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Complete: Chapter Problem SetRack to AssignmentAttempts:Kecp thc Highest:application the distribution aample meunsPeople suffcring (rom hypertension hean sease kidn...

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Complete: Chapter Problem SetRack to AssignmentAttempts:Kecp thc Highest:application the distribution aample meunsPeople suffcring (rom hypertension hean sease kidney problers public health need departments Wimnit tncir Intakes some U.S. states and sodlum. The Canadian provinccs thelr customers require cornmunlty water systems the sodium concentration anineina nouly Wei Lxce Ads example the notification desiqnaled Icvel Iimit: Connecticut; (or mg/L (milligrams per liter) . Suppose that the cours

Complete: Chapter Problem Set Rack to Assignment Attempts: Kecp thc Highest: application the distribution aample meuns People suffcring (rom hypertension hean sease kidney problers public health need departments Wimnit tncir Intakes some U.S. states and sodlum. The Canadian provinccs thelr customers require cornmunlty water systems the sodium concentration anineina nouly Wei Lxce Ads example the notification desiqnaled Icvel Iimit: Connecticut; (or mg/L (milligrams per liter) . Suppose that the course particular vear the mean concentaton s0iium system dnnaling Connecticut is 25,8 mG/L dter water anothc standard deviatlon mo/l; Imagine that LVa ccoaumnent selecs simple random carnple 3 water specimens ycar- Ecch specimen coumc this antor tesuno end of the Ycarthe water departmcnt corputes Uhe concentration dCnoee mcan epecincns the mean excceds 28 mq/L; Ihc water Cepartmcnl nolifles recommeno? people viho are pubiIc and Eodi ~restricted dicta informn ttci phyicians drinking water the sodium cantent treir Use tre Distributions toob ansverthe following question (Hint: Stan ( sctuina IeanAmdl ~andurd devlallon pdrmeter the exnected Tean and standard error for the distnbuton aamnole Meam concentralions ) Nomnai Distnounon Nean 75.5 StarDard Dnvijocn ct concentratiot eodium the dtinuino Wute erinn emlall M Fibhinthc lirnnt thcro / uatot damirment 0,.0o63 Lyee thoueh



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Filling Bottles A certain brand of apple juice is supposed "to have 64 ounces of juice. Because the punishment for underfilling bottles is severe, the target mean amount of juice is 64.05 ounces. However, the filling machine is not precise, and the exact amount of juice varies from bottle to bottle. The quality-control manager wishes to verify that the mean amount of juice in each bottle is 64.05 ounces so that she can be sure that the machine is not over- or underfilling. She randomly samples 22 bottles of juice and measures the content and obtains the following data: $$\begin{array}{llllll} 64.05 & 64.05 & 64.03 & 63.97 & 63.95 & 64.02 \\ \hline 64.01 & 63.99 & 64.00 & 64.01 & 64.06 & 63.94 \\ \hline 63.98 & 64.05 & 63.95 & 64.01 & 64.08 & 64.01 \\ \hline 63.95 & 63.97 & 64.10 & 63.98 & & \end{array}$$ (a) Because the sample size is small, she must verify that the amount of juice is normally distributed and the sample does not contain any outliers. The normal probability plot and boxplot are shown. Are the conditions for testing the hypothesis satisfied? (b) Should the assembly line be shut down so that the machine can be recalibrated? Assume $\sigma=0.06$ ounces and use a 0.01 level of significance. (c) Explain why a level of significance of $\alpha=0.01$ might be more reasonable than $\alpha=0.1 .$ [Hint: Consider the consequences of incorrectly rejecting the null hypothesis.] FIGURE CANT COPY

What we want to conduct A p D. T. A pair differences tests at the alpha equals 1% confidence level. Testing the claim that the population means A bar next pr are not equal. We have the data for A and B. Given below assuming amount shapes, not your distribution on the right. I've already calculated D. Bar noted that an equal seven and calculated SD as 70.47 So we proceeded to do the five steps listed below to solve this first. We evaluate the requirements and hypotheses. So the requirements to use the students distribution have been met because the distribution shape we have degree of freedom and minus one equals six. Are null hypothesis is mute equals zero. Or alternative is beauty does not equal zero. And we're testing at alpha equals 00.1 confidence Next will compute the test statistic and P value. So the statistic is T equals D. Bar divided by SD over. Uh huh. 2.083 from a tea table. This puts p between .1.05. So we can conclude that P is greater than alpha, which means we fail to reject the null hypothesis, which means that we lack evidence that beauty does not equal to about.

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

All right. We have a sample of size and equals 196. And in that sample we see 29 objects satisfying a certain observation we want to observe that is R equals 29 out of 10 equals 196. We want to use this data to test the claim that P is greater than 092 with alpha equals 0.5 Or confidence level 5%. Now that we've identified the confidence level, we can proceed in the following procedural steps in order to conduct this hypothesis test first. Is it appropriate to use the normal distribution? Yes, it is. Because N. P and Q. P. R both greater than five secondly what hypotheses are retesting? We're testing H and R P equals 50.92 H. A. P greater than 0.92 Which means we're conducting a right tailed or one tailed test. Next compute P. A. And the test statistic he had simply are over end or 0.148 plugging that as well as P. Q and N. Into rz stat formula on the right. Give Z equals 0.271 or 2.71 next let's compute the P value based on our Z equals 2.71 We can use this table to identify the P value. The P value is simply the area under the normal curve to the right of the Z score. Since this is the right tool test from the table, we get P equals 0.34 We've illustrated this in the graph on the right next. We use this P value to reject H. Not. Yes, we do because he is the alpha. And we interpret this to mean that we have evidence that P is greater than 0.92

