattle. So there are six categories. And if you go to stat tests and then we're gonna go to the Unova test and then you just type in your columns separated by commas remember there were six columns, six data columns that we used and we need to make sure that all of them are in there and last one and then also you know make sure you separate those by commons, otherwise it's going to read it wrong. So then um that gives us everything we need. So the F. Is the F statistic, that's the third step. So we're looking at this it's about 2.281 as our F. Value. So two point 281 is our f statistic Which is actually barely in the non rejection region 2.281. So that means we fail to reject. Okay and also we can verify that with this p value here. So the p values 0.7 which is a pretty small p value, but it's still in this case greater than the alpha value. So the alpha value remembers point oh five, so it's barely greater than the alpha value. And whenever it's greater than the alpha value, uh we failed to reject, I should probably put H not there, so we failed to reject H not whenever the P values greater than the alpha. Okay. So then the last step is to summarize everything with actual words. So what does this all mean? It just means that there is not sufficient evidence, there is not sufficient. I guess you could say statistical evidence to suggest that the mean salaries from the different metropolitan areas are different. Okay. And that's the five step process for an Innova one way and over test

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?

All right. We have a sample with N equals 317. And of those 317 sample members are equal 61 Members satisfy a certain condition we want to observe. We want to test the claim that the population proportion p does not equal 610.261 at a confidence level of 1% or alpha equals 0.1 Now that we've identified the confidence level, we follow the following procedural steps to conduct a hypothesis test. First is appropriate to the normal distribution. Yes, it is because N P and Q P R both greater than five. What are the hypotheses were testing are known? Is that P equals 50.261 Are alternative is that he does not equal 0.261 Ak We're conducting a two tailed test computer we had at our test statistic. Next Piatt is simply are over N or 20.192 Rz stat takes as input P p Q and N. As we see in the formula on the right, giving us Z equals negative 2.78 For this problem. Next to be the P value associated with the sea stat. We use a Z table to find out the area outside the positive and negative Z score as we've shown in yellow and the graph on the right is P equals 2.54 Next do we use this P value to reject? Asian on? Yes, because P is less than ideal to alpha, which we interpret to me that we have evidence that he is not equal 0.26


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