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Q3pharmaceutical company producing vitamin capsules desires proportion of calcium content between 40 and 55 ppm. random sample of 20 capsules chosen from the output...

Question

Q3pharmaceutical company producing vitamin capsules desires proportion of calcium content between 40 and 55 ppm. random sample of 20 capsules chosen from the output yields sample mean calcium content ot 44 ppm wIth standard devation of - tudd Find the natural tolerance limits of the process (613p) . Ifthe process control at the present values of its parameters what proportion of the output will be nonconforming (defects) , assuming normal distnbution of the characteristic ! Find the process capa

Q3 pharmaceutical company producing vitamin capsules desires proportion of calcium content between 40 and 55 ppm. random sample of 20 capsules chosen from the output yields sample mean calcium content ot 44 ppm wIth standard devation of - tudd Find the natural tolerance limits of the process (613p) . Ifthe process control at the present values of its parameters what proportion of the output will be nonconforming (defects) , assuming normal distnbution of the characteristic ! Find the process capability index Cpk; and comment On process performance. If the target value is 47.5 ppm . find the Cpm; and Cpmk indices and comment on their valucs.



Answers

The U.S. Food and Drug Administration recommends that individuals consume $1000 \mathrm{mg}$ of calcium daily. The International Dairy Foods Association (IDFA) sponsors an advertising campaign aimed at male teenagers. After the campaign, the IDFA obtained a random sample of 50 male teenagers and found that the mean amount of calcium consumed was $1081 \mathrm{mg}$, with a standard deviation of $426 \mathrm{mg}$. Conduct a test to determine if the campaign was effective. Use the $\alpha=0.05$ level of significance.

All right. We want to find our sample mean X bar, our sample standard deviation S and construct a 99.9% confidence interval about the population, meaning you given the data below assuming the population is normally distributed, we then want to interpret our findings to start off with let's find our X bar and R. S using appropriate definitions. X bar is simply the some of the data divided by the number of data points. And in this case that's 9.95 and s is the square root of the sum of deviations about the mean square divided by n minus one. In this case, 1.2 Now, to construct the confidence interval, we have to start by identifying the critical TC value associated with its confidence interval and synthesize so to identify the T score, we're going to use a tea table A T T. What can you find on google in a textbook and takes two things as input. The degree of freedom in this case is nine and the confidence level in this case 90.999 To obtain the correct you score. Doing so we obtain TC was 4.78 next? What we want to do is identify what the margin of error is, which is given by the formula on the left here. Plugging in our T C. R. S and R. N. Gives E equals 1.54 for this problem. And now we can talk into our confidence interval to obtain the correct interval using the formula expire minus E. Is that some U. Is that's an X. Y plus C. Doing so we obtain our interval 8.41 is less than you is less than 11.49 which we can interpret as meaning we are 99.9% confident that mu is between 8.41 and 11.49 Based on our sample data.