In question 1 11. It talks about how external pressure must be applied more than the osmotic pressure in order to allow solvent from a hi concentrated solution on one side of a semi permeable membrane to send its water molecules through the membrane to the other side. That might be pure water. This osmotic pressure obeys the laws similar to the ideal gas law where P. B equals Nrt. It then shows you that if we replace P with osmotic pressure that we get an over V. Which is related to more clarity. So we have osmotic pressure being equal to polarity, multiplied by RT. With this equation, it asks you to determine the osmotic pressure at 25 degrees Celsius of a 250.20 molar sucrose solution. Well, sucrose only separates, well, it doesn't separate into anything one unit of sucrose creates uh one mole of sucrose. Well, one molecule of sucrose doesn't separate like ionic compounds do. So it's polarity is it's osmolarity. So when using the equation, osmotic pressure will be equal to the straight up polarity of the sucrose solution. 0.20 Multiplied by are multiplied by T. The r value is 0.8 to 06 Leader atmospheres per kelvin mole. So we'll end up getting pressure in units of atmospheres when using this equation. T however, needs to have units of kelvin because our our value has units of kelvin innit leader atmospheres per kelvin mole. Therefore will convert the 25 degrees Celsius. That is provided into kelvin temperature by adding to 73 and we get an osmotic pressure for the solution of 0.49 atmospheres. For part B. It talks about seawater containing 34 g of salts. For every leader of solution that wants you to assume that the salute consists entirely of sodium chloride and to calculate the osmotic pressure of seawater 25 degrees Celsius. We'll do the same thing as we did before. Osmotic pressure will be equal to the polarity multiplied by are multiplied by T. But this needs to be the osmolarity referring to the total polarity of all the salute ions in solution. So if we have 3.4 g of sodium chloride per liter, we can convert the gram sodium chloride into molds sodium chloride by dividing by the molar mass sodium chloride. So we will then have the grams will cancel will have units of mold sodium chloride per liter sodium chloride. That will be the more clarity sodium chloride. But the osmolarity of the solution because sodium chloride separates completely into sodium ions and chloride and ions will be two times that. So this whole thing right here will be the osmolarity or the end value. Well then multiply that by R. Value. And then the temperature again 25 degrees Celsius plus 2 73 to get our kelvin temperature and we get 285 atmospheres for part C. It states that the average osmotic pressure of blood is 7.7 atmospheres at 25 degrees Celsius. So what concentration of glucose will be isotonic with blood? Well, we'll use the same osmotic pressure equation but now we'll be solving for the osmolarity of a glucose solution. Knowing the pressure, if the pressure is 7.7 atmospheres, that'll be set equal to the osmolarity of the glucose solution. Multiplied by the R value. Same as always 0.8 to 06 Multiplied by the kelvin temperature. Which will be the 25 plus 2 73 or 2 98 kelvin rearrangement of this equation gives us an osmolarity of 315 Mueller for glucose because glucose is not separate into ions. It's osmolarity. Is it straight up polarity now for part D. Why says I'm is an enzyme that breaks bacterial cell walls. The solution that contains 0.150 g of the enzyme in 212 mL of solution has an osmotic pressure a 2120.953 tour at 25 degrees Celsius. And for this question, it asks you to calculate the molar mass of the list design. Well, if we go back up here, you'll notice where we put in the molar mass, we put in the molar mass right here. So if we know the pressure and we don't know the molar mass, then this is where our unknown variable will be found in our calculation. It'll look something like this. We'll take the pressure and tour and converted into atmospheres by dividing by 7 60. This will then be osmotic pressure and atmospheres. We'll set that equal to the mass that we have multiplied by one over the molar mass as we did up here. One over the molar mass. Well then divide that whole thing by the leaders of the solution. The two 10 mL can be converted into leaders by dividing by 1000. So now we have this right here. If we knew molar mass would give us the polarity of the license. I'm solution because the licensee does not separate into ions which was not necessarily obvious in the question. But if it's not a common ionic compound, you can assume that the more clarity is the osmolarity and it does not separate this whole thing right here then is the polarity. We will multiply that by the R. Value and then by the T. Value. And all we have now is an algebraic equation for which we can solve for the molar mass. So I'll save you some space here and not take you through all the algebra. But the molar mass ends up becoming 13,900 g per mole. Then for part E. Osmotic pressure of acquis solution of a certain protein was measured in order to determine the protein smaller mass. The solution contained 3.50 mg of dissolved protein in five mL. The osmotic pressure at 25 degrees Celsius was 1.542 Or what is the molar mass. So this will be another calculation very similar to what we did up here. We'll take the tour converted into atmospheres by dividing by 7 16. We'll set that equal to the mass, multiplied by one over the unknown molar mass in order to get moles. Well then divide the moles by the volume. In order to get more clarity. five mL can be converted into leaders by dividing by 1000. Mhm. This whole thing here then is the polarity. Well multiply that by there are value 0.8 to 06 Multiply that by the temperature. And then again, all we have left is an algebraic expression for which we will solve the molar mass. So without going through all the algebra. Hopefully, you know, enough to get through the algebra, you just need to isolate this essentially what we're doing is we're gonna multiply both sides by mm mm is going to come over here and then we'll divide both sides by this number right here. When we do that, we have mm equal to all of this over here. Except that part with this being divided on both sides. And we get 8910 g per mole for answer


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