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Tota piasma vdiumeImpartantcerermininc the required asma comjonent blood replacement Inerapy persan unqerqaino surcer' Plasma voiume influenced by the overa he...

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Tota piasma vdiumeImpartantcerermininc the required asma comjonent blood replacement Inerapy persan unqerqaino surcer' Plasma voiume influenced by the overa health and physical activily al an individual. Suppose thal Tandom samplc 50 male firelighters are tested and Lhat Lhey have plasma volume sample mean of * 37.5 mlkg (milliliters plasma kilogram body weight)_ Assume that 7,70 ml/kg for the distribution of blood plasma_ (a) Find 99ou confidence interva for the population Mean blood plasm

Tota piasma vdiume Impartant cerermininc the required asma comjonent blood replacement Inerapy persan unqerqaino surcer' Plasma voiume influenced by the overa health and physical activily al an individual. Suppose thal Tandom samplc 50 male firelighters are tested and Lhat Lhey have plasma volume sample mean of * 37.5 mlkg (milliliters plasma kilogram body weight)_ Assume that 7,70 ml/kg for the distribution of blood plasma_ (a) Find 99ou confidence interva for the population Mean blood plasma volume male firefighters _ What the margin of error? (Round your answers to two decimae places ) lower Ilmlt upper limit margin of ertor (b} what condilions arge neclssary your calculations? (Select all Lhat apply.) 0s unknour the distrbution weights unifon the distrbution weights Mdunc Ois <nown (c} Interpret vour results the context of this problem. 391 the intervals created using this methad will contain the true uverade plcod plzsMa Kolume Male firefignters. The prabability tnat this interval contains the true average plcod plasma volume Male firetighters 0.01. 15 af the intervals created using this Method contain the true average Dlaag plasma valume male tiretighters_ The prababilily tnat this inlerval contains Lhe Lrue cvcrajc Llood plasma volume male firelighlers (d} Find the sampl necessany number.) Inale firefighters 99%0 confidence leve with maxima margin of eitor 2,50 for the mean plasma volume male firefighters _ (Round up the nearest whole



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Total plasma volume is important in determining the required plasma component in blood replacement therapy for a person undergoing surgery. Plasma volume is influenced by the overall health and physical activity of an individual. (Reference: See Problem 16.) Suppose that a random sample of 45 male firefighters are tested and that they have a plasma volume sample mean of $\bar{x}=37.5 \mathrm{ml} / \mathrm{kg}$ (milliliters plasma per kilogram body weight). Assume that $\sigma=7.50 \mathrm{ml} / \mathrm{kg}$ for the distribution of blood plasma. (a) Find a $99 \%$ confidence interval for the population mean blood plasma volume in male firefighters. What is the margin of error? (b) What conditions are necessary for your calculations? (c) Interpret your results in the context of this problem. (d) Sample Size Find the sample size necessary for a $99 \%$ confidence level with maximal margin of error $E=2.50$ for the mean plasma volume in male firefighters.

All right. We are observing a population with mean 4.8 and we want to calculate the sample mean and standard deviation for the following sample data obtain to do this. We we know the definition of expire and S for a sample which are given here, X is equal to some of the data divided by n or 4.4, and s is equal to the sum of the deviations about the mean square, divided by n minus one or 0.28 Next we want to implement a left tailed test on this particular population, so we want to test the population is actually less than 4.8. Using a significance level output equals 0.5 Where we are told that X is approximately normal, that is this distribution is approximately normal. So we have to answer the following questions. In order to complete this test. To start off with, what are the significance of hypotheses? Alpha equals 0.5 No hypothesis H is new at this 0.4 point eight and H. A. Is that me was less than 4.8. It's the left tail test. What distribution When we used to be the test statistic, we're going to use a student's T distribution because sigma is unknown. We know that we can do so because the shape is symmetrical and bell curved. Given this, we want to calculate the T stat which is given by this formula. The T stat here equates a negative 3.499 Now, given this T stat, we want to compute the P P interval and then sketch the associated students distribution for P. Since we have degree of freedom, uh six minus one equals five. We have the R. P interval is between 50.5 point 01 We can grasp this as the area to the left of our T stat highlighted here in yellow and Marcus P. From this, we can conclude since P is less than legal to alpha, we have statistically significant findings and we can reject R. H. Saw, which we can interpret to ultimately mean that we have sufficient evidence suggesting our population means is less than no means for pointing.

We want to construct a 95% confidence interval for the population. Mean you with sample with N equals 27 expire at the 22.46 sigma equals 4.1. We want to note ian interpret it and afterwards identify the end required To keep you at the same confidence interval for alpha equals .99 and equals 1.55. Start off with let's identify the Z score needed to construct this confidence interval. That's the score is simply 1.96 which we can find an easy table on Google and textbook. Next calculate the margin of error. E given by this formula plugging in our zc our cinema and r N gives us equals 1.55 We interpret this to mean that we are 95% confident the population mean falls between within 1.55 of the mean. Or rather the simple mean. Next construct your confidence interval which is experiments is less than you. Is that the next plus e plugging in our export NRE gives 20.91 is less than you is less than 24.01. And finally, let's calculate the end needed for our final conditions. Remember that the formula calculate this end is simply end is equal to Z time segment over a all square. We plug in our Z for alpha equals .99, which is 2.576. We plug in our margin of error 1.55 and our populations of the deviation to get 46.43, which we have to round up to get an equals 47.

Number 24 Christian A. The level of significance is mentioned is all for equal 0.1 final hypothesis with the state that the population menu is equal to the value mentioned in the claim. So it's nude to new equal 28 million later per gram per kilogram. The alternative hypothesis stated the opposite off the whole hypothesis, according to the claim, plus using different or not equal so each one to mu that equal 28 mL per kilogram. If the alternative hypotheses uses less, then then the test is left. IND. Question. If the alternative hypothesis uses bigger, then then the test is right field. If the alternative hypothesis use is not equal, then the test is toe field, so the answer will be toe still questionable. Be the level of significance is mentioned is also equal. 0.1 on our hypothesis would state that that the population menu is equal to the value mentioned in the claim, so it's no to new equal 11 percent. If the alternative hypothesis uses less, then when the test is left field, it's alternative. High prices. Use is bigger then, then the test is right field. If the alternative hypothesis is not equal, then the test is toe field in the answer will be to till question number C result. Port A and B. It's no team. You two new equal 28 million liters per kilogram, each one to you not equal 28 million liters per kilogram and it's equal 2.62 The B value is a probability off, obtaining a value more extreme or equal to the standard test. Statics then didn't mind the probability using table e, the prosperity equal be off. It is listened minus 2.62 Or that is bigger than 2.6 to equal to multiply boy probability off that is less than minus 2.62 Equal toe multiplied by 0.44 Equal 0.88 Question and move de given all for equal 0.1 and result port see three equal 30.0 Eat, eat. If the value is smaller than the significance level, all for the internal hypothesis is rejected. So be is listen points you one. So, Richard, it sure if we rejected on on high processing the data is a statically significant level Alpha question number E result aboard E It snowed to mu equal 28 million liters kilograms each, one to mu not equal 28 millimeter per kilogram. Result. Port de reject each node. There is sufficient evidence to support the claim that Ruger's Red Claude a lot cell volume is different from new equal 28 mL peer kill again.

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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