This is problem number 73 from section 9.3. Uh, this problems talking about healthy bones and that they're the RD a recommended daily allowance of women there recommended you that women between ages of 18 and 24 get 1200 milligrams of calcium. Uh, however, they believe that this intake is actually less, uh, what really happens. So they did a study. They took a random sample of 36 women. And they put all that dad in the mini tab and you have the outputs there actually ran the test, um, for you in the second box. And then they ask us three questions So question letter a. Determine whether there's any out, Lars, and this is sort of a review from way earlier in this book. So if you remember, you have an outlier rule. The outlier rule is 1.5 times the Iike you are like you are that is Q three minus. Q. Once arrange Q three minus Q. One. So we're going to go 1090 0.5 minus 632 0.3, and I will get an I q. R of 458 0.2. Then find out my, uh, 1.5 times i q r rule. We're going to go 458.2. We're gonna take the Ike you are. We're gonna multiply that by 1.5 and I get 687.3 now, to find out if there's any out liars, we're gonna take that value. And we are going to subtract from Q one to get a low point. So anything below that mark is an outlier, and then we're gonna take the 687.3. We're going to add it to Q three and say, Hey, any value above that one is going to be an outlier. So let's start off here with q one is 632.3. We're gonna subtract out at 1.5 rule, so 687 0.3 and we end up with negative 55. Now they tell me my minimum value is 374. So there's no lo outlier, no lo outlier. And then we're going to do the same thing for the high end. We're gonna take you three and we're gonna add in that 1.5 rule and this is going to give me 1777.8. My maximum value here is 1425. You seat in a box there, so there's no hi out liars. There's no outliers here. So now moving on the part beat Interpret the P value in context. Now they tell me in that middle box at second box at the P value all the way. At the end, there is 0.0 It's approximately zero. So to interpret that what a P value means is, if the no hypothesis was true, if the no hypothesis is true, the probability of getting the sample result that we got is the P value. So here we go. If the true calcium intake for 18 to 24 year old women is 1200 milligram. So if that's true and we take a sample size 36 the probability of getting the result we got of 865.2 milligrams is approximately. The P value is approximately zero. So you could see my Senate, Sir. It's a true calcium intake of 18 and 25 year old women is 1200 milligrams. The probability of taking a sample size 36 from the population to get a sample result of 865.2 milligrams is approximately zero. This is one of the sentences that I have my students memorize, which is some blanks in here so they can fill stuff in, And then part see, do these daddy give convincing evidence to support the researchers? Sufficient suspicion carry out attacks. So we're gonna do a state plan. Do conclude here. So I'm going to do this a little quicker here. We've already done a couple of these. So state will perform a one sample T test again. Name of the test, given my awful value to see if there is convincing evidence of support that women have a lower calcium intake than the recommended daily allowance. So I'm sort of writing that Exactly how the questions asked I need to define my mu mu equals a true mean daily intake of calcium for women 18 to 24 year old Uh, my no hypothesis. And I forgot something on here. I need to put, um you in here? Mu equals 1200 milligrams. My alternate hypothesis Mu is less than 1200 milligrams. Okay, so our d a wants us have 1200. We believe there's less than 12 hundreds. That's my ace. Abo and H today plan three things random 36 women were randomly selected, 10% Rule 36 women is definitely less than 10% of all young women and large counts. Thank goodness we took a sample bigger and 30. So I can use the central limit theorem here and say that we can perform a T distribution. The do part and all these values are ah in those charts there. So we know that X bar, my mean, is 856.2 milligrams. I know My standard deviation is 306.7. My t stat value is massive negative 6.73 were over six standard deviations away, which gives me a P value of approximately zero conclusion again. Since the P value is approximately zero, which is less than 00.5 which equals Alfa, we reject the no hypothesis. There is convincing evidence towards the alternate. There is convincing evidence that 18 to 24 year old women consume less than 1200 milligrams of calcium per day

So we've been given a balanced equation to work with here. Um And so we've got some potassium permanganate is gonna react with our calcium oxalate And we've got the polarity and the volume of our committal four. So it seems like a good place to start. You've got 0946 moller and 0- 6 to leaders. So that will give us .002 48 malls of our permanganate. So we want to find out how much calcium oxalate there is. So we're gonna go ahead and get them all ratio here. So moles of permanganate to malls of our calcium obsolete. And then we'll multiply by its smaller mass. It will change moles to grams. So our mole ratio is 5-2 and are given balanced equation And one more is 1 28.1 g. So that's going to give us 0.794 rams of our calcium oxalate. So the trick now is to find out how much calcium that would provide us with. So if we have .794 grams of calcium oxalate, we can figure out how much calcium in there is just by multiplying by the percentage of calcium. So the grams of calcium per grams of calcium oxalate. So calcium is 40.08 In the Mauler mass of the calcium oxalate as we saw before is 1 28.1. So this will give us .248 g of calcium. So that's going to be 248 mggg of our calcium ion, which is within our range of 100 two, mg. So, yes, the sample is within the normal range.

All right. We want to find our sample mean X bar, our sample standard deviation S and construct a 99.9% confidence interval about the population, meaning you given the data below assuming the population is normally distributed, we then want to interpret our findings to start off with let's find our X bar and R. S using appropriate definitions. X bar is simply the some of the data divided by the number of data points. And in this case that's 9.95 and s is the square root of the sum of deviations about the mean square divided by n minus one. In this case, 1.2 Now, to construct the confidence interval, we have to start by identifying the critical TC value associated with its confidence interval and synthesize so to identify the T score, we're going to use a tea table A T T. What can you find on google in a textbook and takes two things as input. The degree of freedom in this case is nine and the confidence level in this case 90.999 To obtain the correct you score. Doing so we obtain TC was 4.78 next? What we want to do is identify what the margin of error is, which is given by the formula on the left here. Plugging in our T C. R. S and R. N. Gives E equals 1.54 for this problem. And now we can talk into our confidence interval to obtain the correct interval using the formula expire minus E. Is that some U. Is that's an X. Y plus C. Doing so we obtain our interval 8.41 is less than you is less than 11.49 which we can interpret as meaning we are 99.9% confident that mu is between 8.41 and 11.49 Based on our sample data.


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