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Accuracy & Frecision WorksheetAcrac: refers to how clace meanurement ksto true accepted or target vahie Frecision: Refers t0 the reproducibility of 4 ceries of ...

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Accuracy & Frecision WorksheetAcrac: refers to how clace meanurement ksto true accepted or target vahie Frecision: Refers t0 the reproducibility of 4 ceries of measurementeThe folloning meusurcments "rre madc ta dcterinine the density of & material wbose ralue was according t0 the Handhaok of Chcmistry and Fhygics 124H/ml Trial 1.20 &/mL Trial +2 122 g/mL Triel *3 1.22 &lmL make general comment on the accuracy of these resulsmake general comment on the precision of these res

Accuracy & Frecision Worksheet Acrac: refers to how clace meanurement ksto true accepted or target vahie Frecision: Refers t0 the reproducibility of 4 ceries of measuremente The folloning meusurcments "rre madc ta dcterinine the density of & material wbose ralue was according t0 the Handhaok of Chcmistry and Fhygics 124H/ml Trial 1.20 &/mL Trial +2 122 g/mL Triel *3 1.22 &lmL make general comment on the accuracy of these resuls make general comment on the precision of these results what may have caused these results? Tbc following mcasurements were made to dctermnine the density ofa material wbose vzlue was according to the handbook of Chemistry and Physics, 1.15 g/ml Trial #1 0.95 B/mL Trial *2 116 &ml Trial +3 1.26 &/mL make general comment on the accuracy of these results make- general comment on the precdsion ofthese resulty what Inay have = caused these results? Tbe following mcasurements were made t0 determine the density of material wbooe value wal, according t0 the handbook of Chemistry = aud _ Phyeics, 3.75 g/oL Tral 4.75 B/mL Trial 476 NmL Trial 43 4.74 B/inL make . general comment On the accuracy of theba regulta Inake genera) comment o0 the preckalon of these results



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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.

In this question. We haven't irregularly shaped chunk of metal and were asked to determine the density. Now we know this metal has a volume of 74.1 to 2 grams, and we know that's when it's placed and the graduated cylinder. The water level rises from 28.2 minute leaders to 36.7, so we could find the volume by taking the difference between these waters. So we know 36.7, minus 28.2 is equal to 8.5. So here we have a volume. We also know density is equal to mass over volume. We have the mass, and we just calculated the volume so we can find the density fairly easily. Let's go ahead and do that. So we find the dis equals eight point seven to grants per cubic centimeters. Now we're also given a list of metals that could be that this chunk can be related to given its density. However, it's not possible to identify the middle because the closest density that we have is 8.65 which relates to cadmium. But there's also another one cobalt that's around 8.90 because it doesn't exactly hit or

So Chevy Jebbie Shelves rule is 100%. Are 100 times 1 -1 of her case squared percent of the data values is within K standard deviations where K is greater than one. So our solution is the first time the Z score, which is x minus mu divided by sigma. Ah this is equal to 17.28 -14, divided by 0.4. This equals 8.2. So using Chevy chiefs inequality or sorry, the rule, we'll just plug that in 100 times one minus 1/8 10.2 squared and we get approximately mhm 98 51%. With the other weights are at least the 17 oz.

All right. We want to test the hypothesis. New equals 64.5 versus the null hypothesis or alternative hypothesis does not equal 64.5 for population standard deviation 0.6 given the following data at alpha equals 0.1 confidence. This question is testing your knowledge of how to conduct a hypothesis test for population renew. When the standard deviation sigma is known. We see through the five steps listed here to solve first, we verify normality. So from the normal probability bar on the left, in the box, on the right, we conclude that the data is normal without outliers. Thus we can proceed to solve so and be we calculate that sample size and X five simple meat. This is N equals 22 X 5 64.7 Thus you see we can calculate our test statistic, this is zero equals experiments were not over segment about 10, which gives negative 38.53 for alpha equals 0.1 We continue the critical value in step D. For two tailed tests using a Z. Table where the area H 2.5 are critical value is plus or minus 2.58 Thus we can conclude in step E the following first. Let's rewrite our critical Z score correctly. Since our Xena is to the left of our negative 2.58 we can conclude that zero is in the critical region. So we reject H. Not because you know it's a critical region. We also address the fact that ALPHA equals 00.1 is more reasonable than Alpha equals 0.1 because there are severe punishments for making the type point ever in this problem I. E. There's a severe punishment for making an area generally. So we want to keep the 0.0.1 stringent criteria.


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